L16-Problems

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chapter6PROBLEMS(全英文版设置)

The worst case pull-up resistance occurs whenever a single path exists from the output node to Vdd. Examples of vectors for the worst case are ABCDEFG=1111100 and 0101110. The best case pull-up resistance occurs when ABCDEFG=0000000.
Solution The logic function is : Y = (A + B)CD . The transistor sizes are given in the figure
above.
b. What are the input patterns that give the worst case tpHL and tpLH. State clearly what are the initial input patterns and which input(s) has to make a transition in order to achieve this maximum propagation delay. Consider the effect of the capacitances at the internal nodes.
Solution No
c. Will these two circuits’ rise and fall times always be equal to each other? Why or why not?
Solution
No. Circuit B appears optimized for the case where the transistor with input E is on the

AlbertEinstein课文汉译例句讲解练习答案

Unit ThreeText I: My Friend, Albert EinsteinBan esh Hoffma nnI.Pre-reading Brainstorming:1.What do you know about Einstein? What was he like?The text is mai nly about Ein ste in pers on ality and his in comparable con tributi ons to scie nee. It is in evitable that men tio n must be made of Ein stein ' theory of relativity and his other achieveme nts in mathematics and physics which, however, are very difficult for layman （外行人）to understand and explain. For this reason, only very brief notes are given to the technical terms. What is more important about the text is a description of Ein stein concerning his kn ack for going in st in ctively to the heart of a matter ” （凭本能抓住事物本质的技巧）（Line 5-6）, his utter naturalness” （Line, 17-18）, the fhntastic intensity and depth of his concentration "（考虑问题的强度和深度都是奇妙无比）（Line, 46）, the revolutionary ideas "（Line, 95） about mathematics and physics, etc. There is much to lear n from the Ian guage of the text, too.His main achieveme nts: theory of relativity; E=mc （en ergy equals mass times the speed of light squared）（能量=质量x光速）；the qua ntum theory of light （光量子理论）A very famous scientist, a scientific genius with a lock of long, graying hair.2.How do you think Hoffmann describes Einstein as his friend?Hoffma nn takes a differe nt perspective. He tries to reveal some of the less well-known aspects of Einstein personality, traits that characterize him more as a man than as a scientific genius.II.Related In formatio n1.Wolfgang Amadeus Mozart: （1756-1791）, Austrian composer, one of the world ' s greatest mus igeniuses wrote masterpieces in every branch of music. During his short life, Mozart composed a great volume of music. His 789 compositi ons in clude operas, symph oni es, con cert（协奏曲）,quartets（四重奏）for the piano and for stringed instruments and sonatas for both piano and violin. His music has delicate beauty and is always fresh and pleas ing to the ear.2.Ludwig van Beethoven: （1770-1827）, German composer, was one of music' greatestgeniuses.His works have a rare of originality, emotional depth, and expressive power. He was known for his nine symph oni es, pia nos essenee concertos and sonatas, and string quartets (弦乐四重奏) .Most ofBeethoven's compositi ons were writte n in the classical forms established by his predecessors Mozart and Hayd n, so he is sometimes con sidered the last great composer in the classical traditi on. But he also remolded and expa nded the old forms and in fused them with a highlypers onal inten sity of emoti on, so he is also referred to as the first of the Roma ntics. 3. The Nobel Prize: Alfred Bernhard Nobel (1838-1896), a distinguished Swedish chemist and industrialist, provided for the award of Nobel Prize. He experime nted with differe nt kinds of explosives such as n itroglyceriR 硝酸甘油) and dynamite (黄色炸药)，both deadly explosives. However, he was a pacifist and he feared that his inventions might further warfare. In his will he left about $9,000,000 in a fund to reward those who did most for his fellow man in scienee, literature, and peace. In his will, he specified that the interest accrued by the fund be annually distributed in the form of prizes to those who during the preceding year, shall have conferred the greatest ben efit on mankind ” in the field of physics, chemistry, physiology or medicine, literature, and peace, regardless of nationality.4. The Nazis: Nazism is a political doctrine of racial supremacy, nationalism, and dictatorship. Nazi is an abbreviation of German word for National Socialist Party founded by Hitler. He defined the Germanic people as a race, called Aryans superior to other races. He blamed Germany's troubles on Jewish capitalism, Communism, and the heavy reparation payments Germany was required to make to the victorious Allies by the Treaty of Versailles (June 28, 1919)that en ded the First World War. III. Text Comprehension1) Main Idea :This profile (short, vivid biography, briefly outlining a person ' most outstanding characteristics: his ability, personality, or career 人物简介)is mai nly about Ei nstei n person ality and his incomparable con tributio ns to scie nee. 2) Purpose of writing and Tone ：The purpose is to illustrate with an ecdotessome characteristic features of Ein ste in both as a man and as a scie ntist.3) Organi zati on and Developme nt: Introduction (P1):Using the word simplicity " to begin the illustration of EinsteinBody (P2-19):P2-4: About his modesty ;P5-7: Einstein ' s brief life history and his two great theories;P8-11: About his concentration on workP12-13: About his love of natural simplicity;P14-16: About his academic courageP17-18: About his sense of justiceP19: About his youthful innocenceConclusio n (P20)Summing up what it means to have known Ein ste in and his work.4)Comprehension Questions1.Which phrase in the first paragraph explains the abstract notion of“ simplicity ”？---"goi ng in sti nctively to the heart of a matter"2.From the two an ecdotes related in para. 2-4, what impressi on of Ein stei n have you got?---He was a very modest pers on, n ever thinking himself any superior to or more authoritative tha n others because of his fame and achieveme nts as a great scie ntist of the time.3.What, according to the author, is Einstein 'most outstanding trait as a scie ntist?---Concentration. Refer to the first sentence of para. 9. (The intensity anddepth of his concen trati on were fan tastiQ4.Why did Ein stei n in sist on worki ng hard whe n he was so badly shake n by his wife ' s death?---Work ing hard requires concen trati on, which would help him to dispel the feeli ng of sorrow.5.How do you in terpret the sentence in para. 11: “ To help him, I steered thediscussion away from routine matters into more difficult theoreticalproblems ”？---Tackli ng more difficult theoretical problems requires greater concen trati on and absorpti on.This would help him temporarily forget the sad ness caused by his wife's death.6.What revelation is made through Einstein comment on Beethoven and Mozart ' works?---As a simple man, Ei nstei n takes it that beauty exists in the Uni verse. Such beauty is n atural, pure, and simple. Beauty found is eve n greater and more admirable tha n beauty created.7.How did Ein ste in feel about the destructive effect produced as a result of the applicati on of his E=mc formula?---This is someth ing he had not expected. He was greatly dismayed by the devastat ing effect his formula produced once it was put into applicati on.8.Do you think the an ecdote related in para. 19 aims to illustrate Ein ste in “whimsicality ” ?If not, whatonality trait other than being whimsicalityis revealed here?---He was not really a whimsical man. If he could be called a whimsical man, then his whimsicality came from the young heart and childlike innocence which he had man aged to reta in.5)Difficult Sentences for Paraphrasing:1.This kn ack for going in sti nctively to the heart of a matter was the secret of his major scientific discoveries --- this and his extraordinary feeling for beauty. (Para. 1)---This n atural ability of in tuitively gett ing to the esse nee of a subject was the key to the great discoveries made by him in scie nee. This n atural gift and his unu sual aware ness of beauty.2.The inten sity and depth of his concen trati on were fan tastic. Whe n battl ing a recalcitra nt problem, he worried it as an ani mal worries its prey.(P-9)---His en grossme nt in ideas was in credibly intense and deep. When attack ing a problem difficult to solve, he kept attempting to deal with it with great effort, just as an ani mal chases and bites a weaker ani mal it preys upon un til the latter gives in.3.A dreamy, faraway and yet in ward look would come over his face. There was no appearanceof concentration, no furrowing of the blow --- only a placid inner commu nio n. (P-10)他脸上会露出心不在焉、恍恍惚惚却又像在内心思索的神色。

英中文术语对照表(统计学习理论的本质-张学工译)

统计学习理论的本质：英中文术语对照表来源：张学工译, VN Vapnik原著, 统计学习理论的本质, 清华大学出版社, 2000使用范围：南京师范大学计算机科学与技术学院研究生。

声明：任何人在其出版物使用或者上载到互连网都必须得到译者及出版社的许可。

AdaBoost algorithm （AdaBoost(自举)算法）163admissible structure （容许结构） 95algorithmic complexity （算法复杂度） 10annealed entropy （退火熵） 55ANOVA decomposition （ANOVA分解） 199a posteriori information （后验信息） 120a priori information （先验信息） 120approximately defined operator （近似定义的算子） 230 approximation rate （逼近速率） 98artificial intelligence （人工智能） 13axioms of probability theory （概率理论的公理） 60back propagation method （后向传播方法） 126basic problem of probability theory （概率论的基本问题） 62basic problem of statistics （统计学的基本问题） 63Bayesian approach （贝叶斯方法） 119Bayesian inference （贝叶斯推理） 34bound on the distance to the smallest risk （与最小风险的距离的界） 77 bound on the values of achieved risk （所得风险值的界） 77bounds on generalization ability of a learning machine （学习机器推广能力的界） 76canonical separating hyperplanes （标准分类超平面） 132capacity control problem （容量控制问题） 116cause-effect relation （因果关系） 9choosing the best sparse algebraic polynomial （选择最佳稀疏多项式）117choosing the degree of polynomial （选择多项式阶数） 116 classification error （分类错误） 19codebook （码本） 106complete (Popper's) nonfalsifiability （完全(波普)不可证伪性） 52 compression coefficient （压缩系数） 107consistency of inference （推理的一致性） 36constructive distribution-independent bound on the rate of convergence （构造性的不依赖于分布的收敛速度界） 69convolution of inner production （内积回旋） 140criterion of nonfalsifiability （不可证伪性判据） 47data smoothing problem （数据平滑问题） 209decision-making problem （决策选择问题） 296decision trees （决策树） 7deductive inference （演绎推理） 47density estimation problem （密度估计问题）：parametric(Fisher-Wald) setting（参数化(Fisher-Wald)表示） 20nonparametric setting （非参数表示） 28discrepancy （差异） 18discriminant analysis （判别分析） 24discriminant function （判别函数） 25distribution-dependent bound on the rate of convergence （依赖于分布的收敛速度界） 69distribution-independent bound on the rate of convergence （不依赖于分布的收敛速度界） 69ΔΔ-margin separating hyperplane （间隔分类超平面） 132 empirical distribution function （经验分布函数） 28empirical processes （经验过程） 40empirical risk functional （经验风险泛函） 20empirical risk minimization inductive principle （经验风险最小化归纳原则） 20ensemble of support vector machines （支持向量机的组合） 163 entropy of the set of functions （函数集的熵） 42entropy on the set of indicator functions （指示函数集的熵） 42 equivalence classes （等价类） 292estimation of the values of a function at the given points （估计函数在给定点上的值） 292expert systems （专家系统） 7ε-insensitivity （ε不敏感性） 181ε-insensitive loss function （ε不敏感损失函数） 181feature selection problem （特征选择问题） 118function approximation （函数逼近） 98function estimation model （函数估计模型） 17Gaussian （高斯函数） 26generalized Glivenko-Cantelli problem （广义Glivenko-Cantelli问题）66generalized growth function （广义生长函数） 85generator random vectors （随机向量产生器） 17Glivenko-Cantelli problem （Glivenko-Cantelli问题） 66growth function （生长函数） 55Hamming distance （汉明距离） 104handwritten digit recognition （手写数字识别） 146hard threshold vicinity function （硬限邻域函数） 103hard vicinity function （硬领域函数） 269hidden Markov models （隐马尔可夫模型） 7hidden units （隐结点） 101Huber loss function （Huber损失函数） 183ill-posed problems （不适定问题）： 9solution by variation method （变分方法解） 236solution by residual method （残差方法解） 236solution by quasi-solution method （拟解方法解） 236 independent trials （独立试验） 62inductive inference （归纳推理） 50inner product in Hilbert space （希尔伯特空间中的内积） 140 integral equations （积分方程）：solution for exact determined equations （精确确定的方程的解）237solution for approximately determined equations （近似确定的方程的解） 237kernel function （核函数） 27Kolmogorov-Smirnov distribution （Kolmogorov-Smirnov分布） 87 Kulback-Leibler distance （Kulback-Leibler距离） 32Kuhn-Tücker conditions （库恩-塔克条件） 134Lagrangian multiplier （拉格朗日乘子） 133Lagrangian （拉格朗日函数） 133Laplacian （拉普拉斯函数） 277law of large number in the functional space （泛函空间中的大数定律）41law of large numbers （大数定律） 39law of large numbers in vector space （向量空间中的大数定律） 41 Lie derivatives （Lie导数） 20learning matrices （学习矩阵） 7least-squares method （最小二乘方法） 21least-modulo method （最小模方法） 182linear discriminant function （学习判别函数） 31linearly nonseparable case （线性不可分情况） 135local approximation （局部逼近） 104local risk minimization （局部风险最小化） 103locality parameter （局部性参数） 103loss-function （损失函数）：for AdaBoost algorithm （AdaBoost算法的损失函数） 163for density estimation （密度估计的损失函数） 21for logistic regression （逻辑回归的损失函数） 156for pattern recognition （模式识别的损失函数） 21for regression estimation （回归估计的损失函数） 21 madaline（Madaline自适应学习机） 7main principle for small sample size problems （小样本数问题的基本原则） 28maximal margin hyperplane （最大间隔超平面） 131maximum likehood method （最大似然方法） 24McCulloch-Pitts neuron model （McCulloch-Pitts神经元模型） 2 measurements with the additive noise （加性噪声下的测量） 25 metric ε-entropy （ε熵度量） 44minimum description length principle （最小描述长度原则） 104 mixture of normal densities （正态密度的组合） 26National Institute of Standard and Technology (NIST) digit database （美国国家标准技术研究所(NIST)数字数据库） 173neural networks （神经网络） 126non-trivially consistent inference （非平凡一致推理） 36 nonparametric density estimation （非参数密度估计） 27normal discriminant function （正态判别函数） 31one-sided empirical process （单边经验过程） 40optimal separating hyperplane （最优分类超平面） 131overfitting phenomenon （过学习现象） 14parametric methods of density estimation （密度估计的参数方法） 24 partial nonfalsifiability （部分不可证伪性） 51Parzen's windows method （Parzen窗方法） 27pattern recognition problem （模式识别问题） 19perceptron （感知器） 1perceptron's stopping rule （感知器迭代终止规则） 6polynomial approximation of regression （回归的多项式逼近） 116 polynomial machine （多项式机器） 143potential nonfalsifiability （潜在不可证伪性） 53probability measure （概率测度） 59probably approximately correct (PAC) model （可能近似正确(PAC)模型） 13problem of demarcation （区分问题） 49pseudo-dimension （伪维） 90quadratic programming problem （二次规划问题） 133quantization of parameters （参数的量化） 110quasi-solution （拟解） 112radial basis function machine （径向基函数机器） 144random entropy （随机熵） 42radnom string （随机串） 10randomness concept （随机性概念） 10regression estimation problem （回归估计问题） 19regression function （回归函数） 19regularization theory （正则化理论） 9regularized functional （正则化泛函） 9reproducing kernel Hilbert space （再生核希尔伯特空间） 244 residual principle （残差原则） 236rigorous (distribution-dependent) bounds （严格(依赖于分布的)界） 85 risk functional （风险泛函） 18risk minimization from empirical data problem （基于经验数据最小化风险的问题） 20robust estimators （鲁棒估计） 26robust regression （鲁棒回归） 26Rosenblatt's algorithm （Rosenblatt算法） 5set of indicators （指示器集合） 73set of unbounded functions （无界函数集合） 77σ-algebra （σ代数） 60sigmoid function （S型(sigmoid)函数） 125small samples size （小样本数） 93smoothing kernel （平滑核） 100smoothness of functions （函数的平滑性） 100soft threshold vicinity function （软阈值领域函数） 103soft vicinity function （软领域函数） 269soft-margin separating hyperplane （软间隔分类超平面） 135spline function （样条函数）：with a finite number of nodes （有限结点的样条函数） 194with an infinite number of nodes （无穷多结点的样条函数） 195 stochastic approximation stopping rule （随机逼近终止规则） 34 stochastic ill-posed problems （随机不适定问题） 113strong mode estimating a probability measure （强方式概率度量估计）63structural risk minimization principle （结构风险最小化原则） 94 structure （结构） 94structure of growth function （生长函数的结构） 79supervisor （训练器） 17support vector machines （支持向量机） 137support vectors （支持向量） 134support vector ANOVA decomposition （支持向量ANOVA分解） 199 SVM n approximation of the logistic regression （逻辑回归的SVM n逼近） 155SVM density estimator （SVM密度估计） 246SVM conditional probability estimator （SVM条件概率估计） 257 tails of distribution （分布的尾部） 78tangent distance （切距） 149training set （训练集） 18transductive inference （转导推理） 293Turing-Church thesis （Turing-Church理论） 177two layer neural networks machine （两层神经网络机器） 145two-sided empirical process （双边经验过程） 40U.S. Postal Service digit database （美国邮政数字数据库） 173 uniform one-sided convergence （一致单边收敛） 39uniform two-sided convergence （一致双边收敛） 39VC dimension of a set of indictor functions （指示函数集的VC维） 79 VC dimension of a set of real functions （实函数集的VC维） 81VC entropy （VC熵） 44VC subgraph （VC子图） 90vicinal risk minimization method（领域风险最小化） 268vicinity kernel（领域核）：273one-vicinal kernel （单领域核） 273two-vicinal kernel （双领域核） 273VRM method （VRM方法）：for pattern recognition （模式识别的VRM方法） 273for regression estimation （回归估计的VRM方法） 282for density estimation （密度估计的VRM方法） 284for conditional probability estimation （条件概率估计的VRM方法） 285for conditional density estimation （条件密度估计的VRM方法）286weak mode estimating a probability measure （弱方式概率度量估计）63weight decay procedure （权值衰减过程） 102。

Problem_Solving问题处理方法

WHO HOW FAR
Exercise 1
15
HOW HOW Much When WHO
PD Critical Thinking Process
Action Plan
RCCM
角色扮演：我们公司发生过这样的事情吗？
•
Sales: Our Sales numbers are low. Our sales team is losing business to our competitor. They have a much more advantaged product. 销售：我们的销售额下降了。由于竞争对手上市了更优质的产品，抢走了我们的业务 Product Development: Yes, I know, I‟ve received at least 3 calls this month from sales reps who are losing business to the competition. 研发：知道了。关于此事，这个月销售代表已经来过3个电话了 Sales: We need to do something about this. Can we develop a faster, cheaper product? 销售：需要采取行动。我们能够研发出成本更低、生产更快的产品吗？ Product Development: Hold on now, let me do a bit of research to verify if their product actually is faster and cheaper than ours. I‟ll get one of their units and have our team of engineers test it. This is no easy task, it will take about a month. 研发：采取行动前，我得调查一下他们的产品确实比我们的产品更便宜、更好。先让工程师测试竞争对手的产品。该项测试有一定难度，需要一个月才能出结果 Sales: If we get a faster, cheaper product, our sales will increase drastically. 销售：如果有成本更低、生产更快的产品，我们的销售额将迅速增长

人教新课标必修二英语Unit 3Computers【教学设计】

Unit 3 Computers本堂课包括Warming up 和Reading 两部分。

他们是语言知识和文化意识的输入过程；是语言技能、情感态度和学习策略的培养过程，也是语言输出过程。

在整个教学过程中，它是培养学生阅读能力的主要过程，教学重点是阅读技能的培养、交际能力及情感态度的提升。

Warming up部分根据每个人对电脑的看法不同让学生就“Do you like computer?”进行小组探究和讨论，从自身实际出发谈论自己的看法，有助于激发学生对于整个单元的学习兴趣。

Reading部分（Computers）主要讲述的是电脑的发展史，通过一系列问题的设置让学生能够在全面理解文章的基础上提高信息处理、加工和学习的能力。

Knowledge objectives:掌握以下词语和句型：Calculate,sum, solve, simplify, artificial, intelligence, revolution,reality, totally, application, finance, in common, over time, as a result, from···on, so···that. Ability objectives:训练学生一定的阅读技巧，使他们掌握一些有效的学习策略，从而提高阅读速度和篇章理解的准确性，并养成一定的自主学习能力；培养学生快速阅读的能力、捕捉信息的能力及运用语言进行交际的能力。

Emotion objectives:通过对“computer”的学习，启发学生思考电脑给我们带来的影响，从而以一种正确的方式利用电脑和网络资源；另外通过电脑的发展历程让学生意识到科技对生活的影响。

1. 教学重点：1.了解计算机的发展过程，进一步认识计算机的演变以及对人类的影响。

2.训练学生的阅读技巧，提高学生阅读速度和理解能力。

OpenProblemsList

Open Problems ListArising from MathsCSP Workshop,Oxford,March2006Version0.3,April25,20061Complexity and Tractability of CSPQuestion1.0(The Dichotomy Conjecture)Let B be a relational structure.The problem of deciding whether a given relational structure has a homomorphism to B is denoted CSP(B).For which(ﬁnite)structures is CSP(B)decidable in polynomial time?Is it true that for anyﬁnite structure B the problem CSP(B)is either decidable in polynomial time or NP-complete?Communicated by:Tomas Feder&Moshe Vardi(1993) Question1.1A relational structure B is called hereditarily tractable if CSP(B )is tractable for all substructures B of B.Which structures B are hereditarily tractable?Communicated by:Pavol Hell Question1.2A weak near-unanimity term is deﬁned to be one that satisﬁes the following identities:f(x,...,x)=x and f(x,y,....y)=f(y,x,y,....y)=...=f(y,...,y,x).Is CSP(B)tractable for any(ﬁnite)structure B which is preserved by a weak near-unanimity term?Communicated by:Benoit Larose,Matt Valeriote Question1.3A constraint language1S is called globally tractable for a problem P,if P(S)is tractable,and it is called(locally)tractable if for everyﬁnite L⊆S,P(L)is tractable.These two notions of tractability do not coincide in the Abduction problem(see talk by Nadia Creignou).•For which computational problems related to the CSP do these two notions of tractability coincide?•In particular,do they coincide for the standard CSP decision problem?Communicated by:Nadia Creignou 1That is,a(possibly inﬁnite)set of relations over someﬁxed set.1Question1.4(see also Question3.5)It has been shown that when a structure B has bounded pathwidth duality the corresponding problem CSP(B)is in the complexity class NL (see talk by Victor Dalmau).Is the converse also true(modulo some natural complexity-theoretic assumptions)?Communicated by:Victor Dalmau Question1.5Is there a good(numerical)parameterization for constraint satisfaction problems that makes themﬁxed-parameter tractable?Question1.6Further develop techniques based on delta-matroids to complete the com-plexity classiﬁcation of the Boolean CSP(with constants)with at most two occurrences per variable(see talk by Tomas Feder).Communicated by:Tomas Feder Question1.7Classify the complexity of uniform Boolean CSPs(where both structure and constraint relations are speciﬁed in the input).Communicated by:Heribert Vollmer Question1.8The microstructure graph of a binary CSP has vertices for each variable/value pair,and edges that join all pairs of vertices that are compatible with the constraints.What properties of this graph are suﬃcient to ensure tractability?Are there properties that do not rely on the constraint language or the constraint graph individually?2Approximability and Soft ConstraintsQuestion2.1Is it true that Max CSP(L)is APX-complete whenever Max CSP(L)is NP-hard?Communicated by:Peter Jonsson Question2.2Prove or disprove that Max CSP(L)is in PO if the core of L is super-modular on some lattice,and otherwise this problem is APX-complete.The above has been proved for languages with domain size3,and for languages contain-ing all constants by a computer-assisted case analysis(see talk by Peter Jonsson).Develop techniques that allow one to prove such results without computer-assisted analysis.Communicated by:Peter Jonsson Question2.3For some constraint languages L,the problem Max CSP(L)is hard to approximate better than the random mindless algorithm on satisﬁable or almost satisﬁable instances.Such problems are called approximation resistant(see talk by Johan Hastad).Is a single random predicate over Boolean variables with large arity approximation resistant?What properties of predicates make a CSP approximation resistant?What transformations of predicates preserve approximation resistance?Communicated by:Johan Hastad2Question2.4Many optimisation problems involving constraints(such as Max-Sat,Max CSP,Min-Ones SAT)can be represented using soft constraints where each constraint is speciﬁed by a cost function assigning some measure of cost to each tuple of values in its scope.Are all tractable classes of soft constraints characterized by their multimorphisms?(see talk by Peter Jeavons)Communicated by:Peter Jeavons 3AlgebraQuestion3.1The Galois connection between sets of relations and sets of operations that preserve them has been used to analyse several diﬀerent computational problems such as the satisﬁability of the CSP,and counting the number of solutions.How can we characterise the computational goals for which we can use this Galois connection?Communicated by:Nadia Creignou Question3.2For any relational structure B=(B,R1,...,R k),let co-CSP(B)denote the class of structures which do not have a homomorphism to B.It has been shown that the question of whether co-CSP(B)is deﬁnable in Datalog is determined by P ol(B),the polymorphisms of the relations of B(see talk by Andrei Bulatov).Let B be a core,F the set of all idempotent polymorphisms of B and V the variety generated by the algebra(B,F).Is it true that co-CSP(B)is deﬁnable in Datalog if and only if V omits types1and2(that is,the local structure of anyﬁnite algebra in V does not contain a G-set or an aﬃne algebra)?Communicated by:Andrei Bulatov Question3.3Does every tractable clone of polynomials over a group contain a Mal’tsev operation?Communicated by:Pascal Tesson Question3.4Classify(w.r.t.tractability of corresponding CSPs)clones of polynomials of semigroups.Communicated by:Pascal Tesson Question3.5Is it true that for any structure B which is invariant under a near-unanimity operation the problem CSP(B)is in the complexity class NL?Does every such structure have bounded pathwidth duality?(see also Question1.4)Both results are known to hold for a2-element domain(Dalmau)and for majority operations(Dalmau,Krokhin).Communicated by:Victor Dalmau,Benoit Larose3Question3.6Is it decidable whether a given structure is invariant under a near-unanimity function(of some arity)?Communicated by:Benoit Larose Question3.7Let L be aﬁxedﬁnite lattice.Given an integer-valued supermodular func-tion f on L n,is there an algorithm that maximizes f in polynomial time in n if the function f is given by an oracle?The answer is yes if L is a distributive lattice(see“Supermodular Functions and the Complexity of Max-CSP”,Cohen,Cooper,Jeavons,Krokhin,Discrete Applied Mathemat-ics,2005).More generally,the answer is yes if L is obtained fromﬁnite distributive lattices via Mal’tsev products(Krokhin,Larose–see talk by Peter Jonsson).The smallest lattice for which the answer is not known is the3-diamond.Communicated by:Andrei Krokhin Question3.8Find the exact relationship between width and relational width.(It is known that one is bounded if and and only if the other is bounded.)Also,what types of width are preserved under natural algebraic constructions?Communicated by:Victor Dalmau 4LogicQuestion4.1The(basic)Propositional Circumscription problem is deﬁned as fol-lows:Input:a propositional formulaφwith atomic relations from a set S,and a clause c.Question:is c satisﬁed in every minimal model ofφ?It is conjectured(Kirousis,Kolaitis)that there is a trichotomy for this problem,that it iseither in P,coNP-complete or inΠP2,depending on the choice of S.Does this conjecturehold?Communicated by:Nadia Creignou Question4.2The Inverse Satisfiability problem is deﬁned as follows: Input:aﬁnite set of relations S and a relation R.Question:is R expressible by a CNF(S)-formula without existential variables?A dichotomy theorem was obtained by Kavvadias and Sideri for the complexity of this problem with constants.Does a dichotomy hold without the constants?Are the Schaefer cases still tractable?Communicated by:Nadia Creignou4Question4.3Let LFP denote classes of structures deﬁnable inﬁrst-order logic with a least-ﬁxed-point operator,let HOM denote classes of structures which are closed under homomorphisms,and let co-CSP denote classes of structures deﬁned by not having a homomorphism to someﬁxed target structure.•Is LFP∩HOM⊆Datalog?•Is LFP∩co-CSP⊆Datalog?(forﬁnite target structures)•Is LFP∩co-CSP⊆Datalog?(forω-categorical target structures)Communicated by:Albert Atserias,Manuel BodirskyQuestion4.4(see also Question3.2)Deﬁnability of co-CSP(B)in k-Datalog is a suﬃcient condition for tractability of CSP(B),which is sometimes referred to as having width k. There is a game-theoretic characterisation of deﬁnability in k-Datalog in terms of(∃,k)-pebble games(see talk by Phokion Kolaitis).•Is there an algorithm to decide for a given structure B whether co-CSP(B)is deﬁnable in k-Datalog(for aﬁxed k)?•Is the width hierarchy strict?The same question when B isω-categorical,but not necessarilyﬁnite?Communicated by:Phokion Kolaitis,Manuel BodirskyQuestion4.5Find a good logic to capture CSP with“nice”(e.g.,ω-categorical)inﬁnite templates.Communicated by:Iain Stewart 5Graph TheoryQuestion5.1The list homomorphism problem for a(directed)graph H is equivalent to the problem CSP(H∗)where H∗equals H together with all unary relations.•It is conjectured that the list homomorphism problem for a reﬂexive digraph is tractable if H has the X-underbar property(which is the same as having the bi-nary polymorphism min w.r.t.some total ordering on the set of vertices),and NP-complete otherwise.•It is conjectured that the list homomorphism problem for an irreﬂexive digraph is tractable if H is preserved by a majority operation,and NP-complete otherwise. Do these conjectures hold?Communicated by:Tomas Feder&Pavol Hell5Question5.2“An island of tractability?”Let A m be the class of all relational structures of the form(A,E1,...,E m)where each E i is an irreﬂexive symmetric binary relation and the relations E i together satisfy the following‘fullness’condition:any two distinct elements x,y are related in exactly one of the relations E i.Let B m be the single relational structure({1,...,m},E1,...,E m)where each E i is the symmetric binary relation containing all pairs xy except the pair ii.(Note that the relations E i are not irreﬂexive.)The problem CSP(A m,B m)is deﬁned as:Given A∈A m,is there a homomorphism from A to B m?When m=2,this problem is solvable in polynomial time-it is the recognition problem for split graphs(see“Algorithmic Graph Theory and Perfect Graphs”,M.C.Golumbic, Academic Press,New York,1980)When m>3,this problem is NP-complete(see“Full constraint satisfaction problems”,T.Feder and P.Hell,to appear in SIAM Journal on Computing).What happens when m=3?Is this an“island of tractability”?Quasi-polynomial algorithms are known for this problem(see“Full constraint satisfaction problems”,T. Feder and P.Hell,,to appear in SIAM Journal on Computing,and“Two algorithms for list matrix partitions”,T.Feder,P.Hell,D.Kral,and J.Sgall,SODA2005).Note that a similar problem for m=3was investigated in“The list partition problem for graphs”, K.Cameron,E.E.Eschen,C.T.Hoang and R.Sritharan,SODA2004.Communicated by:Tomas Feder&Pavol Hell Question5.3Finding the generalized hypertree-width,w(H)of a hypergraph H is known to be NP-complete.However it is possible to compute a hypertree-decomposition of H in polynomial time,and the hypertree-width of H is at most3w(H)+1(see talk by Georg Gottlob).Are there other decompositions giving better approximations of the generalized hypertree-width that can be found in polynomial time?Communicated by:Georg Gottlob Question5.4It is known that a CSP whose constraint hypergraph has bounded fractional hypertree width is tractable(see talk by Daniel Marx).Is there a hypergraph property more general than bounded fractional hypertree width that makes the associated CSP polynomial-time solvable?Are there classes of CSP that are tractable due to structural restrictions and have unbounded fractional hypertree width?Communicated by:Georg Gottlob,Daniel Marx Question5.5Prove that there exist two functions f1(w),f2(w)such that,for every w, there is an algorithm that constructs in time n f1(w)a fractional hypertree decomposition of width at most f2(w)for any hypergraph of fractional hypertree width at most w(See talk by Daniel Marx).Communicated by:Daniel Marx6Question5.6Turn the connection between the Robber and Army game and fractional hypertree width into an algorithm for approximating fractional hypertree width.Communicated by:Daniel Marx Question5.7Close the complexity gap between(H,C,K)-colouring and (H,C,K)-colouring (see talk by Dimitrios Thilikos)Find a tight characterization for theﬁxed-parameter tractable(H,C,K)-colouring problems.•For the(H,C,K)-colouring problems,ﬁnd nice properties for the non-parameterisedpart(H−C)that guaranteeﬁxed-parameter tractability.•Clarify the role of loops in the parameterised part C forﬁxed-parameter hardnessresults.Communicated by:Dimitrios Thilikos6Constraint Programming and ModellingQuestion6.1In a constraint programming system there is usually a search procedure that assigns values to particular variables in some order,interspersed with a constraint propagation process which modiﬁes the constraints in the light of these assignments.Is it possible to choose an ordering for the variables and values assigned which changes each problem instance as soon as possible into a new instance which is in a tractable class? Can this be done eﬃciently?Are there useful heuristics?Question6.2The time taken by a constraint programming system toﬁnd a solution toa given instance can be dramatically altered by modelling the problem diﬀerently.Can the eﬃciency of diﬀerent constraint models be objectively compared,or does it depend entirely on the solution algorithm?Question6.3For practical constraint solving it is important to eliminate symmetry,in order to avoid wasted search eﬀort.Under what conditions is it tractable to detect the symmetry in a given problem in-stance?7Notes•Representations of constraints-implicit representation-eﬀect on complexity•Unique games conjecture-structural restrictions that make it false-connectionsbetween deﬁnability and approximation•MMSNP-characterise tractable problems apart from CSP7•Migrate theoretical results to tools•What restrictions do practical problems actually satisfy?•Practical parallel algorithms-does this align with tractable classes?•Practically relevant constraint languages(”global constraints”)•For what kinds of problems do constraint algorithms/heuristics give good results?8。

leetcode 牛顿法解方程英文

leetcode 牛顿法解方程英文Leetcode is a popular platform for coding challenges, and one of the problems it offers is solving equations using Newton's method. Newton's method is an iterative numerical method used to find the roots of a function. In the context of Leetcode, it is often used to solve equations that cannot be easily solved algebraically.To understand how Newton's method can be applied to solve equations, let's consider a simple example. Suppose we want to find the root of the equation f(x) = x^2 - 4. We can start by guessing an initial value for x, let's say x0 = 2. Then, we can use the formula x1 = x0 - f(x0)/f'(x0), where f'(x) represents the derivative of f(x). In our example, f'(x) = 2x, so x1 = 2 - (2^2 - 4)/(2*2) = 1.5. We can repeat this process until we converge to a solution.One advantage of Newton's method is its fast convergence. With each iteration, the approximation of the root becomes more accurate. However, it is important tonote that Newton's method may not always converge or may converge to a local minimum instead of the desired root. Therefore, it is crucial to carefully choose the initial guess and ensure that the function is well-behaved.In the context of Leetcode, solving equations using Newton's method often involves coding a function that implements the iterative process described above. This requires knowledge of programming languages and algorithms. It also requires an understanding of the problem at hand and the ability to translate it into code.Implementing Newton's method to solve equations on Leetcode can be a challenging but rewarding experience. It allows you to apply mathematical concepts and algorithms to real-world problems. It also helps develop problem-solving skills and improves your programming abilities.Moreover, solving equations using Newton's method on Leetcode can be a great way to practice and improve your coding skills. It requires logical thinking, attention to detail, and the ability to break down complex problems intosmaller, manageable steps. It also provides an opportunity to learn and explore different programming languages and techniques.In conclusion, Leetcode offers the opportunity to solve equations using Newton's method, which is a powerful numerical method for finding roots of equations. It requires a combination of mathematical understanding, coding skills, and problem-solving abilities. Implementing Newton's method on Leetcode can be a challenging but rewarding experience that helps improve your programming skills and problem-solving capabilities.。

ch1-6 problem solutions

Solution
The behaviour is similar to preemption in the computer world, although preemtion is initiated by a third party while in the example a blocked 'participant' solved the deadlock. In the computer world a solution like the one described is difficult because either all locked processes would signal at once, and thus dont solve the lock, or the behavior of the program would be not deterministic.
Solution
a, d
25. Figure 1-23 shows that a number of UNIX system calls have no Win32 API equivalents. For each of the calls listed as having no Win32 equivalent, what are the consequences for a programmer of converting a UNIX program to run under Windows?

23. Give a sketch of how an operating system that can disable interrupts could implement semaphores.

Tutorial.Problems[1]

The payoﬀ matrix of the column player is given in the table below: Paper Scissors Rock Paper 0 1 −1 Scissors −1 0 1 Rock 1 −1 0 (a) Show that the value of this game is zero. (b)∗ A game is called symmetric if B is a square matrix and B + B T = 0. Prove that every symmetric game has value zero. 7. Scotland vs China. An investor wishes to invest a portion q of her money in a Scottish fund and the remaining portion into a Chinese fund. The income on the investment depends on whether the pound devalues or not. The probability of devaluation is p. The table below shows for each of these possible economic outcomes the proﬁt which will be obtained if all the money is invested in a single fund:
a pure strategy giving her the same payoﬀ as when using s∗ 2 . So she will also not have any incentive to deviate. Conclusion: there always is a NE pair with both strategies pure. This, however, cannot be true, as we have seen in the Matching Pennies game. 5. Conservative strategies. In each of the zero-sum games below (the matrices correspond to the payoﬀs of the column player), ﬁnd the mixed strategy of each player which maximizes their worst-case expected payoﬀ. What are the expected payoﬀs? In each case ﬁnd a Nash equilibrium. [ ] [ ] [ ] 2 8 4 −3 −3 3 (a) (b) (c) 5 4 0 2 5 4 6. Rock-Paper-Scissors. Everybody knows this game: Two players simultaneously shake their ﬁsts a couple times and at the third shake they transform their hands into one of three objects: rock, paper or scissors. If both choose the same object, the game ends in a draw: no one wins anything. Rock always wins over scissors (by breaking), scissors over paper (by cutting), and paper over rock (by packing).

Sunnen Honing Guide

HONING GUIDEH e l p f u l T i p s F o r S o l v i n g B o r e S i z i n g & F i n i s h i n g P r o b l e msPRINTED IN U.S.A. 0410©COPYRIGHT SUNNEN ®PRODUCTS COMPANY 2004, ALL RIGHTS RESERVED X-SP-5061PSUNNEN PRODUCTS COMPANY7910 Manchester Ave., St. Louis, MO 63143 U.S.A.Phone: 314-781-2100 Fax: 314-781-2268U.S.A. Toll-Free Sales and Service –Automotive: 1-800-772-2878 • Industrial: 1-800-325-3670International Division Fax: 314-781-6128 e-mail:*****************GENERAL HONE CORP (A Sunnen Company)471 US 250 East, Ashland, OH 44805 U.S.A.Phone:419-289-3000 Fax: 419-281-0700U.S.A. Toll-Free Sales and Service: 1-800-837-1999SUNNEN PRODUCTS LIMITED No. 1 Centro, Maxted RoadHemel Hempstead, Herts HP2 7EF ENGLANDPhone:++ 44 1442 39 39 39 Fax: ++ 44 1442 39 12 12SUNNEN AG Fabrikstrasse 18586 Ennetaach-Erlen, SwitzerlandPhone:++ 41 71 649 33 33 Fax: ++ 41 71 649 34 34Sunnen®reserves the right to change or revise specifications and product design in connection with any feature of our products contained herein. Such changes do not entitle the buyer to corresponding changes,improvements, additions, or replacements for equipment, supplies or accessories previously sold. Information contained herein is considered to be accurate based on available information at the time of printing.Should any discrepancy of information arise,Sunnen recommends that user verify discrepancy with Sunnen before proceeding.SHANGHAI SUNNEN MECHANICAL CO., LTD.889 Kang Qiao East Road, PuDong Shanghai 201319, P.R. ChinaPhone:86 21 5 813 3322 Fax: 86 21 5 813 2299SUNNEN ITALIA S.R.L.Viale Stelvio 12/1520021 Ospiate di Bollate (MI) ItalyPhone:39 02 383 417 1 Fax: 39 02 383 417 50“SUNNEN AND THE SUNNEN LOGO ARE REGISTERED TRADEMARKS OF SUNNEN PRODUCTS COMPANY.”Standard Sunnen Tooling and Procedures will satisfy all ordinary honing requirements. If problems are encountered, they can usually be solved by following procedure suggested below. For detailed information on difficult or unusual honing problems contact your local Sunnen Field Engineer or Sunnen Customer Service Department in St. Louis, Missouri, USA.CONDITION TO BE CORRECTED STEP1STEP2STEP3STEP4STEP5STONE GLAZED*Sharpen A or J stone Increase cutting Increase stroking Use a softer stone Check oil to be STONE Stone surface looks with MAN-700pressure speed(one with a lower sure you are using NOT clean but cutting diamond dresser; use hardness number)Sunnen Industrial CUTTING grains are dulledLBN-700 dressing stick Honing Oil*on all other stones (Honing dial STONE LOADED*Clean stone with Increase stroking Use a softer stone Use a coarser grit Check oil to be needle moves Stone surface looks LBN-700 dressing pressure (one with a lower stone (one with a sure you are using too slowly)smeared and stickhardness number)lower grit number)Sunnen Industrial clogged with chipsHoning Oil*SLOW STOCK REMOVAL*Increase spindle speedIncrease cutting Check oil to be Use a softer stone Use a coarser grit (Honing dial needle moves too slowly)pressuresure you arc using (one with a Lower stone (one with a Sunnen Industrial hardness number)lower grit number)Honing Oil*POOR STONE LIFE* Decrease cutting Use faster spindle Use harder stone Use coarser grit stone Using Sunnen (Honing dial needle moves too fast)pressur e spe ed (higher hardness number) (lower g rit number) Industrial Honing Oil/Coolant*BELLMOUTH True stone and shoes Use softer stone If Bore is LONGERShorten STONE only If bellmouth persists with truing sleeve (OK with lower than 2/3 stone length:(or row of stones)shorten stones still If part is short or hardness number)slightly on both ends more but do not shorten unbalanced, shorten shoes any further stroke length If Bore is SHORTER Shorten STONES and CAUTION:than 2/3 stone length:SHOES equally to Overcorrection of1-1/2 times bore bellmouth will lead length to barrel conditionBARREL True stone and shoes Use finer grit stone Use longer stone or Use mandrel with CAUTION:with truing sleeve (one with higher grit shorten guide shoes longer stone and shoe Overcorrection ofnumber)on both ends barrel will leadto bellmouth conditionTAPER IN True stone and shoes Change stroke so tight Reverse work on If power stroking, make OPEN HOLE with truing sleeveend of bore is stroked mandrel more oftensure spindle and stroker over stone further are in alignment TAPER IN Shorten stone and True stone and shoes If hole has insufficient Provide sufficient Provide adequate oil BLIND HOLE shoes to about 3/4frequently with truing or no relief at bottom,relief at bottom of flow at bottom of hole shorten stone more sleeveuse hard tip stone holeto wash cutting out if taper persists OUT-OF-ROUNDMake sure honing Thoroughly true stone If thin wall part,If stone stops cutting If power stroking tool is recommended and shoes to exact decrease cutting at decreased pressure make sure spindle size for diameter to hole diameter pressure use stone with lower and stroker are in be honedhardness numberalignmentWAVINESS Use honing tool with sufficient stone length to bridge waviness (or Iands and ports in bore)RAINBOWUse L, BL, or multi-stone mandrel. Stone Use shorter stroke Use stone with lower length should be at least 1-1/2 times the length hardness number to length of bore for best bow correction (less overstroke)avoid part flexing FINISH TOO ROUGH*Decrease cutting Use finer grit stone Check oil to be sure Thoroughly true shoes For extremely fine pressure(one with higher grit you are using Sunnen to exact hole diameterfinishes in soft or number)Industrial Honing Oil*exotic material, use bronze mandrel or bronze shoes RANDOM SCRATCHESDecrease cutting Use finer grit stone Use softer stone If hard steel mandrel is Check oil to be pressure(one with higher grit (one with lower being used, change to sure you are using number)hardness number)soft steel mandrel. If Sunnen Industrial soft steel mandrel or Honing Oil*shoes are being used,change to bronze mandrel or shoes*Many honing problems, such as poor cutting action, poor stone life, and rough finish are caused by wrong honing oil, insufficient honing oil, dirty honing oil, or contaminated honing oil. Use only clean, full-strength Sunnen Industrial Honing Oil. Make sure that honing oil is neither diluted or "cut" with other oils. Keep solvents and cleaning fluids away from honing machine.{{FOR FAST STOCK REMOVAL IN DE-BURRED, BORED, REAMED OR GROUND HOLESFOR FINE FINISHING SOFT STEELHARDENED STEELIN PREVIOUSLHONED HOLES DIAMETER HONING FOR INCLUDIES: CAST STEELS HARD SOFT STEEL,SPINDLE OF BOREUNITDE-BURRING & STAINLESS STEELS FOR OCCASIONALFOR SOFT CAST IRON,CARBIDE GLASSS STEEL,BRONZE,CARBIDE,GLASSSPEED IN ROUGH (USE BRONZE MANDREL WORKPRODUCTION BRASSBRONZE &SOFT ALUMINUM CERAMICRPM HOLESOR SHOES IN WORK RUNS&CERAMIC * * *BRASSCAST IRON(MAX)STAINLESS STEEL)ALUMINUM * * *ALL FOR FOR TRY THIS IF FIRSTCHOICE MATERIALS OCCASIONALPRODUCTION STONE DOES NOT CUT,* *!!mm INCHES WORK WORK FIRST USE THIS STONE1,52- 2,03(0.060-0.080)D2D6-A67D6-A67D6-NM69D6-A65D6-A63D6-NM69D6-J63D6-J67D6-DM57D6-DM57D6-J93D6-J95D6-DR07D6-DR0730002,03- 2,54(0.080-0.100)D2D8-A67D8-A67D8-NM69D8-A65D8-A63D8-NM69D8-J63D8-J67D8-DM57D8-DM57D8-J93D8-J95D8-DR07D8-DR0730002,54- 3,05(0.100-0.120)K3K3-A611K3-A67K3-NM69K3-A65K3-A63K3-NM69K3-J63K3-J67K3-DM57K3-DM57K3-J93K3-J95K3-DMO7K3-DMO73000BU L3-A611L3-A67L3-NM69L3-A65L3-A63L3-NM69L3-J63L3-J67L3-DM57L3-DM57L3-J93L3-J95L3-DMO7L3-DMO73,05- 3,81(0.120-0.150)K4K4-A611K4-A67K4-NM69K4-A65K4-A63K4-NM69K4-J63K4-J67K4-DM57K4-DM57K4-J93K4-J95K4-DMO7K4-DMO73000BL4L4-A611L4-A67L4-NM69L4-A65L4-A63L4-NM69L4-J63L4-J67L4-DM57L4-DM57L4-J93L4-J95L4-DMO7L4-DMO73,81- 4,70(0.150-0.185)K5K5-A611K5-A67K5-NM69K5-A65K5-A63K5-NM69K5-J63K5-J57K5-DM57K5-DM57K5-J93K5-J95K5-DMO7K5-DMO73000BL5L5-A611L5-A67L5-NM69L5-A65L5-A63L5-NM69L5-J63L5-J57L5-DM57L5-DM57L5-J93L5-J95L5-DMO7L5-DMO74,70- 6,22(0.185-0.245)K6, JK6K6-A415K6-A57K6-NM69K6-A55K6-A63K6-NM69K6-J63K6-J57K6-DM57K6-DM57K6-J93K6-J95K6-DMO7K6-DMO73000L6, BL6L6-A413L6-A57L6-NM69L6-A55L6-A63L6-NM69L6-J63L6-J57L6-DM57L6-DM57L6-J93L6-J95L6-DMO7L6-DMO7K8, JK8K8-A413K8-A57K8-NM55K8-A55K8-A63K8-NM55K8-J63K8-J57K8-DM55K8-DM55K8-J93K8-J95K8-DMO5K8-DM056,22- 7,82(0.245-0.308)L8, BLS L8-A413L8-A57L8-NM55L8-A55L8-A63L8-NM55L8-J63L8-J57L8-DM55L8-DM55L8-J93L8-J95L8-DM05L8-DMO53000Y8Y8-A49Y8-A57'Y8-A55Y8-A63*Y8-J63Y8-J57****Y8-J93Y8-J95****K10, JK10K10-A413K10-A57K10-NM55K10-A55K10-A63K10-NM55K10-J63K10-J57K10-DM55K10-DM55K10-J93K10-J95K10-DM05K10-DMO57,82- 9,40(0.308-0.370)L10, BL10L10-A413L10-A57L10-NM55L10-A55L10-A63L10-NM55L10-J63L10-J57L10-DM55L10-DM55L10-J93L10-J95L10-DM05L10-DMO52000Y10Y10-A49Y10-A57*Y10-A55Y10-A63*Y10-J63Y10-J57****Y10-J93Y10-J95****K12, JK12K12-A413K12-A57K12-NM55K12-A55K12-A63K12-NM55K12-J63K12-J57K12-DM55K12-DM55K12-J93K12-J95K12-DM05K12-DMO516009,40- 12,57(0.370-0.495)L12, BL12L12-A413L12-A57L12-NM55L12-A55L12-A63L12-NM55L12-J63L12-J57L12-DM55L12-DM55L12-J93L12-J95L12-DMO5L12-DMO5to Y12Y12-A49Y12-A57*Y12-A55Y12-A63*Y12-J63Y12-J57****Y12-J93Y12-J95****2000K16, JK16K16-A413K16-A57K16-NM55K16-A55K16-A63K16-NM55K16-J63K16-J57K16-DM55K16-DM55K16-J93K16-J95K16-DMO5K16-DMO512,57- 15,72(0.495-0.619)L16, BL16L16-A413L16-A57L16-NM55L16-A55L16-A63L16-NM55L16-J63L16-J57L16-DM55L16-DM55L16-J93L16-J95L16-DMO5L16-DMO51270Y16Y16-A49Y16-A57"Y16-A55Y16-A63*Y16-J63Y16-J57**Y16-J93Y16-J95K20, JK20K20-A413K20-A57K20-NM55K20-A55K20-A63K20-NM55K20-J63K20-J57K20-DM55K20-DM55K20-J93K20-J95K20-DMO5K20-DMO515,72- 18,90(0.619-0.744)L20, BL20L20-A413L20-A57L20-NM55L20-A55L20-A63L20-NM55L20-J63L20-J57L20-DM55L20-DM55L20-J93L20-J95L20-DMO5L20-DMO51000P20P20-A413P20-A57P20-NM55P20-A55P20-A63P20-NM55P20-J63P20-J57P20-DM55P20-DM55P20-J93P20-J95P20-DMO5P20-DMO5Y20Y20-A411Y20-A57Y20-A55Y20-A63*Y20-J63Y20-J57""**Y20-J93Y20-J95AK20, JAK20K20-A413K20-A57K20-NM55K20-A55K20-A63K20-NM55K20-J63K20-J57K20-DM55K20-DM55K20-J93K20-J95K20-DMO518,90- 25,40(0.744-1.00)AL20, BAL20L20-A413L20-A57L20-NM55L20-A55L20-A63L20-NM55L20-J63L20-J57L20-DM55L20-DM55L20-J93L20-J95L20-OM05800P20P20-A413P20-A57P20-NM55P20-A55P20-A63P20-NM55P20-J63P20-J57P20-DM55P20-DM55P20-J93P20-J95P20-DMO5AAY20N20Y20-A411Y20-A57*Y20-A55Y20-A63*Y20-J63Y20-J57****Y20-J93Y20-J95**AK20, JAK20K20-A413K20-A57K20-NM55K20-A55K20-A63K20-NM55K20-J63K20-J57K20-DM55K20-DM55K20-J93K20-J95K20-DM05K20-DMO525,40- 26,19(1.000-1.031)AL20, BAL20L20-A413L20-A57L20-NM55L20-A55L20-A63L20-NM55L20-J63L20-J57L20-DM55L20-DM55L20-J93L20-J95L20-DMO5L20-DMO5640P28P28-A413P28-A57P28-NM55P28-A55P28-A63P28-NM55P28-J63P28-J57P28-DM55P28-DM55P20-J93P20-J95P20-DMO5P20-DMO5R28R28-A413R28-A57R28-NM55R28-A55R28-A63R28-NM55R28-J63R28-J57R28-DM55R28-DM55Y20-J93Y20-J95..AK20K20-A413K20-A57K20-NM55K20-A55K20-A63K20-NM55K20-J63K20-J57K20-DM55K20-DM55K20-J93K20-J95K20-DMO5K20-DMO526,19- 31,75(1.031-1.250)P28P28-A413P28-A57P28-NM55P28-A55P28-A63P28-NM55P28-J63P28-J57P28-DM55P28-DM55L20-J93L20-J95L20-DMO5L20-DMO5640R28R28-A413R28-A57R28-NM55R28-A55R28-A63R28-NM55R28-J63R28-J57R28-DM55R28-DM55P28-J93P28-J95P28-DMO5P28-DMO531,75- 92,08(1.250-3.625)P28P28-A413P28-A57P28-NM55P28-A55P28-A63P28-NM55P28-J63P28-J57P28-DM55P28-DM55P28-J93P8-J95P28-DMO5P28-DMO5640 to R28R28-A413R28-A57R28-NM55R28-A55R28-A63R28-NM55R28-J63R28-J57R28-DM55R28-DM55R28-J93R8-J95R28-DMO5R28-DMO550092,08-152,40(3.625-6.000)P28P28-A413P28-A57P28-NM55P28-A55P28-A63P28-NM55P28-J63P28-J57P28-DM55P28-DM55P28-J93P28-J95P28-DMO5P28-DMO5200R28R28-A413R28-A57R28-NM55R28-A55R28-A63R28-NM55R28-J63R28-J57R28-DM55R28-DM55R28-J93R28-J95R28-DMO5R28-DM0525,20- 34,93(0.992-1.375)Y32Y32-A49Y32-A57*Y32-A55Y32-A63*Y32-J63Y32-J57****Y32-J93Y32-J95****640 to 50034,67- 44,45(1.365-1.750)Y44Y44-A49Y44-A57*Y44-A55Y44-A63*Y44-J63Y44-J57****Y44-J93Y44-J95****500 to 40044,20- 57,15(1.740-2.250)Y56Y56-A49Y56-A57*Y56-A55Y56-A63*Y56-J63Y56-J57****Y56-J93Y56-J95****400 to 32056,90- 69,85(2.240-2.750)Y72Y72-A49Y72-A57*Y72-A55Y72-A63*Y72-J63Y72-J57****Y72-J93Y72-J95****320 to 25069,60- 82,55(2.740-3.250)Y88Y88-A49Y88-A57*Y88-A55Y88-A63*Y88-J63Y88-J57****Y88-J93Y88-J95****25082,30- 98,42(3.240-3.875)Y104Y104-A49Y104-A57*Y104-A55Y104-A63*Y104-J63Y104-J57****Y104-J93Y104-J95****20063,50- 68,58(2.500-2.700)AN-600G25-A47G25-A45*G25-A45G25-A63*G25-J63G25-J45G25-DV57G25-DV57G25-J85G25-J87G25-DV87**G25-DV87**25068,58-104,14(2.700-4.100)AN-600M27-A47M27-A45*M27-A45M27-A63*M27-J63M27-J45M27-DV57M27-DV57M27-J85M27-J87M27-DV87**M27-DV87**20088,90-139,70(3.500-5.500)AN-600N37-A47N37-A45*N37-A45N37-A63*N37-J63N37-J45N37-DV57N37-DV57N37-J85N37-J87N37-DV87**N37-DV87**20063,50- 83,82(2.500-3.300)AN-600GY25-A47GY25-A45*GY25-A45GY25-A63*GY25-J45GY25-J45****GY25-J85GY25-J85****250GG25-A47GG25-A45*GG25-A45GG25-A63*GG25-J45GG25-J45****GG25-J85GG25-J85****83,82-106,68(3.300-4.200)AN-600MY33-A47MY33-A45*MY33-A45MY33-A63*MY33-J45MY33-J45****MY33-J85MY33-J85****200MM33-A47MM33-A45*MM33-A45MM33-A63*MM33-J45MM33-J45****MM33-J85MM33-J85****101,60-142,24(4.000-5.600)AN-600NY40-A47NY40-A45*NY40-A45NY40-A63*NY40-J45NY40-J45****NY40-J85NY40-J85****200NN40-A47NN40-A45*NN40-A45NN40-A63*NN40-J45NN40-J45****NN40-J85NN40-J85****ABRASIVE TYPESGRIT SIZEHARDNESSRECOMMENDED STONESGRIT SIZEMATERIAL ABRASIVE TYPE 801001502202803204005006001200HARD ALUMINUM OXIDE 25 / 0,65–20 / 0,5018 / 0,4512 / 0,3010 / 0,25 5 / 0,12 3 / 0,08 1 / 0,03–STEEL SILICON CARBIDE CBN/BORAZON ®–55 / 1,4045 / 1,1540 / 1,0028 / 0,70–20 / 0,50–7 / 0,18 2 / 0,05SOFT ALUMINUM OXIDE 80 / 2,00–35 / 0,9025 / 0,6520 / 0,5016 / 0,407 / 0,18 4 / 0,10 2 / 0,05–STEEL SILICON CARBIDE CBN/BORAZON ®–65 / 1,60–70 / 1,75––25 / 0,65–16 / 0,40 5 / 0,12CAST SILICON CARBIDE 100 / 2,50–30 / 0,7520 / 0,5012 / 0,30– 6 / 0,15 5 / 0,12 3 / 0,08–IRON DIAMOND –––80 / 2,00––50 / 1,27–20 / 0,5012 / 0,30ALUMINUM,SILICON CARBIDE 170 / 4,30–80 / 2,0055 / 1,4033 / 0,8527 / 0,7016 / 0,4012 / 0,30 2 / 0,05–BRASS, BRONZE CARBIDE DIAMOND ––30 / 0,7520 / 0,50––7 / 0,18– 3 / 0,08 1 / 0,03CERAMIC DIAMOND ––50 / 1,2740 / 1,00––20 / 0,50–15 / 0,4010 / 0,25GLASSDIAMOND––95 / 2,4070 / 1,80––30 / 0,75–15 / 0,408 / 0,20SUNNEN STONE GUIDEAPPROXIMATE SURFACE FINISH IN MICROINCHES R A / MICROMETERS (MICRONS) R A* CBN/Borazon stones available on special order. !CBN/Borazon stones available on special order. ** CBN/Borazon stones available on special order.*** CBN/Borazon stones available on special order.FINISH FIGURES BASD ON USE OF SUNNEN HONING OIL. IN GENERAL, USE OF WATER BASED COOLANTS RESULT IN FINER SURFACE FINISH AND REDUCED METAL REMOVAL RATES.A—ALUMINUM OXIDE C —SILICON JCARBIDE DM DR —DIAMONDDV NM CBN /NR —NVBORAZON1—702—803—1004—1505—2206—2807—3208—4009—5000—60090—90000—12001—SOFT2—3—5—7—11—15—HARDK8 –A57SERIES ABRASIVE GRIT HARD-TYPE SIZE NESS}}}HONING GUIDEInformation shown is applicable to (conventional) stones, which must be used with honing oil, and NM & DM stones (metal bond)which may be used with either honing oil or water based coolant. The process fluid is very important. Many honing problems,such as poor cutting action, poor stone life, and rough finish are traceable to the process fluid. In general, honing performance will be reduced when water based coolants are used.。

FloTHERM基础培训教程PPT课件

7
热设计的基本要求
满足设备可靠性的要求满足设备预期工作的热环境的要求满足对冷却系统的限制要求
热设计工程师 —— 与EE, ME, Layout等项目
相关人员紧密配合，力求提高产品各方面性能并降低成本
8
了解散热性能的方法
实验研究
— 优点：直观，可靠 — 缺点：昂贵，周期长
数值仿真(CFD)
Table 数据表窗口提供输入输出参数的数据表输出
19
FloTHERM文件结构
索引文件库文件区项目文件
20
FloTHERM文件结构
首先FLOTHERM软件借助四个目录管理文件管理每个项目文件
项目文件夹
千万别去尝试去修改项目文件中名中的数字串
21
定义一个新项目
定义项目名称定义散热环境以及散热方式定义求解域
Step2：点击monitor point
也可以不选择元件，直接建立监控点并把位置设置到关心的地方
35
网格定义
36
求解器设置
设置求解方式设置迭代次数附加选项
37
错误检查与初始化
错误检查
— Error：Data error interrupting solution — Warning：flags set up problems such as incorrect location of
总部：英国伦敦
分公司：
英国、美国、俄罗斯匈牙利、法国、德国意大利、瑞典、日本中国、印度、新加坡
研发中心：
伦敦、波士顿、硅谷圣迭戈、法兰克福、布达佩斯、莫斯科、班加罗尔
代理商：
以色列、韩国、日本、台湾、澳大利亚、巴西

01_introduction_problems

Module:Introduction(Week2out of5)Course:Algorithmic Toolbox(Course1out of6)Specialization:Data Structures and Algorithms Programming Assignment1:IntroductionRevision:September10,2016IntroductionWelcome to your first programming assignment of the Algorithmic Toolbox class!It consists of seven algorithmic problems.The first three problems require you just to implement carefully the algorithms covered in the lectures.The remaining four problems will require you to first design an algorithm and then to implement it.For all the problems,we provide starter solutions in C++,Java,and Python3.These solutions implement straightforward naive algorithms that usually work only for small inputs.To verify this, you may want to submit these solutions to the grader.This will usually give you a“time limit exceeded”message for Python starter files and either“time limit exceeded”or“wrong answer”message for C++and Java solutions(the reason for wrong answer being an integer overflow issue).Your goal is to replace a naive algorithm with an efficient one.In particular,you may want to use the naive implementation for stress testing your efficient implementation.In this programming assignment,the grader will show you the input data if your solution fails on any of the tests.This is done to help you to get used to the algorithmic problems in general and get some experience debugging your programs while knowing exactly on which tests they fail.However,for all the following programming assignments,the grader will show the input data only in case your solution fails on one of the first few tests(please review the questions9.4and9.5in the FAQ section for a more detailed explanation of this behavior of the grader).1Learning OutcomesUpon completing this programming assignment you will be able to:1.See the huge difference between a slow algorithm and a fast one.2.Play with examples where knowing something interesting about a problem helps to design an algorithmthat is much faster than a naive one.3.Implement solutions that work much more faster than straightforward solutions for the following com-putational problems:(a)compute a small Fibonacci number;(b)compute the last digit of a large Fibonacci number;(c)compute a huge Fibonacci number modulo m;(d)compute the last digit of a sum of Fibonacci numbers;(e)compute the last digit of a partial sum of Fibonacci numbers;(f)compute the greatest common divisor of two integers;(g)compute the least common multiple of two integers.4.Implement the algorithms covered in the lectures,design new algorithms.5.Practice implementing,testing,and debugging your solution.In particular,you will find out how inpractice,when you implement an algorithm,you bump into unexpected questions and problems not covered by the general description of the algorithm.You will also check your understanding of the algorithm itself and most probably see that there are some aspects you did not think of before you had to actually implement it.You will overcome all those complexities,implement the algorithms,test them,debug,and submit to the system.Remember the advice(link)we gave you in the first module about testing your programs and feel free to return to those videos or parts of them again while working on your assignment.In the end of this document you will find also general recommendations on solving algorithmic problems.Passing Criteria:3out of7Passing this programming assignment requires passing at least3out of7code problems from this assignment. In turn,passing a code problem requires implementing a solution that passes all the tests for this problem in the grader and does so under the time and memory limits specified in the problem statement.2Contents1Problem:Small Fibonacci Number4 2Problem:Last Digit of a Large Fibonacci Number6 3Problem:Greatest Common Divisor8 4Problem:Least Common Multiple10 5Advanced Problem:Huge Fibonacci Number modulo m12 6Advanced Problem:Sum of Fibonacci Numbers14 7Advanced Problem:Partial Sum of Fibonacci Numbers16 8General Instructions and Recommendations on Solving Algorithmic Problems188.1Reading the Problem Statement (18)8.2Designing an Algorithm (18)8.3Implementing Your Algorithm (18)8.4Compiling Your Program (18)8.5Testing Your Program (20)8.6Submitting Your Program to the Grading System (20)8.7Debugging and Stress Testing Your Program (20)9Frequently Asked Questions219.1I submit the program,but nothing happens.Why? (21)9.2I submit the solution only for one problem,but all the problems in the assignment are graded.Why? (21)9.3What are the possible grading outcomes,and how to read them? (21)9.4How to understand why my program fails and to fix it? (22)9.5Why do you hide the test on which my program fails? (22)9.6My solution does not pass the tests?May I post it in the forum and ask for a help? (23)9.7My implementation always fails in the grader,though I already tested and stress tested it alot.Would not it be better if you give me a solution to this problem or at least the test casesthat you use?I will then be able to fix my code and will learn how to avoid making mistakes.Otherwise,I do not feel that I learn anything from solving this problem.I am just stuck (23)31Problem:Small Fibonacci NumberProblem IntroductionRecall the definition of Fibonacci sequence:F0=0,F1=1,and F i=F i−1+F i−2fori≥2.Your goal in this problem is to implement an efficient algorithm for computingFibonacci numbers.The starter files for this problem contain an implementationof the following naive recursive algorithm for computing Fibonacci numbers in C++,Java,and Python3:Fibonacci(n):if n≤1:return nreturn Fibonacci(n−1)+Fibonacci(n−2)Try compiling and running a starter solution on your machine.You will see that computing,say,F40already takes noticeable time.Another way to appreciate the dramatic difference between an exponential time al-gorithm and a polynomial time algorithm is to use the following visualization byDavid Galles:/~galles/visualization/DPFib.html.Try computing F20by a recursive algorithm by entering“20”and pressing the“Fi-bonacci Recursive”button.You will see an endless number of recursive calls.Now,press“Skip Forward”to stop the current algorithm and call the iterative algorithmby pressing“Fibonacci Table”.This will compute F20very quickly.(Note that the visualization uses a slightly different definition of Fibonacci numbers:F0=1insteadof F0=0.This of course has almost no influence on the running time.)Problem DescriptionTask.Given an integer n,find the n th Fibonacci number F n.Input Format.The input consists of a single integer n.Constraints.0≤n≤45.Output Format.Output F n.Time Limits.language C C++Java Python C#Haskell JavaScript Ruby Scala time in seconds11 1.55 1.52553 Memory Limit.512MB.Sample1.Input:3Output:2Explanation:F3=2.4Sample2.Input:10Output:55Explanation:F10=55.Starter FilesThe starter files for this problem contain an implementation of a naive algorithm for the problem in C++, Java,and Python3.For these programming languages,your goal is to replace the naive algorithm with an efficient one.For other programming languages,you need to implement a solution from scratch.Filename: fibonacciNeed Help?Ask a question or see the questions asked by other learners at this forum thread.52Problem:Last Digit of a Large Fibonacci NumberProblem IntroductionYour goal in this problem is to find the last digit of n-th Fibonacci number.Recall that Fibonacci numbers grow exponentially fast.For example,F200=280571172992510140037611932413038677189525.Therefore,a solution likeF[0]←0F[1]←1for i from2to n:F[i]←F[i−1]+F[i−2]print(F[n]mod10)will turn out to be too slow,because as i grows the i th iteration of the loop computes the sum of longer and longer numbers.Also,for example,F1000does not fit into the standard C++int type.To overcome this difficulty,you may want to store in F[i]not the i th Fibonacci number itself,but just its last digit(that is,F i mod10).Computing the last digit of F i is easy:it is just the last digit of the sum of the last digits of F i−1and F i−2:F[i]←(F[i−1]+F[i−2])mod10This way,all F[i]’s are just digits,so they fit perfectly into any standard integer type,and computing a sum of F[i−1]and F[i−2]is performed very quickly.Problem DescriptionTask.Given an integer n,find the last digit of the n th Fibonacci number F n(that is,F n mod10).Input Format.The input consists of a single integer n.Constraints.0≤n≤107.Output Format.Output the last digit of F n.Time Limits.language C C++Java Python C#Haskell JavaScript Ruby Scalatime in seconds11 1.55 1.52553Memory Limit.512MB.Sample1.Input:3Output:2Explanation:F3=2.6Sample2.Input:331Output:9Explanation:F331=668996615388005031531000081241745415306766517246774551964595292186469. Sample3.Input:327305Output:5Explanation:F327305does not fit into one line of this pdf,but its last digit is equal to5.Starter FilesThe starter files for this problem contain an implementation of a naive algorithm for the problem in C++, Java,and Python3.For these programming languages,your goal is to replace the naive algorithm with an efficient one.For other programming languages,you need to implement a solution from scratch.Filename: fibonacci_last_digitNeed Help?Ask a question or see the questions asked by other learners at this forum thread.73Problem:Greatest Common Divisor Problem IntroductionThe greatest common divisor GCD(a,b)of two non-negative integers a and b (which are not both equal to0)is the greatest integer d that divides both a and b. Your goal in this problem is to implement the Euclidean algorithm for computing the greatest common divisor.Efficient algorithm for computing the greatest common divisor is an important basic primitive of commonly used cryptographic algorithms like RSA.GCD(1344,217) =GCD(217,42) =GCD(42,7)=GCD(7,0)=7Problem DescriptionTask.Given two integers a and b,find their greatest common divisor.Input Format.The two integers a,b are given in the same line separated by space.Constraints.1≤a,b≤2·109.Output Format.Output GCD(a,b).Time Limits.language C C++Java Python C#Haskell JavaScript Ruby Scalatime in seconds11 1.55 1.52553Memory Limit.512MB.Sample1.Input:1835Output:1Explanation:18and35do not have common non-trivial divisors.Sample2.Input:288515381183019Output:17657Explanation:28851538=17657·1634,1183019=17657·67.Starter FilesThe starter files for this problem contain an implementation of a naive algorithm for the problem in C++, Java,and Python3.For these programming languages,your goal is to replace the naive algorithm with an efficient one.For other programming languages,you need to implement a solution from scratch.Filename: gcdWhat To DoTo solve this problem,it is enough to implement carefully the corresponding algorithm covered in the lectures.8Need Help?Ask a question or see the questions asked by other learners at this forum thread.94Problem:Least Common MultipleProblem IntroductionThe least common multiple of two positive integers a and b is the least positiveinteger m that is divisible by both a and b.Problem DescriptionTask.Given two integers a and b,find their least common multiple.Input Format.The two integers a and b are given in the same line separated by space.Constraints.1≤a,b≤2·109.Output Format.Output the least common multiple of a and b.Time Limits.language C C++Java Python C#Haskell JavaScript Ruby Scalatime in seconds11 1.55 1.52553Memory Limit.512MB.Sample1.Input:68Output:24Explanation:Among all the positive integers that are divisible by both6and8(e.g.,48,480,24),24is the smallest one.Sample2.Input:288515381183019Output:1933053046Explanation:1933053046is the smallest positive integer divisible by both28851538and1183019.Starter FilesThe starter files for this problem contain an implementation of a naive algorithm for the problem in C++, Java,and Python3.For these programming languages,your goal is to replace the naive algorithm with an efficient one.For other programming languages,you need to implement a solution from scratch.Filename: lcm10What To DoPlay with small datasets to find out how the least common multiple LCM(a,b)is related to the greatest common divisor GCD(a,b).Do you see?For any two positive integers a and b,LCM(a,b)·GCD(a,b)=a·b. Prove this property and implement an algorithm for computing the least common multiple using your solution for the greatest common divisor.Need Help?Ask a question or see the questions asked by other learners at this forum thread.115Advanced Problem:Huge Fibonacci Number modulo mWe strongly recommend you start solving advanced problems only when you are done with the basic problems (for some advanced problems,algorithms are not covered in the video lectures and require additional ideas to be solved;for some other advanced problems,algorithms are covered in the lectures,but implementing them is a more challenging task than for other problems).Problem IntroductionIn this problem,your goal is to compute F n modulo m,where n may be really huge:up to1018.For such values of n,an algorithm looping for n iterations will not fit into one second for sure.Therefore we need to avoid such a loop.To get an idea how to solve this problem without going through all F i for i from0to n,let’s see what happens when m is small—say,m=2or m=3.i0123456789101112131415F i01123581321345589144233377610F i mod20110110110110110F i mod30112022101120221Take a detailed look at this table.Do you see?Both these sequences are periodic!For m=2,the period is011and has length3,while for m=3the period is01120221and has length8.Therefore,to compute, say,F2015mod3we just need to find the remainder of2015when divided by8.Since2015=251·8+7,we conclude that F2015mod3=F7mod3=1.This is true in general:for any integer m≥2,the sequence F n mod m is periodic.The period always starts with01and is known as Pisano period.Problem DescriptionTask.Given two integers n and m,output F n mod m(that is,the remainder of F n when divided by m). Input Format.The input consists of two integers n and m given on the same line(separated by a space). Constraints.1≤n≤1018,2≤m≤105.Output Format.Output F n mod m.Time Limits.language C C++Java Python C#Haskell JavaScript Ruby Scalatime in seconds11 1.55 1.52553Memory Limit.512MB.Sample1.Input:1239Output:1Explanation:F1mod239=1mod239=1.12Sample2.Input:2391000Output:161Explanation:F239mod1000=39679027332006820581608740953902289877834488152161(mod1000)=161. Sample3.Input:281621358830524Output:10249Explanation:F2816213588does not fit into one page of this file,but F2816213588mod30524=10249.Starter FilesThe starter files for this problem contain an implementation of a naive algorithm for the problem in C++, Java,and Python3.For these programming languages,your goal is to replace the naive algorithm with an efficient one.For other programming languages,you need to implement a solution from scratch.Filename: fibonacci_hugeNeed Help?Ask a question or see the questions asked by other learners at this forum thread.136Advanced Problem:Sum of Fibonacci NumbersWe strongly recommend you start solving advanced problems only when you are done with the basic problems (for some advanced problems,algorithms are not covered in the video lectures and require additional ideas to be solved;for some other advanced problems,algorithms are covered in the lectures,but implementing them is a more challenging task than for other problems).Problem IntroductionThe goal in this problem is to find the last digit of a sum of the first n Fibonacci numbers.Problem DescriptionTask.Given an integer n,find the last digit of the sum F0+F1+···+F n.Input Format.The input consists of a single integer n.Constraints.0≤n≤1014.Output Format.Output the last digit of F0+F1+···+F n.Time Limits.language C C++Java Python C#Haskell JavaScript Ruby Scalatime in seconds11 1.55 1.52553Memory Limit.512MB.Sample1.Input:3Output:4Explanation:F0+F1+F2+F3=0+1+1+2=4.Sample2.Input:100Output:5Explanation:The sum is equal to927372692193078999175,the last digit is5.Starter FilesThe starter files for this problem contain an implementation of a naive algorithm for the problem in C++, Java,and Python3.For these programming languages,your goal is to replace the naive algorithm with an efficient one.For other programming languages,you need to implement a solution from scratch.Filename: fibonacci_sum_last_digitWhat To DoInstead of computing this sum in a loop,try to come up with a formula for this sum.For this,play with small values of n.Then,use a solution for the previous problem.14Need Help?Ask a question or see the questions asked by other learners at this forum thread.157Advanced Problem:Partial Sum of Fibonacci NumbersWe strongly recommend you start solving advanced problems only when you are done with the basic problems (for some advanced problems,algorithms are not covered in the video lectures and require additional ideas to be solved;for some other advanced problems,algorithms are covered in the lectures,but implementing them is a more challenging task than for other problems).Problem IntroductionNow,we would like to find the last digit of a partial sum of Fibonacci numbers:F m+F m+1+···+F n.Problem DescriptionTask.Given two non-negative integers m and n,where m≤n,find the last digit of the sum F m+F m+1+···+F n.Input Format.The input consists of two non-negative integers m and n separated by a space. Constraints.0≤m≤n≤1018.Output Format.Output the last digit of F m+F m+1+···+F n.Time Limits.language C C++Java Python C#Haskell JavaScript Ruby Scalatime in seconds11 1.55 1.52553Memory Limit.512MB.Sample1.Input:37Output:1Explanation:F3+F4+F5+F6+F7=2+3+5+8+13=31.Sample2.Input:1010Output:5Explanation:F10=55.Sample3.Input:10200Output:2Explanation:F10+F11+···+F200=73454486715781809323490890211044929642326216Starter FilesThe starter files for this problem contain an implementation of a naive algorithm for the problem in C++, Java,and Python3.For these programming languages,your goal is to replace the naive algorithm with an efficient one.For other programming languages,you need to implement a solution from scratch.Filename: fibonacci_partial_sumWhat To DoUse a solution for the previous problem to solve this problem.Need Help?Ask a question or see the questions asked by other learners at this forum thread.178General Instructions and Recommendations on Solving Algorith-mic ProblemsYour main goal in an algorithmic problem is to implement a program that solves a given computational problem in just few seconds even on massive datasets.Your program should read a dataset from the standard input and write an answer to the standard output.Below we provide general instructions and recommendations on solving such problems.Before reading them,go through readings and screencasts in the first module that show a step by step process of solving two algorithmic problems:link.8.1Reading the Problem StatementYou start by reading the problem statement that contains the description of a particular computational task as well as time and memory limits your solution should fit in,and one or two sample tests.In some problems your goal is just to implement carefully an algorithm covered in the lectures,while in some other problems you first need to come up with an algorithm yourself.8.2Designing an AlgorithmIf your goal is to design an algorithm yourself,one of the things it is important to realize is the expected running time of your ually,you can guess it from the problem statement(specifically,from the subsection called constraints)as follows.Modern computers perform roughly108–109operations per second.So,if the maximum size of a dataset in the problem description is n=105,then most probably an algorithm with quadratic running time is not going to fit into time limit(since for n=105,n2=1010)while a solution with running time O(n log n)will fit.However,an O(n2)solution will fit if n is up to103=1000, and if n is at most100,even O(n3)solutions will fit.In some cases,the problem is so hard that we do not know a polynomial solution.But for n up to18,a solution with O(2n n2)running time will probably fit into the time limit.To design an algorithm with the expected running time,you will of course need to use the ideas covered in the lectures.Also,make sure to carefully go through sample tests in the problem description.8.3Implementing Your AlgorithmWhen you have an algorithm in mind,you start implementing it.Currently,you can use the following programming languages to implement a solution to a problem:C,C++,C#,Haskell,Java,JavaScript, Python2,Python3,Ruby,Scala.For all problems,we will be providing starter solutions for C++,Java,and Python3.If you are going to use one of these programming languages,use these starter files.For other programming languages,you need to implement a solution from scratch.8.4Compiling Your ProgramFor solving programming assignments,you can use any of the following programming languages:C,C++, C#,Haskell,Java,JavaScript,Python2,Python3,Ruby,and Scala.However,we will only be providing starter solution files for C++,Java,and Python3.The programming language of your submission is detected automatically,based on the extension of your submission.We have reference solutions in C++,Java and Python3which solve the problem correctly under the given restrictions,and in most cases spend at most1/3of the time limit and at most1/2of the memory limit. You can also use other languages,and we’ve estimated the time limit multipliers for them,however,we have no guarantee that a correct solution for a particular problem running under the given time and memory constraints exists in any of those other languages.Your solution will be compiled as follows.We recommend that when testing your solution locally,you use the same compiler flags for compiling.This will increase the chances that your program behaves in the18same way on your machine and on the testing machine(note that a buggy program may behave differently when compiled by different compilers,or even by the same compiler with different flags).∙C(gcc5.2.1).File extensions:.c.Flags:gcc-pipe-O2-std=c11<filename>-lm∙C++(g++5.2.1).File extensions:.cc,.cpp.Flags:g++-pipe-O2-std=c++14<filename>-lmIf your C/C++compiler does not recognize-std=c++14flag,try replacing it with-std=c++0x flag or compiling without this flag at all(all starter solutions can be compiled without it).On Linux and MacOS,you most probably have the required compiler.On Windows,you may use your favorite compiler or install,e.g.,cygwin.∙C#(mono3.2.8).File extensions:.cs.Flags:mcs∙Haskell(ghc7.8.4).File extensions:.hs.Flags:ghc-O∙Java(Open JDK8).File extensions:.java.Flags:javac-encoding UTF-8java-Xmx1024m∙JavaScript(Node v6.3.0).File extensions:.js.Flags:nodejs∙Python2(CPython2.7).File extensions:.py2or.py(a file ending in.py needs to have a first line which is a comment containing“python2”).No flags:python2∙Python3(CPython3.4).File extensions:.py3or.py(a file ending in.py needs to have a first line which is a comment containing“python3”).No flags:python3∙Ruby(Ruby2.1.5).File extensions:.rb.ruby∙Scala(Scala2.11.6).File extensions:.scala.scalac198.5Testing Your ProgramWhen your program is ready,you start testing it.It makes sense to start with small datasets—for example, sample tests provided in the problem description.Ensure that your program produces a correct result.You then proceed to checking how long does it take your program to process a massive dataset.For this,it makes sense to implement your algorithm as a function like solve(dataset)and then implement an additional procedure generate()that produces a large dataset.For example,if an input to a problem is a sequence of integers of length1≤n≤105,then generate a sequence of length exactly105,pass it to your solve()function,and ensure that the program outputs the result quickly.Also,check the boundary values.Ensure that your program processes correctly sequences of size n= 1,2,105.If a sequence of integers from0to,say,106is given as an input,check how your program behaves when it is given a sequence0,0,...,0or a sequence106,106,...,106.Check also on randomly generated data.For each such test check that you program produces a correct result(or at least a reasonably looking result).In the end,we encourage you to stress test your program to make sure it passes in the system at the first attempt.See the readings and screencasts from the first week to learn about testing and stress testing:link.8.6Submitting Your Program to the Grading SystemWhen you are done with testing,you submit your program to the grading system.For this,you go the submission page,create a new submission,and upload a file with your program.The grading system then compiles your program(detecting the programming language based on your file extension,see Subsection8.4) and runs it on a set of carefully constructed tests to check that your program always outputs a correct result and that it always fits into the given time and memory limits.The grading usually takes no more than a minute,but in rare cases when the servers are overloaded it might take longer.Please be patient.You can safely leave the page when your solution is uploaded.As a result,you get a feedback message from the grading system.The feedback message that you will love to see is:Good job!This means that your program has passed all the tests.On the other hand, the three messages Wrong answer,Time limit exceeded,Memory limit exceeded notify you that your program failed due to one these three reasons.Note that the grader will not show you the actual test you program have failed on(though it does show you the test if your program have failed on one of the first few tests;this is done to help you to get the input/output format right).8.7Debugging and Stress Testing Your ProgramIf your program failed,you will need to debug it.Most probably,you didn’t follow some of our suggestions from the section8.5.See the readings and screencasts from the first week to learn about debugging your program:link.You are almost guaranteed to find a bug in your program using stress testing,because the way these programming assignments and tests for them are prepared follows the same process:small manual tests, tests for edge cases,tests for large numbers and integer overflow,big tests for time limit and memory limit checking,random test generation.Also,implementation of wrong solutions which we expect to see and stress testing against them to add tests specifically against those wrong solutions.Go ahead,and we hope you pass the assignment soon!20。

acm递推求解

acm递推求解不管道路多么崎岖坎坷，我永远不停下追逐梦想的脚步！商人，不佩剑；超级楼梯timelimit:2000/1000ms(java/other)memorylimit:65536/32768k(java/other)totalsubmission(s):0acceptedsubmission(s):0problemdescription有一楼梯共m级，刚开始时你在第一级，若每次只能跨上一级或二级，要走上第m级，共有多少种走法？input输入数据首先包含一个整数n，表示测试实例的个数，然后是n行数据，每行包含一个整数m（1<=m<=40）,表示楼梯的级数。

output对于每个测试实例，请输出不同走法的数量sampleinput223sampleoutput12authorlcysource2021实验班长学期考试我的错误代码#include<stdio.h>intmain(){intn,i,j,m;intcount=0;while(scanf("%d",&n)!=eof){scanf("%d",&m);while(n--){for(i=0;i<=m;i++){for(j=0;j<=(int)(m/2);j++){if(i+2*j==m)count++;}}}printf("%d\\n",count);count=0;}return0;}当你抵达第n阶的时候存有两种抵达方式。

在n-1处上1个楼梯。

在n-2处上2个楼梯。

所以上n阶楼梯的情况总数=上n-1的总数+上n-2的总数这样递推公式就出来了。

f（n）=f（n-1）+f（n-2）11111111111#include<stdio.h>ints[41];intmain(){intt,n;scanf("%d",&t);s[1]=s[2]=1;//第一级到第二级只一种，就在第一级当然只一种。

Problem-solvingb...

Neurocomputing44–46(2002)735–742/locate/neucomProblem-solving behavior in a system modelof the primate neocortexAlan H.BondCalifornia Institute of Technology,Mailstop136-93,Pasadena,CA91125,USAAbstractWe show how our previously described system model of the primate neocortex can be extended to allow the modeling of problem-solving behaviors.Speciÿcally,we model di erent cognitive strategies that have been observed for human subjects solving the Tower of Hanoi problem. These strategies can be given a naturally distributed form on the primate neocortex.Further, the goal stacking used in some strategies can be achieved using an episodic memory module corresponding to the hippocampus.We can give explicit falsiÿable predictions for the time sequence of activations of di erent brain areas for each strategy.c 2002Published by Elsevier Science B.V.Keywords:Neocortex;Modular architecture;Perception–action hierarchy;Tower of Hanoi;Problem solving;Episodic memory1.Our system model of the primate neocortexOur model[4–6]consists of a set of processing modules,each representing a corti-cal area.The overall architecture is a perception–action hierarchy.Data stored in each module is represented by logical expressions we call descriptions,processing within each module is represented by sets of rules which are executed in parallel and which construct new descriptions,and communication among modules consists of the trans-mission of descriptions.Modules are executed in parallel on a discrete time scale, corresponding to20ms.During one cycle,all rules are executed once and all inter-module transmission of descriptions occurs.Fig.1depicts our model,as a set of cor-tical modules and as a perception–action hierarchy system diagram.The action of theE-mail address:***************.edu(A.H.Bond).0925-2312/02/$-see front matter c 2002Published by Elsevier Science B.V.PII:S0925-2312(02)00466-6736 A.H.Bond/Neurocomputing44–46(2002)735–742Fig.1.Our system model shown in correspondence with the neocortex,and as a perception–action hierarchy.system is to continuously create goals,prioritize goals,and elaborate the highest priority goals into plans,then detailed actions by propagating descriptions down the action hierarchy,resulting in a stream of motor commands.(At the same time,perception of the environment occurs in a ow of descriptions up the perception hierarchy.Perceived descriptions condition plan elaboration,and action descriptions condition perception.) This simple elaboration of stored plans was su cient to allow is to demonstrate simple socially interactive behaviors using a computer realization of our model.A.H.Bond/Neurocomputing44–46(2002)735–7427372.Extending our model to allow solution of the Tower of Hanoi problem2.1.Tower of Hanoi strategiesThe Tower of Hanoi problem is the most studied,and strategies used by human subjects have been captured as production rule systems[9,1].We will consider the two most frequently observed strategies—the perceptual strategy and the goal recursion strategy.In the general case,reported by Anzai and Simon[3],naive subjects start with an initial strategy and learn a sequence of strategies which improve their performance. Our two strategies were observed by Anzai and Simon as part of this learning sequence. Starting from Simon’s formulation[8],we were able to represent these two strategies in our model,as follows:2.2.Working goalsSince goals are created dynamically by the planning activity,we needed to extend our plan module to allow working goals as a description type.This mechanism was much better than trying to use the main goal module.We can limit the number of working goals.This would correspond to using aÿxed size store,corresponding to working memory.The module can thus create working goals and use the current working goals as input to rules.Working goals would be held in dorsal prefrontal areas,either as part of or close to the plan module.Main motivating topgoals are held in the main goal module corresponding to anterior cingulate.2.3.Perceptual tests and mental imageryThe perceptual tests on the external state,i.e.the state of the Tower of Hanoi apparatus,were naturally placed in a separate perception module.This corresponds to Kosslyn’s[7]image store.The main perceptual test needed is to determine whether a proposed move is legal.This involves(a)making a change to a stored perceived representation corresponding to making the proposed move,and(b)making a spatial comparison in this image store to determine whether the disk has been placed on a smaller or a larger one.With these two extensions,we were able to develop a representation of the perceptual strategy,depicted in Fig.2.3.Episodic memory and its use in goal stackingIn order to represent the goal recursion strategy,we need to deal with goal stacking, which is represented by push and pop operations in existing production rule represen-tations.Since we did not believe that a stack with push and pop operations within a module is biologically plausible,we found an equivalent approach using an episodic memory module.738 A.H.Bond/Neurocomputing44–46(2002)735–742Fig.2.Representation of the perceptual strategy on our brain model.This module creates associations among whatever inputs it receives at any given time, and it sends these associations as descriptions to be stored in contributing modules. In general,it will create episodic representations from events occurring in extended temporal intervals;however,in the current case we only needed simple association. In the Tower of Hanoi case,the episode was simply taken to be an association between the current working goal and the previous,parent,working goal.We assume that these two working goals are always stored in working memory and are available to the plan module.The parent forms a context for the working goal.The episode description is formed in the episodic memory module and transmitted to the plan module where it is stored.The creation of episodic representations can proceed in parallel with the problem solving process,and it can occur automatically or be requested by the plan module.Rules in the plan module can retrieve episodic descriptions usingA.H.Bond/Neurocomputing44–46(2002)735–742739the current parent working goal,and can replace the current goal with the current parent,and the current parent with its retrieved parent.Thus the working goal context can be popped.This representation is more general than a stack,since any stored episode could be retrieved,including working goals from episodes further in the past. Such e ects have,in fact,been reported by Van Lehn et al.[10]for human subjects. With this additional extension,we were able to develop a representation of the goal recursion strategy,depicted in Fig.3.Descriptions of episodes are of the form con-text(goal(G),goal context(C)).goal(G)being the current working goal and goal context(C)the current parent working goal.Theÿgure shows a slightly more general version,where episodes are stored both in the episodic memory module and the plan module.This allows episodes that have not yet been transferred to the cortex to be used.We are currently working on extending our model to allow the learning a sequence of strategies as observed by Anzai and Simon.This may result in a di erent representation of these strategies,and di erent performance.740 A.H.Bond/Neurocomputing44–46(2002)735–742during perceptual analysis during movementP MFig.4.Predictions of brain area activation during Tower of Hanoi solving.4.Falsiÿable predictions of brain area activationFor the two strategies,we can now generate detailed predictions of brain area acti-vation sequences that should be observed during the solution of the Tower of Hanoi ing our computer realization,we can generate detailed predictions of activa-tion levels for each time step.Since there are many adjustable parameters and detailed assumptions in the model,it is di cult toÿnd clearly falsiÿable predictions.However, we can also make a simpliÿed and more practical form of prediction by classifying brain states into four types,shown in Fig.4.Let us call these types of states G,E,P and M,respectively.Then,for example,the predicted temporal sequences of brain state types for3disks are:A.H.Bond/Neurocomputing44–46(2002)735–742741For the perceptual strategy:G0;G;E;P;G;E;P;G;E;P;E;M;P;G;E;P;G;E;P;E;M;P;G;E;P;G;E;P;E;M;P;G;E;P;E;M;P;G;E;P;G;E;P;E;M;P;G;E;P;E;M;P;G;E;P;E;M;P;G0:and for the goal recursion strategy:G0;G;E;P;G+;E;P;G+;E;P;E;M;P;G∗;E;P;E;M;P;G∗;E;P;G+;E;P;E;M;P;G∗;E;P;E;M;P;G;E;P;G+;E;P;E;M;P;G∗;E;P;E;M;E;G;E;P;E;M;P;G0: We can generate similar sequences for di erent numbers of disks and di erent strate-gies.The physical moves of disks occur during M steps.The timing is usually about 3:5s per physical move,but the physical move steps probably take longer than the average cognitive step.If a physical move takes1:5s,this would leave about300ms per cognitive step.The perceptual strategy used is an expert strategy where the largest disk is always selected.We assume perfect performance;when wrong moves are made,we need a theory of how mistakes are made,and then predictions can be generated.In the goal recursion strategy,we assume the subject is using perceptual tests for proposed moves, and is not working totally from memory.G indicates the creation of a goal,G+a goal creation and storing an existing goal(push),and G∗the retrieval of a goal(pop). Anderson et al.[2]have shown that pushing a goal takes about2s,although we have taken creation of a goal to not necessarily involve pushing.For us,pushing only occurs when a new goal is created and an existing goal has to be stored.G0is activity relating to the top goal.It should be noted that there is some redundancy in the model,so that,if a mismatch to experiment is found,it would be possible to make some changes to the model to bring it into better correspondence with the data.For example,the assignment of modules to particular brain areas is tentative and may need to be changed.However, there is a limit to the changes that can be made,and mismatches with data could falsify the model in its present form.AcknowledgementsThis work has been partially supported by the National Science Foundation,Informa-tion Technology and Organizations Program managed by Dr.Les Gasser.The author would like to thank Professor Pietro Perona for his support,and Professor Steven Mayo for providing invaluable computer resources.References[1]J.R.Anderson,Rules of the Mind,Lawrence Erlbaum Associates,Hillsdale,NJ,1993.[2]J.R.Anderson,N.Kushmerick,C.Lebiere,The Tower of Hanoi and Goal structures,in:J.R.Anderson(Ed.),Rules of the Mind,Lawrence Erlbaum Associates,Hillsdale,New Jersey,1993,pp.121–142.742 A.H.Bond/Neurocomputing44–46(2002)735–742[3]Y.Anzai,H.A.Simon,The theory of learning by doing,Psychol.Rev.86(1979)124–140.[4]A.H.Bond,A computational architecture for social agents,Proceedings of Intelligent Systems:ASemiotic Perspective,An International Multidisciplinary Conference,National Institute of Standards and Technology,Gaithersburg,Maryland,USA,October20–23,1996.[5]A.H.Bond,A system model of the primate neocortex,Neurocomputing26–27(1999)617–623.[6]A.H.Bond,Describing behavioral states using a system model of the primate brain,Am.J.Primatol.49(1999)315–388.[7]S.Kosslyn,Image and Brain,MIT Press,Cambridge,MA,1994.[8]H.A.Simon,The functional equivalence of problem solving skills,Cognitive Psychol.7(1975)268–288.[9]K.VanLehn,Rule acquisition events in the discovery of problem-solving strategies,Cognitive Sci.15(1991)1–47.[10]K.VanLehn,W.Ball,B.Kowalski,Non-LIFO execution of cognitive procedures,Cognitive Sci.13(1989)415–465.Alan H.Bond was born in England and received a Ph.D.degree in theoretical physics in1966from Imperial College of Science and Technology,University of London.During the period1969–1984,he was on the faculty of the Computer Science Department at Queen Mary College,London University,where he founded and directed the Artiÿcial Intelligence and Robotics Laboratory.Since1996,he has been a Senior Scientist and Lecturer at California Institute of Technology.His main research interest concerns the system modeling of the primate brain.。

leetcode算法题库

leetcode算法题库LeetCode is a popular online platform that is dedicated to helping individuals practice and improve their coding skills. It offers a vast collection of algorithmic problems that are designed to test the problem-solving abilities of programmers.Here are some of the popular algorithmic problem categories available on LeetCode:1. Arrays and Strings: This category covers problems related to array manipulation, string manipulation, and pattern matching.2. Linked Lists: This category focuses on problems related to linked lists, such as reversing a linked list, merging two linked lists, and detecting cycles in a linked list.3. Binary Trees: Problems in this category involve binary tree traversal, binary tree construction, and binary tree manipulation.4. Dynamic Programming: This category contains problems that require the use of dynamic programming techniques to optimize solutions to subproblems.5. Sorting and Searching: Problems in this category emphasize sorting and searching algorithms, such as merge sort, quicksort, binary search, and more.6. Graphs: This category covers problems related to graph traversal, graph algorithms (such as Breadth-First Search and Depth-First Search), and graph manipulation.7. Backtracking: Problems in this category often require exploring all possible solutions by iteratively building and exploring potential solutions.8. Math: Math-related problems include topics like prime numbers, number theory, combinatorics, and geometry.9. Bit Manipulation: This category covers problems that require bitwise manipulation, such as finding the single number in an array, counting bits, and performing bitwise operations.10. Design: Design problems encompass designing data structures and algorithms to solve specific problems efficiently.Each category contains a variety of different problem difficulties, ranging from easy to hard. LeetCode also provides a discussion forum where users can share their approaches and discuss problem solutions with each other.By practicing problems on LeetCode, users can enhance their problem-solving skills, learn new algorithms and data structures, and prepare for coding interviews.。

l-shape算法例题

l-shape算法例题英文回答：L-shape algorithm is a popular method used in computational geometry to solve the problem of finding the intersection of two line segments. It is named after the shape that the line segments form when they intersect at a right angle. The algorithm is based on the observation that if two line segments intersect, then there must be an L-shaped region formed by the endpoints of the line segments.To apply the L-shape algorithm, we first check if the bounding boxes of the two line segments intersect. If they do not intersect, then we can conclude that the line segments do not intersect. If the bounding boxes do intersect, we proceed to check if the line segmentsactually intersect.The algorithm works by considering the relative positions of the endpoints of the line segments. We startby checking if the endpoints of one line segment are on opposite sides of the other line segment. If they are, then the line segments intersect. If not, we continue to check if the endpoints of the other line segment are on opposite sides of the first line segment. If they are, then the line segments intersect. If neither of these conditions are met, then the line segments do not intersect.Let's take an example to illustrate the L-shape algorithm. Suppose we have two line segments: AB and CD. The coordinates of the endpoints are as follows:A(1, 1)。

Problem1011（c++）

Problem1011（c++）/*1:越长的木棍对后面木棍的约束力越大,因此要把小木棍排序, 按木棍长度从大到小搜索,这样就能在尽可能靠近根的地方剪枝。

:当出现加上某根木棍恰好能填满一根原始木棍,但由在后面的搜索中失败了,就不必考虑其他木棍了,直接退出当前的枚举。

:考虑每根原始木棍的第一根木棍,如果当前枚举的木棍长度无法得出合法解,就不必考虑下一根木棍了，当前木棍一定是作为某根原始木棍的第一根木棍的，现在不行,以后也不可能得出合法解。

也就是说每根原始木棍的第一根小木棍一定要成功,否则就返回。

:剩下一个通用的剪枝就是跳过重复长度的木棍，当前木棍跟它后面木棍的无法得出合法解，后面跟它一样长度的木棍也不可能得到合法解,因为后面相同长度木棍能做到的,前面这根木棍也能做到。

*/#include<stdio.h>#include<string.h>#include<stdlib.h>int cmp(const void*p1,const void*p2){return *(int*)p2-*(int*)p1;}int a[100],v[100];int n,len,s,tot,min;bool dfs(int k,int mi,int left){int i;if(left==min)for(i=k;i<=n;i++)if(!v[i]&&a[i]<=mi){v[i]=1;if(a[i]==mi&&dfs(1,len,left-a[i])) return 1;else if(dfs(i+1,mi-a[i],left-a[i])) return 1;v[i]=0;if(a[i]==min)return 0;if(left==tot)return 0;if(mi==len) return 0;while(a[i+1]==a[i])i++;}return 0;}int main(){int i,res;while(scanf("%d",&n),n){tot=0;for(i=1;i<=n;i++){scanf("%d",&a[i]);tot+=a[i];}qsort(a+1,n,sizeof(a[0]),cmp); len=a[1];memset(v,0,sizeof(v));res=tot;for(;len<tot;len++)if(tot%len==0&&dfs(1,len,tot)) {res=len;break;}printf("%d\n",res);}return 0;}。

真空碳热还原固相合成磷酸铁锂的正交实验

真空碳热还原固相合成磷酸铁锂的正交实验周环波;林丽;程凡;龚春丽;王丽【摘要】采用真空条件下的碳热还原固相方法,以Fe2O3、LIOH和P2O5为原料、活性炭为还原剂,合成了锂离子电池正极材料LiFePO4;选择真空度、原料配比、反应温度、恒温保持时间(反应时间)和升温速率等5个影响因素,设计了一组5因素4水平(L1645)的正交实验,较系统地研究了LiFePO4的合成工艺参数及电化学性能.研究结果表明:真空碳还原合成LiFePO4工艺参数因素对材料电化学性能影响的大小顺序为:反应温度>原料配比>反应时间>真空度>升温速率;反应温度控制在600-650 ℃时所合成的LiFePO4材料的晶体结构和电性能较好;原料LiOH、Fe2O3、P2O5和活性炭比例为1.05:1:1:1.8、反应时间为12～18 h、真空度控制10-1～10 Pa时,所得LiFePO4材料的综合电化学性能较好.【期刊名称】《电源技术》【年(卷),期】2010(034)004【总页数】4页(P367-370)【关键词】磷酸铁锂;正交实验;碳热还原;真空【作者】周环波;林丽;程凡;龚春丽;王丽【作者单位】孝感学院化学与材料科学学院,湖北,孝感,432100;孝感学院新技术学院,湖北,孝感,432100;孝感学院化学与材料科学学院,湖北,孝感,432100;孝感学院化学与材料科学学院,湖北,孝感,432100;孝感学院化学与材料科学学院,湖北,孝感,432100【正文语种】中文【中图分类】TM912.9磷酸铁锂(LiFePO4)作为锂离子电池正极材料具有良好的发展前景，该材料具有极好的热稳定性，制造的电池具有极好的安全性。

LiFePO4过充电时的分解反应温度最宽、在充放电过程中该材料的体积变化较小(约为9.6%)，而且这种变化刚好与碳负极在充放电过程所发生的体积膨胀相当[1]；LiFe-PO4具有较高的理论比容量和较高的工作电压；合成LiFePO4的原料丰富、价格便宜、无环境污染。

U2知识点总复习

1. I always feel relaxed when I lie in bed,
listening to music.
躺
2.
You should never to lie
yaolwuar yfr撒sieb谎nedsh.onest
and
3. Cuckoos(布谷下鸟蛋) lay eggs in other
cutting tress
Any more?
water pollutio
n
problems
air pollutio
n
rubbish/ waste
economy 经济, 节约
economical 节约的, 经济的
economically 经济地, 经济学地 economics 经济学
economicsmism 经济主义
adjust to a new environment 适应环境 beautify the environment 美化环境
inhabited environment
居住环境
New words and phrases:
ecotourist ,ecosystem, environmental protection solar energy, fossil fuels, eco-friendly clear-cut logging, opencast mining slash, ozone layer, acid rain , global warming
e.g. In addition, I have one more suggestion to make. 另外，我还有个建议要提。
In addition to the tips mentioned above, students should pay attention to one more tip. 除了上所提到的注意事项，学生们还得注意一点。

燃气管道设计中管道转角有关问题的探讨

燃气管道设计中管道转角有关问题的探讨2010-2-20李华琴李永威邹涛分享到： QQ空间新浪微博开心网人人网摘要：在燃气管道设计中对管道转角的处理，可以采用机制弯管、热煨弯头、冷弯弯管、弹性敷设。

针对各种管道转角处理方式，给出了规范要求、应用条件、计算公式和工程实例。

关键词：管道转角；热煨弯管；冷弯弯管；机制弯头；弹性敷设Discussion on Problems Concerning Pipeline Corners in Design of Gas PipelineLI Hua-qin，LI Yong-wei，ZOU TaoAbstract：The mechanical elbow，hot-bending bend，cold-bending bend and elastic installation can be used to handle pipeline corners in design of gas pipeline. Aimed at the various handling modes of pipeline corn ers，the specifications，application conditions，calculation formula an d engineering example are given.Key words：pipeline comer；hot-bending bend；cold-bending bend；mechani cal elbow； elastir installation1 问题的提出在进行城市燃气管道设计时，经常会遇到管道转角的问题，需要合理解决。

目前有关这方面的研究不多，而且在工程设计中也经常被我们忽略。

然而管道转角的问题非常重要，主要包括以下4个方面：① 在管道设计中冷弯弯管[1]、热煨弯管[2]、机制弯头[3、4]及弹性敷设[1]的适用范围及确定原则。

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Problems
1. Using equation )1(822
+=J J I h E r π calculate the rotational energy
levels, in joules, for J=0 to 2 for 12C 16O and 13C 16O. Convert these to units of cm -1. Given r e =0.1128nm for CO molecule on the ground state; h=6.626×10-34Js; m u =1.66054×10-27kg, 1J=5.03411×1022 cm -1.
2. Given that the vibration wavenumbers of the molecules HCl, SO and
PN are 2991, 1149 and 1337 cm -1 respectively, calculate from
2121⎪⎪⎭⎫ ⎝⎛=μπνk their force constants and hence comment on the
comparative bond strengths. Known m u =1.66054×10-27kg.
3. Derive possible terms arising from an s and a d electron in both LS and
jj coupling approximations.
4. Why is the He 2 molecule unstable in its ground electronic state? Give
the electron configurations for two excited states of low energy that would be expected to be stable.
5. Would the ground state of the He 2+ ion be expected to be stable?
6. Which should have the larger equilibrium bond length in the ground
state (a) the N2 molecule or the N2+ ion (b) the Li2 or the K2 molecule?
7. The Raman spectrum of gaseous hydrogen bromide was excited by the
radiation from a mercury at 435.83, 404.44 and 407.78nm, Raman lines were found at 451.39, 455.27 and 490.51nm, How can theses results be analyzed?
8. Discuss the appearance of the electronic emission spectrum resulting
from a transition between a stable upper electronic state of a diatomic molecule and a low state which is (a) unstable at all internuclear distances, and (b) stable but predissociated at large internuclear distances.
9. Cl atom has 17 electrons, the ground state electron configuration of Cl
is and the ground state is
10. Derive electron configuration for ground state B2 molecule, possible
terms and then determine the ground state according Hund’s rule. 11. Draw an angular momentum coupling picture for diatomic molecule
under Hund Case (a) and Hund Case (b)coupling (the coupling of molecular rotation and electron angular momentum) condition and give a brief words description.
12. Not considering centrifugal distortion, calculate approximate
rotational constant of both upper state B’and lower state B”for RhF molecule according to observed lines listed below and determine band center ν0.
J R(J)/cm-1 P(J)/cm-1
2 19296.0159 19293.0728
3 19296.7362
13. From the following separations of vibrational levels in the ground
electronic state of CO , calculate values for ωe andωexe
ν'-ν”1-0 2-1
[G(ν+1)-G(ν)]/cm-1 2138 2115
14. Explain the Franck-Condon principle
15. Draw a potential curve picture to explain what is dissociation and
predissociation.
16. Number of normal vibrational modes is for
nonlinear N atoms molecule and is for linear N atoms molecule.
17. The electron configuration of ground state He atom is 1s2, if lift one
electron from 1s to 2p orbital, work out the electron configuration and
electronic states from this configuration using LS coupling. What transitions are allowed from these excited states to the ground state?
18. Derive electron configuration for ground state B2 molecule, possible
terms and then determine the ground state according Hund’s rule.。