c数值算法程序大全c15-2
C程序经典算法50例
C程序经典算法50例1.二分查找算法:在有序数组中查找指定元素。
2.冒泡排序算法:通过不断比较相邻元素并交换位置,将较大的元素向后冒泡。
3.快速排序算法:通过选择一个基准元素,将数组分割为左右两部分,并递归地对两部分进行快速排序。
4.插入排序算法:将数组划分为已排序和未排序两部分,每次从未排序中选择一个元素插入到已排序的合适位置。
5.选择排序算法:遍历数组,每次选择最小元素并放置在已排序部分的末尾。
6.希尔排序算法:将数组按照一定间隔进行分组并分别进行插入排序,然后逐步减小间隔并重复这个过程。
7.归并排序算法:将数组递归地划分为两部分,然后将两个有序的部分进行合并。
8.桶排序算法:将元素根据特定的映射函数映射到不同的桶中,然后对每个桶分别进行排序。
9.计数排序算法:统计每个元素的出现次数,然后根据计数进行排序。
10.基数排序算法:从低位到高位依次对元素进行排序。
11.斐波那契数列算法:计算斐波那契数列的第n项。
12.阶乘算法:计算给定数字的阶乘。
13.排列问题算法:生成给定数组的全排列。
14.组合问题算法:生成给定数组的所有组合。
15.最大连续子序列和算法:找出给定数组中和最大的连续子序列。
16.最长递增子序列算法:找出给定数组中的最长递增子序列。
17.最长公共子序列算法:找出两个给定字符串的最长公共子序列。
18.最短路径算法:计算给定有向图的最短路径。
19.最小生成树算法:构建给定连通图的最小生成树。
20.汉诺塔算法:将n个圆盘从一个柱子移动到另一个柱子的问题。
21.BFS算法:广度优先算法,用于图的遍历和查找最短路径。
22.DFS算法:深度优先算法,用于图的遍历和查找连通分量。
23.KMP算法:字符串匹配算法,用于查找一个字符串是否在另一个字符串中出现。
24.贪心算法:每次都选择当前情况下最优的方案,适用于求解一些最优化问题。
25.动态规划算法:将一个大问题划分为多个子问题,并通过子问题的解求解整个问题,适用于求解一些最优化问题。
数值计算方法程序
数值计算方法程序本文将介绍几种常见的数值计算方法,包括二分法、牛顿法和迭代法。
二分法是一种通过不断将区间一分为二来逼近根的方法。
具体步骤如下:1.确定一个区间[a,b],使得函数在这个区间内有一个根。
2.计算区间的中点c=(a+b)/2,并计算函数在c处的值f(c)。
3.若f(c)为0,则c为函数的一个根;若f(c)>0,则表明根在区间[a,c]内,将b设为c,然后重复步骤2;若f(c)<0,则表明根在区间[c,b]内,将a设为c,然后重复步骤2牛顿法是一种通过不断迭代来逼近根的方法。
具体步骤如下:1.选择一个初始点x0。
2.计算函数在x0处的导数f'(x0)。
3.计算切线的斜率k=f'(x0),并求得切线与x轴的交点x1=x0-f(x0)/f'(x0)。
4.若x1与x0的差值小于给定的精度,则x1为函数的一个根;否则,将x1设为新的初始点,然后重复步骤2迭代法是一种通过不断迭代逼近函数的不动点的方法。
具体步骤如下:1.选择一个初始点x0。
2.计算x1=f(x0)。
3.若x1与x0的差值小于给定的精度,则x1为函数的一个不动点;否则,将x1设为新的初始点,然后重复步骤2除了上述介绍的数值计算方法,还有梯度下降法、高斯-赛德尔迭代法等方法可以用来解决数学问题。
在编写数值计算方法的程序时,需要注意以下几点:1.特殊情况的处理:例如,函数在一些点上可能发散或者不存在,需要对这些情况进行判断和处理。
2.收敛性的判断:需要设置一个收敛判据,当求得的近似解与真实解的差值小于给定的误差限时,认为已经达到了收敛。
3.算法的优化:可以通过调整参数、改变迭代步长等方式来提高算法的效率和精度。
4.可视化:可以通过绘制函数图像或者绘制迭代收敛曲线等方式来直观地展示计算过程和结果。
总之,数值计算方法是一种通过近似的方式来求解数学问题的方法,具有较好的适用性和实现性。
编写数值计算方法的程序时需要考虑特殊情况的处理、收敛性的判断、算法的优化和可视化等因素。
c语言二分查找算法代码
c语言二分查找算法代码C语言二分查找算法是一种非常常用的查找算法,也是一种高效的算法,特别是当数据规模非常大时,它的优势体现得更加明显。
通过对待查找的数据进行二分割,然后迭代比对的方式,二分查找算法能够在较短的时间内准确地定位目标所在的索引位置。
本文将介绍C语言二分查找算法的代码实现,以及如何通过该算法查找并定位数据目标。
1.基本思想C语言二分查找算法是利用了待查找的数据有序的特性,通过将数据逐渐逼近目标值,最终定位目标所在的位置。
具体实现过程如下:(1)首先确定数据的中间位置。
(2)将待查找的目标值与中间位置的值进行比较。
(3)如果目标值等于中间位置的值,则返回中间位置的索引。
(4)如果目标值小于中间位置的值,则将数据的范围缩小到第一项到中间项之间的数据。
(5)如果目标值大于中间位置的值,则将数据的范围缩小到中间项到最后一项之间的数据。
(6)重复以上步骤,直到找到目标值。
二分查找算法的核心思想就是不断缩小数据的范围,将数据迭代划分为两部分,缩小搜索的范围,以快速定位目标值所在的索引位置。
2.代码实现C语言实现二分查找算法虽然简单,但是需要考虑很多边界条件,才能保证算法的正确性和鲁棒性。
下面,我们将介绍二分查找算法的代码实现。
(1)基本二分查找算法```c int binary_search(int *arr, int num, intvalue) { int low = 0, high = num - 1, mid;while(low <= high) { mid = (low + high)/ 2; if(arr[mid] == value)return mid; else if(arr[mid] > value) high = mid - 1; else low = mid+ 1; } return -1; } ```该算法的基本思路已经在上面进行了分析。
该代码实现中,arr为待查找的有序数据,num为数据的个数,value为待查找的目标值。
C程序代码大全
//根据半径计算圆的周长和面积#include <iostream.h>const float PI=3.1416; //声明常量(只读变量)PI为3.1416float fCir_L(float); //声明自定义函数fCir_L()的原型float fCir_S(float); //声明自定义函数fCir_S()的原型//以下是main()函数main(){float r,l,s; //声明3个变量cout<<"r="; //显示字符串cin>>r; //键盘输入l=fCir_L(r); //计算圆的周长,赋值给变量ls=fCir_S(r); //计算圆的面积,赋值给变量scout<<"l="<<l; //显示计算结果cout<<"\ns="<<s;}//定义计算圆的周长的函数fCir_L()float fCir_L(float x){float z=-1.0; //声明局部变量if (x>=0.0) //如果参数大于0,则计算圆的周长z=2*PI*x;return(z); //返回函数值}//定义计算圆的面积的函数fCir_S()float fCir_S(float x){float z=-1.0; //声明局部变量if (x>=0.0) //如果参数大于0,则计算圆的面积z=PI*x*x;return(z); //返回函数值}/* Program: P1-2.CPPWritten by: HapDate written: 02:11:10*/#include <iostream.h>void main(void){double s1,s2,s3;s1=1.5; /* 对变量s1赋值*/cout<<"s1="<<s1<<endl;/* 对变量s2赋值*/ s2=2.5;cout<<"s2="<<s2<<endl;s3= /* 对变量s3赋值*/ 3.5;cout<<"s3="<<s3<<endl;cout<<"s1+s2+s3="<<s1+s2+s3<<endl; //计算并显示//计算并显示cout<<"s1+s2+s3="<<s1+s2+s3<<endl;}#include <iostream.h>main(){double r=1.0;cout<<"r="<<r<<endl;double l;l=2*3.1416*r; //计算圆的周长,赋值给变量l cout<<"l="<<l<<endl; //显示圆的周长double s=3.1416*r*r; //计算圆的面积,赋值给变量s cout<<"s="<<s<<endl; //显示圆的面积cout<<"r="; //显示提示输入的信息cin>>r; //键盘输入l=2*3.1416*r; //计算圆的周长,赋值给变量l cout<<"l="<<l<<endl; //显示圆的周长s=3.1416*r*r;cout<<"s="<<s<<endl; //显示圆的面积}#include <iostream.h> //包含iostream.h头文件void main(){//输出字符常量、变量和字符串char c1='A';cout<<'W';cout<<c1<<endl;cout<<"This is a test."<<endl;cout<<"------------------"<<endl;//输出整型常量、变量和表达式int n=100;cout<<10;cout<<n;cout<<2*n<<endl; //输出整型表达式cout<<"------------------"<<endl;//输出浮点型常量、变量和表达式double pi=3.1415926,r=10.0,s=pi*r*r;cout<<pi<<endl;cout<<r;cout<<s;cout<<2*r*pi<<endl; //输出浮点型表达式cout<<"------------------"<<endl;//一个cout可以输出多项数据cout<<'W'<<" "<<c1<<endl;cout<<"This is a test."<<endl;cout<<"pi="<<pi<<" r="<<r<<" s="<<s<<endl;}#include <iostream.h> //包含iostream.h头文件main(){//输入输出字符char c;cin>>c;cout<<"c="<<c<<endl;//输入输出整型数据int n;cin>>n;cout<<"n="<<n<<endl;//输入输出浮点型数据double x;cin>>x;cout<<"x="<<x<<endl;//输入提示cout<<"n=";cin>>n;cout<<"n="<<n<<endl;//多项输入cout<<"c n x"<<endl;cin>>c>>n>>x;cout<<"c="<<c<<" n="<<n<<" x="<<x<<endl; }#include <iostream.h> //包含iostream.h头文件main(){//声明整型变量int a,b;//从键盘上为整型变量赋值cout<<"a=";cin>>a;cout<<"b=";cin>>b;//整型数的算术运算cout<<a<<"+"<<b<<"="<<a+b<<endl;cout<<a<<"-"<<b<<"="<<a-b<<endl;cout<<a<<"*"<<b<<"="<<a*b<<endl;cout<<a<<"/"<<b<<"="<<a/b<<endl;cout<<a<<"%"<<b<<"="<<a%b<<endl;//测试溢出short n=32767,m; //n取short类型的最大值cout<<"n="<<n<<endl;m=n+1; //引起溢出cout<<"n+1="<<m<<endl;}#include <iostream.h> //包含iostream.h头文件main(){//声明变量,并初始化int a=010,b=10,c=0X10;//以十进制形式显示数据cout<<"DEC:";cout<<" a="<<a;cout<<" b="<<b;cout<<" c="<<c<<endl;//以八进制形式显示数据cout<<"OCT:";cout<<oct; //指定八进制输出cout<<" a="<<a;cout<<" b="<<b;cout<<" c="<<c<<endl;//以十六进制形式显示数据cout<<"HEX:";cout<<hex; //指定十六进制输出cout<<" a="<<a;cout<<" b="<<b;cout<<" c="<<c<<endl;//八、十和十六进制数混合运算并输出cout<<"a+b+c=";cout<<dec; //恢复十进制输出cout<<a+b+c<<endl;//测试八、十和十六进制输入cout<<"DEC:a="; cin>>a;cout<<"OCT:b="; cin>>b;cout<<"HEX:a="; cin>>c;cout<<"DEC:"<<dec<<endl; //指定十进制输出cout<<"a="<<a<<endl;cout<<"b="<<b<<endl;cout<<"c="<<c<<endl;}#include <iostream.h> //包含iostream.h头文件#include<iomanip.h> // iomanip.h头文件包含setprecision()的定义main(){//float型变量的声明、输入、计算和输出float fx,fy;cout<<"fx=";cin>>fx;cout<<"fy=";cin>>fy;cout<<fx<<"+"<<fy<<"="<<fx+fy<<endl;cout<<fx<<"-"<<fy<<"="<<fx-fy<<endl;cout<<fx<<"*"<<fy<<"="<<fx*fy<<endl;cout<<fx<<"/"<<fy<<"="<<fx/fy<<endl<<endl;//cout<<fx<<"%"<<fy<<"="<<fx%fy<<endl; Error!//double型变量的声明、输入、计算和输出float dx,dy;cout<<"dx=";cin>>dx;cout<<"dy=";cin>>dy;cout<<dx<<"+"<<dy<<"="<<dx+dy<<endl;cout<<dx<<"-"<<dy<<"="<<dx-dy<<endl;cout<<dx<<"*"<<dy<<"="<<dx*dy<<endl;cout<<dx<<"/"<<dy<<"="<<dx/dy<<endl<<endl;//cout<<fx<<"%"<<fy<<"="<<fx%fy<<endl; Error!//测试float和double类型数据的有效位fx=10.0;fy=6.0;float fz=fx/fy;dx=10.0;dy=6.0;double dz=dx/dy;cout<<"fz=";cout<<setprecision(20)<<fx<<"/"<<fy<<"="<<fz<<endl;cout<<"dz=";cout<<setprecision(20)<<dx<<"/"<<dy<<"="<<dz<<endl<<endl;;//float型溢出float x=3.5e14;cout<<"x="<<x<<endl;cout<<"x*x="<<x*x<<endl;cout<<"x*x*x="<<x*x*x<<endl;}#include <iostream.h> //包含iostream.h头文件main(){//字符类型变量的声明char c1='A';char c2;//字符数据的运算及输出c2=c1+32;cout<<"c1="<<c1<<endl;cout<<"c2="<<c2<<endl;//输出字符及ASCII码cout<<c1<<" : "<<int(c1)<<endl;cout<<c2<<" : "<<int(c2)<<endl;cout<<'$'<<" : "<<int('$')<<endl;//输入字符cout<<"c1 c2"<<endl;cin>>c1>>c2;cout<<"c1="<<c1<<" c2="<<c2<<endl;}#include <iostream.h> //包含iostream.h头文件main(){char c1='\a',TAB='\t';//阵铃一声cout<<c1<<endl;//使用水平制表符cout<<1<<TAB<<2<<TAB<<3<<TAB<<4<<endl;//使用双引号cout<<"He said \"Thank you\"."<<endl;//使用回车换行cout<<"abc\n"<<"def"<<'\n';}#include <iostream.h> //包含iostream.h头文件main(){//声明bool变量,并初始化bool flag1=false,flag2=true;//输出布尔常量和变量cout<<"false:"<<false<<endl;cout<<"true: "<<true<<endl;cout<<"flag1="<<flag1<<endl;cout<<"flag2="<<flag2<<endl;//布尔变量的赋值和输出int x=1;flag1=x>0; //存放关系运算结果cout<<"flag1="<<flag1<<endl;flag2=flag1; //bool类型变量相互赋值cout<<"flag2="<<flag2<<endl;//布尔变量超界处理flag1=100;cout<<"flag1="<<flag1<<endl;flag2=-100;cout<<"flag2="<<flag2<<endl;}#include <iostream.h>const double PI=3.1416; //声明常量(const变量)PI为3.1416 main(){//声明3个变量double r,l,s;//输入圆的半径cout<<"r=";cin>>r;//计算圆的周长l=2*PI*r;cout<<"l="<<l<<endl;//计算圆的面积s=PI*r*r;cout<<"s="<<s<<endl;}#include<iostream.h>main(){//定义枚举类型,并指定其枚举元素的值enum color {RED=3,YELLOW=6,BLUE=9};//声明枚举变量a和b,并为枚举变量a赋初值enum color a=RED;color b; //合法,与C语言不同// 输出枚举常量cout<<"RED="<<RED<<endl;cout<<"YELLOW="<<YELLOW<<endl;cout<<"BLUE="<<BLUE<<endl;//枚举变量的赋值和输出b=a;a=BLUE;cout<<"a="<<a<<endl;cout<<"b="<<b<<endl;//a=100; 错误!//a=6 也错误!//枚举变量的关系运算b=BLUE; // 枚举变量的赋值运算cout<<"a<b="<<(a<b)<<endl;}#include <iostream.h>const double PI=3.1416; //声明常量(const变量)PI为3.1416 main(){//声明3个变量double r=3,l,s;//计算圆的周长l=2*PI*r;cout<<"l="<<l<<endl;//计算圆的面积s=PI*r*r;cout<<"s="<<s<<endl;//验证赋值误差int il,is;il=l;is=s;cout<<"il="<<il<<endl;cout<<"is="<<is<<endl;}#include <iostream.h>main(){//变量声明char c;double x,y;//测试自增cout<<"++E and E++ :"<<endl;c='B';cout<<"c="<<++c<<endl; //输出c=Cc='B';cout<<"c="<<c++<<endl; //输出c=Bx=1.5;y=5+ ++x; //加号后的空格不能少cout<<"y="<<y<<endl; //输出y=7.5x=1.5;y=5+x++;cout<<"y="<<y<<endl; //输出y=6.5cout<<"--------------------"<<endl;//测试自减cout<<"--E and E-- :"<<endl;c='B';cout<<"c="<<--c<<endl; //输出c=Ac='B';cout<<"c="<<c--<<endl; //输出c=Bx=1.5;y=5+--x;cout<<"y="<<y<<endl; //输出y=5.5x=1.5;y=5+x--;cout<<"y="<<y<<endl; //输出y=6.5}#include <iostream.h>main(){int a=3, b=2;//输出关系表达式cout<<a<b<<endl;cout<<(a<b)<<(a>b)<<(a>=b)<<(a==b)<<(a!=b)<<endl;bool flag=2*a<b+10;cout<<"flag="<<flag;}#include <iostream.h>main(){float a=3.5,b=2.1,c=0;cout<<"a="<<a<<" b="<<b<<" c="<<c<<endl;//与运算cout<<"a&&b="<<(a&&b)<<endl;//输出1cout<<"a&&c="<<(a&&c)<<endl;//输出0//或运算cout<<"a||b="<<(a||b)<<endl;//输出1cout<<"a||c="<<(a||c)<<endl;//输出1//非运算cout<<"!a="<<!a<<endl<<"!c="<<!c<<endl;//输出0 1//关系运算和逻辑运算bool flag=a>=0 && a<=5; //变量a在[0,5]区间内cout<<"a=>0 && a<=5="<<flag<<endl;//输出1//算术运算、关系运算和逻辑运算cout<<"a+5>2*b+2||a<b+3="<<(a+5>2*b+2||a<b+3)<<endl;//输出1 }#include <iostream.h>main(){//按位与运算cout<<"24&12="<<(24&12)<<endl;//按位异或运算cout<<"24^12="<<(24^12)<<endl;//按位或运算cout<<"24|12="<<(24|12)<<endl;//按位取反运算cout<<"~24="<<(~24)<<endl;//左移位运算cout<<"5<<3="<<(5<<3)<<endl;cout<<"-5<<3="<<(-5<<3)<<endl;//右移位运算cout<<"5>>3="<<(5>>3)<<endl;cout<<"-5>>3="<<(-5>>3)<<endl;}#include <iostream.h>main(){int a=1,b=1,c=3;//显示a,b,c的值cout<<"a="<<a<<" b="<<b<<" c="<<c<<endl;//计算显示(1) b+=a+2*c%5; 的结果b+=a+2*c%5; //相当于表达式语句b=b+(a+2*c%5);cout<<"(1) b="<<b<<endl;//计算显示(2) a<<=c-2*b; 的结果a=1,b=1,c=3;a<<=c-2*b; // 相当于表达式语句a=a<<(c-2*b);cout<<"(2) a="<<a<<endl;//计算显示(3) a*=b=c=3;的结果a=1,b=1,c=3;a*=b=c=3; //相当于语句组c=3;b=c;a=a*b;cout<<"(3) a="<<a<<" b="<<b<<" c="<<c<<endl;//计算显示(4) a+=b+=c;的结果a=1,b=1,c=3;a+=b+=c; //相当于语句组b=b+c; a=a+b;cout<<"(4) a="<<a<<" b="<<b<<" c="<<c<<endl;//计算显示(5) a-=b=++c+2;的结果a=1,b=1,c=3;a-=b=++c+2; //相当于语句组++c;b=b+c+2;a=a-b;cout<<"(5) a="<<a<<" b="<<b<<" c="<<c<<endl;}#include <iostream.h>main(){//用sizeof 计算各类种常量的字节长度cout<<"sizeof('$')="<<sizeof('$')<<endl;cout<<"sizeof(1)="<<sizeof(1)<<endl;cout<<"sizeof(1.5)="<<sizeof(1.5)<<endl;cout<<"sizeof(\"Good!\")="<<sizeof("Good!")<<endl;//用sizeof 计算各类型变量的字节长度int i=100;char c='A';float x=3.1416;double p=0.1;cout<<"sizeof(i)="<<sizeof(i)<<endl;cout<<"sizeof(c)="<<sizeof(c)<<endl;cout<<"sizeof(x)="<<sizeof(x)<<endl;cout<<"sizeof(p)="<<sizeof(p)<<endl;//用sizeof 计算表达式的字节长度cout<<"sizeof(x+1.732)="<<sizeof(x+1.732)<<endl;//用sizeof 计算各类型的字节长度cout<<"sizeof(char)="<<sizeof(char)<<endl;cout<<"sizeof(int)="<<sizeof(int)<<endl;cout<<"sizeof(float)="<<sizeof(float)<<endl;cout<<"sizeof(double)="<<sizeof(double)<<endl;//用sizeof 计算数组的字节长度char str[]="This is a test.";int a[10];double xy[10];cout<<"sizeof(str)="<<sizeof(str)<<endl;cout<<"sizeof(a)="<<sizeof(a)<<endl;cout<<"sizeof(xy)="<<sizeof(xy)<<endl;//用sizeof 计算自定义类型的长度struct st {short num;float math_grade;float Chinese_grade;float sum_grade;};st student1;cout<<"sizeof(st)="<<sizeof(st)<<endl;cout<<"sizeof(student1)="<<sizeof(student1)<<endl; }#include <iostream.h>main(){//声明变量语句中使用顺序运算int x, y;//计算中使用顺序运算x=50;y=(x=x-5, x/5);cout<<"x="<<x<<endl;cout<<"y="<<y<<endl;}#include <iostream.h>main(){//测试表达式类型的转换int n=100,m;double x=3.791,y;cout<<"n*x="<<n*x<<endl;//赋值类型转换m=x;y=n;cout<<"m="<<m<<endl;cout<<"y="<<y<<endl;//强制类型转换cout<<"int(x)="<<int(x)<<endl;cout<<"(int)x="<<(int)x<<endl;cout<<"int(1.732+x)="<<int(1.732+x)<<endl;cout<<"(int)1.732+x="<<(int)1.723+x<<endl;cout<<"double(100)="<<double(100)<<endl;}#include <iostream.h>main(){float a,b,s;cout<<"a b"<<endl;cin>>a>>b; //利用cin从键盘上为变量a,b 赋值s=a;if (a<b) {s=b; //if语句中只有这一个语句,可省略花括号}s=s*s; //变量s中保存a,b中较大的一个数的平方cout<<"s="<<s;}#include <iostream.h>main(){int x,y;cout<<"x=";cin>>x;if (x<=0) { //满足条件执行y=2*x;cout<<"y="<<y; //输出结果}else { //不满足条件执行y=x*x;cout<<"y="<<y; //输出结果}}#include <iostream.h>main(){int a,b,c;int smallest;cout<<"a b c"<<endl;cin>>a>>b>>c;if (a<=b) //外层条件语句{if (a<=c) //内层条件语句smallest=a;elsesmallest=c;}else{if (b<=c) //内层条件语句smallest=b;elsesmallest=c;}cout<<"Smallest="<<smallest<<endl;}#include <iostream.h>main(){int score;//从键盘上输入分数cout<<"score=";cin>>score;//用带else if的条件语句判断处理if (score<0 || score>100){cout<<"The score is out of range!"<<endl;}else if (score>=90)cout<<"Your grade is a A."<<endl;else if (score>=80)cout<<"Your grade is a B."<<endl;else if (score>=70)cout<<"Your grade is a C."<<endl;else if (score>=60)cout<<"Your grade is a D."<<endl;elsecout<<"Your grade is a E."<<endl;}#include <iostream.h>main(){int n;cout<<"n=";cin>>n;if (n>=0 && n<=100 &&n%2==0)cout<<"n="<<n<<endl;elsecout<<"The "<<n<<" is out of range!"<<endl; }#include <iostream.h>main(){int a,b,Max;//输入数据cout<<"a=";cin>>a;cout<<"b=";cin>>b;//找出较大值Max=a>ba:b;cout<<"Max="<<Max<<endl;}#include <iostream.h>main(){int a,b;//输入数据cout<<"a=";cin>>a;cout<<"b=";cin>>b;//除法判断if (b!=0 && a%b==0) {cout<<b<<" divides "<<a<<endl;cout<<"a/b="<<a/b<<endl;}elsecout<<b<<" does not divide "<<a<<endl;}#include <iostream.h>main(){//x,y 为操作数,c为运算符int x,y,z;char c1;cin>>x>>c1>>y; //c1//多路选择语句选择不同表达式计算语句switch(c1) {case '+':cout<<x<<"+"<<y<<"="<<x+y<<endl;break;case '-':cout<<x<<"-"<<y<<"="<<x-y<<endl;break;case '*':cout<<x<<"*"<<y<<"="<<x*y<<endl;break;case '/':cout<<x<<"/"<<y<<"="<<x/y<<endl;break;case '%':cout<<x<<"%"<<y<<"="<<x%y<<endl;break;default :cout<<"Wrong !"<<endl; //当不符合上述情况时执行本子句}}#include<iostream.h>float x=365.5; //声明全局变量main() {int x=1,y=2;double w=x+y;{double x=1.414,y=1.732,z=3.14;cout<<"inner:x="<<x<<endl;cout<<"inner:y="<<y<<endl;cout<<"inner:z="<<z<<endl;cout<<"outer:w="<<w<<endl;cout<<"::x="<<::x<<endl; //访问重名的全局变量}cout<<"outer:x="<<x<<endl;cout<<"outer:y="<<y<<endl;cout<<"outer:w="<<w<<endl;//cout<<"inner:z="<<z<<endl;无效cout<<"::x="<<::x<<endl; //访问重名的全局变量}#include<iostream.h>main() {//显示1,2,3 (10)for(int i=1;i<=10;i++)cout<<i<<" ";cout<<endl;//显示10,9,8 (1)for(int j=10;j>=1;j--)cout<<j<<" ";cout<<endl;//显示1,3,5 (9)for(int k=1;k<=10;k=k+2)cout<<k<<" ";cout<<endl;//显示ABC...Zfor(char c='A';c<='Z';c++)cout<<c;cout<<endl;//显示0,0.1,0.2...1.0for(float x=0;x<=1.0;x=x+0.1)cout<<x<<" ";cout<<endl;//显示0,0.1,0.2...1.0for(float x1=0;x1<=1.0+0.1/2;x1=x1+0.1) cout<<x1<<" ";cout<<endl;//计算s=1+2+3...+100int s=0;for(int n=1;n<=100;n++)s=s+n;cout<<"s="<<s<<endl;}#include<iostream.h>main(){//计算s=1+2+3...+100int s=0,n=1;while(n<=100) {s=s+n;n++;}cout<<"s="<<s<<endl;//累加键盘输入的数据double x,sum=0.0;cout<<"x=";cin>>x;while(x!=0) {sum+=x;cout<<"x=";cin>>x;}cout<<"sum="<<sum<<endl;}#include<iostream.h>main(){//计算s=1+2+3...+100int s=0,n=0;do {n++;s+=n;}while(n<100);cout<<"s="<<s<<endl;//累加键盘输入的数据double x,sum=0.0;do {cout<<"x=";cin>>x;sum+=x;} while(x!=0);cout<<"sum="<<sum<<endl;}#include<iostream.h>main(){//计算和打印打印乘法九九表for (int i=1;i<=9;i++) {cout<<i;for (int j=1;j<=9;j++)cout<<'\t'<<i<<"*"<<j<<"="<<i*j;cout<<endl;}}#include<iostream.h>main(){int x,sum=0;//定义标号L1L1: cout<<"x=";cin>>x;if (x==-1)goto L2; //无条件转移语句,转到L2语句处elsesum+=x;goto L1; //无条件转移语句,转到L1语句处//定义标号L2L2: cout<<"sum="<<sum<<endl;}#include<iostream.h>main(){//累加键盘输入的数据double x,sum=0.0;while(1) {cout<<"x=";cin>>x;if (x<=0) break;sum+=x;}cout<<"sum="<<sum<<endl;}#include<iostream.h>main(){int i;for (i=1;i<=20;i++){if (i%3==0) //能被3 整除的整数,返回进行下次循环continue;cout<<i<<" ";}cout<<endl;}#include<iostream.h>main(){//声明数组和变量int a[5],i,sum;double avg;//从键盘上循环为数组赋值for (i=0;i<5;i++) {cout<<"a["<<i<<"]=";cin>>a[i];}//直接显示数组元素cout<<a[0]<<a[1]<<a[2]<<a[3]<<a[4]<<endl;//利用for循环显示数组各元素的值for (i=0;i<5;i++)cout<<a[i]<<" ";cout<<endl;//计算数组元素之和,并显示计算结果sum=a[0]+a[1]+a[2]+a[3]+a[4];cout<<"sum="<<sum<<endl;//利用循环计算数组的累加和for (sum=0,i=0;i<5;i++)sum+=a[i];//显示累加和及平均值cout<<"sum="<<sum<<endl;avg=sum/5.0;cout<<"avg="<<avg<<endl;}#include<iostream.h>main(){int i,max,index,a[5];//从键盘上为数组赋值for (i=0;i<=4;i++){cout<<"a["<<i<<"]=";cin>>a[i];}// 利用循环遍历数组,找出最大值的元素及其下标max=a[0];for (i=0;i<=4;i++){if (max<a[i]){max=a[i];index=i;}}cout<<"\nMax="<<max<<" index="<<index;}#include<iostream.h>#define size 5main(){//声明变量int i,j;float t,a[size];//从键盘上为数组赋值for (i=0;i<size;i++){cout<<"a["<<i<<"]=";cin>>a[i];}//对数组按从小到大顺序排序for (i=0;i<size-1;i++)for (j=i+1;j<size;j++)if (a[i]>a[j]){t=a[i];a[i]=a[j];a[j]=t;}//显示排序结果for (i=0;i<size;i++)cout<<a[i]<<" ";cout<<endl;//输入要查找的数据int value;int found; //找到为1,否则为0int low,high,mid;for (i=1;i<=3;i++) {cout<<"value=";cin>>value;//二分法查找数组afound=0;low=0;high=size-1;while(low<=high){mid=(high+low)/2;if (a[mid]==value){found=1;break;}if (a[mid]<value)low=mid+1;elsehigh=mid-1;}if (found)cout<<"The valu found at:a["<<mid<<"]="<<a[mid]<<endl;elsecout<<"The "<<value<<" is not found!"<<endl;}}#include<iostream.h>main(){//声明变量int i,j;float t,a[5];//从键盘上为数组赋值for (i=0;i<=4;i++){cout<<"a["<<i<<"]=";cin>>a[i];}//对数组按从大到小顺序排序for (i=0;i<=3;i++)for (j=i+1;j<=4;j++)if (a[i]<=a[j]){t=a[i];a[i]=a[j];a[j]=t;}//显示排序结果for (i=0;i<=4;i++)cout<<a[i]<<" ";}#include<iostream.h>main(){//声明二维数组及变量int a[2][3],i,j;//从键盘上为数组a赋值for (i=0;i<2;i++)for (j=0;j<3;j++){cout<<"a["<<i<<"]["<<j<<"]=";cin>>a[i][j];}//显示数组afor (i=0;i<2;i++) {for (j=0;j<3;j++){cout<<a[i][j]<<" ";}cout<<endl;}//找出该数组的最大元素及其下标int h,l,Max=a[0][0];for (i=0;i<2;i++) {for (j=0;j<3;j++){if (Max<a[i][j]) {Max=a[i][j];h=i;l=j;}}}cout<<"Max:"<<"a["<<h<<"]["<<l<<"]="<<a[h][l]<<endl; }#include<iostream.h>main(){//声明字符数组和变量char str[6];int i;//从键盘上输入字符串cout<<"str=";cin>>str;cout<<str<<endl;//按数组和下标变量两种方式显示字符数组cout<<str<<endl;for (i=0;i<6;i++)cout<<str[i];。
计算机二级-C语言常用算法
一、基本算法1.交换(两量交换借助第三者)例1、任意读入两个整数,将二者的值交换后输出。
main(){int a,b,t;scanf("%d%d",&a,&b);printf("%d,%d\n",a,b);t=a; a=b; b=t;printf("%d,%d\n",a,b);}【解析】程序中加粗部分为算法的核心,如同交换两个杯子里的饮料,必须借助第三个空杯子。
假设输入的值分别为3、7,则第一行输出为3,7;第二行输出为7,3。
其中t为中间变量,起到“空杯子”的作用。
注意:三句赋值语句赋值号左右的各量之间的关系!【应用】例2、任意读入三个整数,然后按从小到大的顺序输出。
main(){int a,b,c,t;scanf("%d%d%d",&a,&b,&c);if(a>b){ t=a; a=b; b=t; }if(a>c){ t=a; a=c; c=t; }if(b>c) { t=b; b=c; c=t; }printf("%d,%d,%d\n",a,b,c);}2.累加累加算法的要领是形如“s=s+A”的累加式,此式必须出现在循环中才能被反复执行,从而实现累加功能。
“A”通常是有规律变化的表达式,s在进入循环前必须获得合适的初值,通常为0。
例1、求1+2+3+……+100的和。
main(){int i,s;s=0; i=1;while(i<=100){s=s+i;i=i+1;}printf("1+2+3+...+100=%d\n",s);}【解析】程序中加粗部分为累加式的典型形式,赋值号左右都出现的变量称为累加器,其中“i = i + 1”为特殊的累加式,每次累加的值为1,这样的累加器又称为计数器。
3.累乘累乘算法的要领是形如“s=s*A”的累乘式,此式必须出现在循环中才能被反复执行,从而实现累乘功能。
c#考试(程序代码大全)
目录2)一元二次方程求解............................................3)1-100之间所有整数的和,能被某数整除的所有整数的和,积......4)闰年的判断..................................................5)水仙花数....................................................6)简单计算器..................................................7)一维数组的声明、初始化、赋值、遍历(求数组所有元素的和,积,最大值,最小值等统计数值)...................................................8)二维数组的声明、初始化、赋值、遍历(求数组所有元素的和,积,最大值,最小值等统计数值)...................................................9)圆台的体积..................................................10)三角形的三边,求三角形的面积...............................11)乘法口诀表.................................................12)字符串的统计(大写,小写字母,数字的个数).................13)数值类型的转换:数值-字符串,字符串-数字,数值类型之间的显示转换,隐式转换.............................................................14)遍历窗体上的组件...........................................15)随机数发生器的用法.........................................16)用判断成绩等级,改变窗体的背景颜色.........................17)用和求1-100之间所有整数的和,积等.........................18)类的创建和使用(类的声明,字段,属性和方法的添加,生成类的实例和使用其成员)...........................................................19)的用法.....................................................1)1( , e)s = (1());(s >= 90 s <= 100)("优秀");(s >= 80 s < 90)("良好");(s >= 70 s < 80)("中");(s >= 60 s < 70)("与格");("不与格");2)一元二次方程求解1( , e)a = (),b = (),c = ();(a 0)("方程的根是: " + ( / b)());= (b, 2) - 4 * a * c;( >= 0)X1 = ( + (, 0.5)) / 2 * a;X2 = ( - (, 0.5)) / 2 * a;= "X1="1()+"\r\2="2();("没有实数根!");3)1-100之间所有整数的和,能被某数整除的所有整数的和,积 1( , e)= 0;( i = 0; i <= 100; i = i + 1)= + i;1 = ();求1~100之内所有偶数的和= 0;( i = 0; i <= 100; i = i + 2)= + i;1 = ();1( , e)= 0;a;( i = 0; i <= 100; )a = i % 7;(a 0)= + i;1 = ();4)闰年的判断1( , e)a = 32(1);((a %40 a %1000) a %400 0)("是闰年");("不是闰年");或办法a = (1 );s = (a % 4 0 ? "该年份是闰年 ": "该年份是平年 " );(s )5)水仙花数1( , e)s = "水仙花数:"+"\r\n";( i = 100; i < 1000; )= i % 10;= i / 10 % 10;= i / 100;(i * * + * * + * * )s () + "\r\n";(s,"你好漂亮");6)简单计算器1( , e)1 = 32(1);2 = 32(3);(2 ("+"))((12)());(2 ("-"))((12)());(2 ("*"))((1*2)());(2 ("/"))((12)());(2 ("%"))((12)());("运算符出错");或办法a = (1);b = (3);s = 2;c = (s "+" ? a + b : (s "-" ? a - b : (s "*" ? a * b : (s "/" ? a / b :a % b))));7)一维数组的声明、初始化、赋值、遍历(求数组所有元素的和,积,最大值,最小值等统计数值)1( , e)1 "原数组为:" + "\r\n";[] a = [5];( i = 0; i < ; )a[i] = (0, 100);1 a[i]() + " ";= a[0], = a[0]; s = 0; = 0;( i = 0; i < ; )(a[i] > )= a[i];(a[i] < )= a[i];s a[i]; = s / ;1 ("\r\n最大值为{0}\r\n最小值为{1}\r\n均值为{2}\r\n和为{3}", , , , s);8)二维数组的声明、初始化、赋值、遍历(求数组所有元素的和,积,最大值,最小值等统计数值)2( , e)2 "原数组为:" + "\r\n";[,] b = [3, 4];( i = 0; i < 3; )( j = 0; j < 4; )b[i, j] = (-100,100);2 b[i, j]() + " , ";= b[0, 0], = b[0, 0]; s = 0; = 0;( i = 0; i < 3; )( j = 0; j < 4; )b[i, j] = (-100, 100);(b[i, j] > ) = b[i, j];(b[i, j] < ) = b[i, j];s b[i, j];= s / (3 * 4);2 ("\r\n最大值为{0}\r\n最小值为a{1}\r\n平均值为{2}\r\n和为{3}", , , , s);9)圆台的体积圆台体积计算公式V=π*h*(R^2 + ^2)/3即:v = *h*(R*R + R* r *r)/31( , e)R = (1),r = (2),h = (3);π = 3.14;X1 = π * h * ((R, 2) + (r, 2) + R * r) / 3;v = "圆台体积是:" + X1();(v);10)三角形的三边,求三角形的面积1( , e)A = (1);B = (2);C = (3);d; f; s;(A + B > C & A + C > B & B + C > A & A - B < C & A - C < B &B -C < A)d = (A * A + C * C - B * B) / (2 * A * C);f = ((1 - d * d), 0.5);s = 0.5 * A * C * f;("三角形面积是:"());("输入数据有误");11)乘法口诀表1( , e)= ""; = 0;199 = "";( i = 1; i < 10; )( j = 1; j <= i; )= i * j;( >= 10)199 + "\n";12)字符串的统计(大写,小写字母,数字的个数)1( , e)a = 1;( t a)(t >= '0' t <='9')(t >= 'a' t <= 'z')(t >= 'A' t <= 'Z')3 = ();4 = ();5 = ();13)数值类型的转换:数值-字符串,字符串-数字,数值类型之间的显示转换,隐式转换类型转换、隐式转换、显示转换、类的用法、方法主要以隐式转换为例(其他的类似求解)补充:1)何时使用隐式和显示转换?答:隐式转换一般发生在数据进行混合运算的情况下,是编译系统自动进行的,不需要加以声明;显示转换,又称为强制类型转换,该方式需要用户明确的指定转换的目标类型,该类型一般形式为: (类型说明符) (需要转换的表达式)二、类型转换隐式转换a; b; c; d; f; g; h; y;1( , e)b = 1;c = b;d = c; f = d; g = f;((), "转换1");2( , e)a = 2; c = a; d = c; f = d; g = f;((), "转换2");显示转换a; b; c; d; f; g; h; y;1( , e)a; b;a = 1.1111f;b = ()a;((), "郭浩);2( , e)a; b;a = 2.3m;b = ()a;3( , e)a; b;a = 1;b = ()a;31( , e)a; b;a = 94;b = (a);类的用法1( , e)a = "123";b = 32(a);2( , e)a = "456";3( , e)a = "789";b = (a);4( , e)a = "0";b = (a);方法a (1)(() );14)遍历窗体上的组件遍历组件代码如下= 0;= 0;= 0;= 0;( c )(c )= + 1;(c )= + 1;(c )= + 1;(c )= + 1;= "的数量:" + + "\r\n"+ "的数量:" + + "\r\n"+ "的数量:" + + "\r\n"+ "的数量:" + + "\r\n";(, "遍历组件", , , 1);运行截图15)随机数发生器的用法求浮点型数组最大元素1 = "生成数组:" + "\r\n";[] a = [5];r = ();( i = 0; i < ; )a[i] = (0, 1000);1 a[i]() + " ";= a[0], = a[0];( i = 0; i < ; )(a[i] > )= a[i];(a[i] < )= a[i];2 "最大值" + ();16)用判断成绩等级,改变窗体的背景颜色 1( , e)(s >= 0 s <= 100)(s / 10)0:1:2:3:4:5:("不与格");6:( "与格");10:( "优秀");("请输入正确的分数"); ( "请输入正确的分数");1( , e)s = 1;(s)"红":"黄":"蓝":"黑":"白":"绿":。
C语言基础简单的数学运算的代码
C语言基础简单的数学运算的代码#include <stdio.h>int main() {// 定义并初始化变量int num1 = 10;int num2 = 5;// 加法运算int sum = num1 + num2;printf("加法运算结果:%d\n", sum);// 减法运算int difference = num1 - num2;printf("减法运算结果:%d\n", difference);// 乘法运算int product = num1 * num2;printf("乘法运算结果:%d\n", product);// 除法运算float quotient = (float)num1 / num2;printf("除法运算结果:%.2f\n", quotient);// 求余运算int remainder = num1 % num2;printf("求余运算结果:%d\n", remainder);return 0;}以上是一个简单的C语言程序,实现了基本的数学运算功能。
程序运行后,会输出每个数学运算的结果。
接下来我会逐行解释代码的含义和执行过程。
首先,在程序的开头我们使用了#include <stdio.h>这行代码,这是为了包含C语言标准库中的输入输出函数,以便后续可以使用printf()函数打印结果。
接着,在main()函数中,我们定义并初始化了两个整型变量num1和num2,分别赋值为10和5。
这两个变量代表了我们要进行数学运算的两个操作数。
然后,我们使用加法运算将num1和num2相加得到sum,并使用printf()函数打印出加法运算的结果。
接着,我们使用减法运算将num1减去num2得到difference,并使用printf()函数打印出减法运算的结果。
C语言常用简单算法
C语言常用简单算法C语言是一门功能强大的编程语言,其算法也是很多的。
下面是一些常用的简单算法:1.二分查找算法:二分查找是一种在有序数组中查找特定元素的算法。
它的基本思想是首先在数组的中间位置找到待查找的元素,如果该元素等于目标值,则查找成功;如果该元素大于目标值,说明目标值在数组的前半部分,则在前半部分继续进行查找;如果该元素小于目标值,则说明目标值在数组的后半部分,则在后半部分继续进行查找。
重复以上步骤,直到找到目标值或者确定目标值不存在。
2.冒泡排序算法:冒泡排序是一种简单直观的排序算法。
它的基本思想是通过反复交换相邻的两个元素,将较大的元素逐渐往后移动,从而实现排序的目的。
具体实现时,每一轮比较都会使最大的元素移动到最后。
3.插入排序算法:插入排序是一种简单直观的排序算法。
它的基本思想是将数组分成已排序部分和未排序部分,每次从未排序部分取出一个元素,然后将该元素插入到已排序部分的合适位置,从而实现排序的目的。
4.选择排序算法:选择排序是一种简单直观的排序算法。
它的基本思想是每次选择一个最小(或最大)的元素放到已排序部分的末尾,从而实现排序的目的。
具体实现时,每一轮选择都通过比较找出未排序部分的最小(或最大)元素。
5.快速排序算法:快速排序是一种高效的排序算法。
它的基本思想是通过选取一个基准元素,将数组分成两个子数组,一个子数组中的元素都小于基准元素,另一个子数组中的元素都大于基准元素,然后对这两个子数组分别进行快速排序,最终实现排序的目的。
6.斐波那契数列算法:斐波那契数列是一列数字,其中每个数字都是前两个数字之和。
常见的斐波那契数列算法有递归算法和迭代算法。
递归算法通过反复调用自身来计算斐波那契数列的值,而迭代算法则通过循环来计算。
7.求最大公约数算法:求两个数的最大公约数是一种常见的问题。
常见的求最大公约数的算法有欧几里得算法和辗转相除法。
欧几里得算法通过不断用较小数除以较大数的余数,直到余数为0,得到最大公约数。
C语言经典算法大全(可编辑)
C语言经典算法大全C语言经典算法大全老掉牙河内塔费式数列巴斯卡三角形三色棋老鼠走迷官一老鼠走迷官二骑士走棋盘八个皇后八枚银币生命游戏字串核对双色三色河内塔背包问题Knapsack Problem数运算蒙地卡罗法求 PIEratosthenes筛选求质数超长整数运算大数运算长 PI最大公因数最小公倍数因式分解完美数阿姆斯壮数最大访客数中序式转后序式前序式后序式的运算关于赌博洗扑克牌乱数排列Craps赌博游戏约瑟夫问题Josephus Problem 集合问题排列组合格雷码Gray Code产生可能的集合m元素集合的n个元素子集数字拆解排序得分排行选择插入气泡排序Shell 排序法 - 改良的插入排序Shaker 排序法 - 改良的气泡排序Heap 排序法 - 改良的选择排序快速排序法一快速排序法二快速排序法三合并排序法基数排序法搜寻循序搜寻法使用卫兵二分搜寻法搜寻原则的代表插补搜寻法费氏搜寻法矩阵稀疏矩阵多维矩阵转一维矩阵上三角下三角对称矩阵奇数魔方阵4N 魔方阵2 2N1 魔方阵 1河内之塔说明河内之塔 Towers of Hanoi 是法国人MClaus Lucas 于1883年从泰国带至法国的河内为越战时北越的首都即现在的胡志明市1883年法国数学家 Edouard Lucas曾提及这个故事据说创世纪时Benares有一座波罗教塔是由三支钻石棒Pag所支撑开始时神在第一根棒上放置64个由上至下依由小至大排列的金盘Disc并命令僧侣将所有的金盘从第一根石棒移至第三根石棒且搬运过程中遵守大盘子在小盘子之下的原则若每日仅搬一个盘子则当盘子全数搬运完毕之时此塔将毁损而也就是世界末日来临之时解法如果柱子标为ABC要由A搬至C在只有一个盘子时就将它直接搬至C当有两个盘子就将B当作辅助柱如果盘数超过2个将第三个以下的盘子遮起来就很简单了每次处理两个盘子也就是A- BA - CB- C这三个步骤而被遮住的部份其实就是进入程式的递回处理事实上若有n个盘子则移动完毕所需之次数为2n - 1所以当盘数为64时则所需次数为264- 1 1XXXXXXXXXX709551615为505390248594782e16年也就是约5000世纪如果对这数字没什幺概念就假设每秒钟搬一个盘子好了也要约5850亿年左右includevoid hanoi int n char A char B char Cif n 1printf "Move sheet d from c to c\n" n A Celsehanoi n-1 A C Bprintf "Move sheet d from c to c\n" n A Chanoi n-1 B A Cint mainint nprintf "请输入盘数"scanf "d" nhanoi n A B Cfn fn-1 fn-2 if n 1fn n if n 0 1includeinclude define N 20 int main voidint Fib[N] 0int i Fib[0] 0Fib[1] 1 for i 2 i N iFib[i] Fib[i-1] Fib[i-2] for i 0 i N i printf "d " Fib[i]printf "\n"return 03 巴斯卡三角形 includedefine N 12long combi int n int rint ilong p 1for i 1 i r ip p n-i1 ireturn pvoid mainint n r tfor n 0 n N nfor r 0 r n rint i 排版设定开始if r 0for i 0 i N-n i printf " "elseprintf " "排版设定结束printf "3d" combi n rprintf "\n"4Algorithm Gossip 三色棋说明三色旗的问题最早由EWDijkstra所提出Dutch Nation Flag Dijkstra为荷兰人 Three-Color Flag来称之假设有一条绳子上面有红白蓝三种颜色的旗子起初绳子上的旗子颜色并没有顺序您希望将之分类并排列为蓝白红的顺序要如何移动次数才会最少注意您只能在绳子上进行这个动作而且一次只能调换两个旗子解法在一条绳子上移动在程式中也就意味只能使用一个阵列而不使用其它的阵列来作辅助问题的解法很简单您可以自己想像一下在移动旗子从绳子开头进行遇到蓝色往前移遇到白色留在中间遇到红色往后移如下所示只是要让移动次数最少的话就要有些技巧如果图中W所在的位置为白色则W1表示未处理的部份移至至白色群组如果W部份为蓝色则B与W的元素对调而B与W必须各1表示两个群组都多了一个元素如果W所在的位置是红色则将W与R交换但R要减1表示未处理的部份减1 注意BWR并不是三色旗的个数它们只是一个移动的指标什幺时候移动结束呢一开始时未处理的R指标会是等于旗子的总数当R的索引数减至少于W的索引数时表示接下来的旗子就都是红色了此时就可以结束移动如下所示 includeincludeinclude define BLUE bdefine WHITE wdefine RED r define SWAP x y char temp \temp color[x] \color[x] color[y] \color[y] temp int main char color[] r w b w wb r b w r \0 int wFlag 0 int bFlag 0int rFlag strlen color - 1int i for i 0 i strlen color iprintf "c " color[i]printf "\n" while wFlag rFlagif color[wFlag] WHITEwFlagelse if color[wFlag] BLUESWAP bFlag wFlagbFlag wFlagelsewhile wFlag rFlag color[rFlag] REDrFlag--SWAP rFlag wFlagrFlag--for i 0 i strlen color i printf "c " color[i]printf "\n" return 05Algorithm Gossip 老鼠走迷官说明老鼠走迷宫是递回求解的基本题型我们在二维阵列中使用2表示迷宫墙壁使用1来表示老鼠的行走路径试以程式求出由入口至出口的路径解法老鼠的走法有上左下右四个方向在每前进一格之后就选一个方向前进无法前进时退回选择下一个可前进方向如此在阵列中依序测试四个方向直到走到出口为止这是递回的基本题请直接看程式应就可以理解includeinclude int visit int int int maze[7][7] 2 2 2 22 2 22 0 0 0 0 0 22 0 2 0 2 0 22 0 0 2 0 2 22 2 0 2 0 2 22 0 0 0 0 0 22 2 2 2 2 2 2 int startI 1 startJ 1 入口int endI 5 endJ 5 出口int success 0 int main voidint i j printf "显示迷宫\n"for i 0 i 7 ifor j 0 j 7 jif maze[i][j] 2printf "█"elseprintf " "printf "\n"if visit startI startJ 0 printf "\n没有找到出口\n"elseprintf "\n\n"for i 0 i 7 ifor j 0 j 7 jif maze[i][j] 2printf "█"else if maze[i][j] 1 printf "◇"elseprintf " "printf "\n"return 0int visit int i int jmaze[i][j] 1 if i endI j endJsuccess 1 if success 1 maze[i][j1]0 visit i j1if success 1 maze[i1][j] 0 visit i1 jif success 1 maze[i][j-1] 0 visit i j-1if success 1 maze[i-1][j] 0 visit i-1 j if success 1maze[i][j] 0 return success6Algorithm Gossip 老鼠走迷官说明由于迷宫的设计老鼠走迷宫的入口至出口路径可能不只一条如何求出所有的路径呢解法求所有路径看起来复杂但其实更简单只要在老鼠走至出口时显示经过的路径然后退回上一格重新选择下一个位置继续递回就可以了比求出单一路径还简单我们的程式只要作一点修改就可以了includeinclude void visit int int int maze[9][9] 2 2 2 22 2 2 2 22 0 0 0 0 0 0 0 22 0 2 2 0 2 2 0 22 0 2 0 0 2 0 0 22 0 2 0 2 0 2 0 22 0 0 0 0 0 2 0 22 2 0 2 2 0 2 2 22 0 0 0 0 0 0 0 22 2 2 2 2 2 2 2 2 int startI 1 startJ 1 入口int endI 7 endJ 7 出口 int main voidint i j printf "显示迷宫\n"for i 0 i 7 ifor j 0 j 7 jif maze[i][j] 2printf "█"elseprintf " "printf "\n"visit startI startJ return 0void visit int i int jint m n maze[i][j] 1 if i endI j endJprintf "\n显示路径\n"for m 0 m 9 mfor n 0 n 9 nif maze[m][n] 2printf "█"else if maze[m][n] 1printf "◇"elseprintf " "printf "\n"if maze[i][j1] 0 visit i j1if maze[i1][j] 0 visit i1 jif maze[i][j-1] 0 visit i j-1if maze[i-1][j] 0 visit i-1 j maze[i][j] 07Algorithm Gossip 骑士走棋盘说明骑士旅游Knight tour[所有的位置include int board[8][8] 0 int main voidint startx startyint i jprintf "输入起始点"scanf "d d" startx starty if travel startx starty printf "\n"elseprintf "\n"for i 0 i 8 ifor j 0 j 8 jprintf "2d " board[i][j]putchar \nreturn 0int travel int x int yint ktmove1[8] -2 -1 1 2 2 1 -1 -2int ktmove2[8] 1 2 2 1 -1 -2 -2 -1 测试下一步的出路int nexti[8] 0int nextj[8] 0记录出路的个数int exists[8] 0int i j k m lint tmpi tmpjint count min tmp i xj yboard[i][j] 1 for m 2 m 64 mfor l 0 l 8 lexists[l] 0 l 0 试探八个方向for k 0 k 8 ktmpi i ktmove1[k]tmpj j ktmove2[k] 如果是边界了if tmpi 0 tmpj 0 tmpi 7 tmpj 7 continue 如果这个方向可走 if board[tmpi][tmpj] 0nexti[l] tmpinextj[l] tmpj可走的方向加一个lcount l如果可走的方向为0个if count 0return 0else if count 1只有一个可走的方向所以直接是最少出路的方向min 0else找出下一个位置的出路数for l 0 l count lfor k 0 k 8 ktmpi nexti[l] ktmove1[k] tmpj nextj[l] ktmove2[k] if tmpi 0 tmpj 0tmpi 7 tmpj 7continueif board[tmpi][tmpj] 0 exists[l]tmp exists[0]min 0从可走的方向中寻找最少出路的方向 for l 1 l count lif exists[l] tmptmp exists[l]min l走最少出路的方向i nexti[min]j nextj[min]board[i][j] mreturn 18Algorithm Gossip 八皇后说明西洋棋中的皇后可以直线前进吃掉遇到的所有棋子如果棋盘上有八个皇后则这八个皇后如何相安无事的放置在棋盘上1970年与1971年 EWDijkstra与NWirth曾经用这个问题来讲解程式设计之技巧解法关于棋盘的问题都可以用递回求解然而如何减少递回的次数在八个皇后的问题中不必要所有的格子都检查过例如若某列检查过该该列的其它格子就不用再检查了这个方法称为分支修剪includeincludedefine N 8 int column[N1] 同栏是否有皇后1int rup[2N1] 右上至左下是否有皇后int lup[2N1] 左上至右下是否有皇后int queen[N1] 0int num 解答编号 void backtrack int 递回求解 intmain voidint inum 0 for i 1 i N icolumn[i] 1 for i 1 i 2N irup[i] lup[i] 1 backtrack 1 return 0void showAnswerint x yprintf "\n解答 d\n" numfor y 1 y N yfor x 1 x N xif queen[y] xprintf " Q"elseprintf " "printf "\n"void backtrack int iint j if i NshowAnswerelsefor j 1 j N jif column[j] 1rup[ij] 1 lup[i-jN] 1 queen[i] j设定为占用column[j] rup[ij] lup[i-jN] 0 backtrack i1column[j] rup[ij] lup[i-jN] 19Algorithm Gossip 八枚银币说明现有八枚银币a b c d e f g h已知其中一枚是假币其重量不同于真币但不知是较轻或较重如何使用天平以最少的比较次数决定出哪枚是假币并得知假币比真币较轻或较重解法单就求假币的问题是不难但问题限制使用最少的比较次数所以我们不能以单纯的回圈比较来求解我们可以使用决策树decision tree使用分析与树状图来协助求解一个简单的状况是这样的我们比较abc与def 如果相等则假币必是g或h我们先比较g或h哪个较重如果g较重再与a比较a是真币如果g等于a则g为真币则h为假币由于h比g轻而 g是真币则h假币的重量比真币轻 include includeinclude void compare int[] int int intvoid eightcoins int[] int main voidint coins[8] 0int i srand time NULL for i 0 i 8 icoins[i] 10 printf "\n输入假币重量比10大或小 "scanf "d" icoins[rand 8] i eightcoins coins printf "\n\n"for i 0 i 8 iprintf "d " coins[i] printf "\n"return 0void compare int coins[] int i int j int kif coins[i] coins[k]printf "\n d 较重" i1elseprintf "\n假币 d 较轻" j1void eightcoins int coins[]if coins[0]coins[1]coins[2]coins[3]coins[4]coins[5]if coins[6] coins[7]compare coins 6 7 0elsecompare coins 7 6 0else if coins[0]coins[1]coins[2]coins[3]coins[4]coins[5]if coins[0]coins[3] coins[1]coins[4]compare coins 2 5 0else if coins[0]coins[3] coins[1]coins[4] compare coins 0 4 1if coins[0]coins[3] coins[1]coins[4]compare coins 1 3 0else if coins[0]coins[1]coins[2]coins[3]coins[4]coins[5]if coins[0]coins[3] coins[1]coins[4]compare coins 5 2 0else if coins[0]coins[3] coins[1]coins[4] compare coins 3 1 0if coins[0]coins[3] coins[1]coins[4]compare coins 4 0 110Algorithm Gossip 生命游戏说明生命游戏game of life1970年由英国数学家J H Conway所提出includeincludeinclude define ROW 10define COL 25define DEAD 0define ALIVE 1int map[ROW][COL] newmap[ROW][COL] void initint neighbors int intvoid outputMapvoid copyMap int mainint row colchar ansinitwhile 1outputMapfor row 0 row ROW rowfor col 0 col COL colswitch neighbors row colcase 0case 1case 4case 5case 6case 7case 8newmap[row][col] DEADbreakcase 2newmap[row][col] map[row][col] breakcase 3newmap[row][col] ALIVEbreakcopyMapprintf "\nContinue next Generation "getcharans toupper getcharif ans Y breakreturn 0void initint row col for row 0 row ROW rowfor col 0 col COL colmap[row][col] DEAD puts "Game of life Program"puts "Enter x y where x y is living cell"printf "0 x d 0 y d\n"ROW-1 COL-1puts "Terminate with x y -1 -1" while 1scanf "d d" row colif 0 row row ROW0 col col COLmap[row][col] ALIVEelse if row -1 col -1breakelseprintf " x y exceeds map ranage"int neighbors int row int colint count 0 c rfor r row-1 r row1 rfor c col-1 c col1 cif r 0 r ROW c 0 c COL continueif map[r][c] ALIVEcountif map[row][col] ALIVEcount--return countvoid outputMapint row colprintf "\n\n20cGame of life cell status\n" for row 0 row ROW rowprintf "\n20c"for col 0 col COL colif map[row][col] ALIVE putchar else putchar -void copyMapint row colfor row 0 row ROW rowfor col 0 col COL colmap[row][col] newmap[row][col]11Algorithm Gossip 字串核对说明今日的一些高阶程式语言对于字串的处理支援越来越强大例如JavaPerl等不过字串搜寻本身仍是个值得探讨的课题在这边以Boyer- Moore法来说明如何进行字串说明这个方法快且原理简洁易懂解法字串搜寻本身不难使用暴力法也可以求解但如何快速搜寻字串就不简单了传统的字串搜寻是从关键字与字串的开头开始比对例如 Knuth-Morris-Pratt 演算法字串搜寻这个方法也不错不过要花时间在公式计算上Boyer-Moore字串核对改由关键字的后面开始核对字串并制作前进表如果比对不符合则依前进表中的值前进至下一个核对处假设是p好了然后比对字串中p-n1至p的值是否与关键字相同如果关键字中有重复出现的字元则前进值就会有两个以上的值此时则取前进值较小的值如此就不会跳过可能的位置例如texture 这个关键字t的前进值应该取后面的3而不是取前面的7 includeincludeinclude void table char 建立前进表int search int char char 搜寻关键字void substring char char int int 取出子字串int skip[256] int main voidchar str_input[80]char str_key[80]char tmp[80] \0int m n pprintf "请输入字串"gets str_inputprintf ""gets str_keym strlen str_inputn strlen str_keytable str_keyp search n-1 str_input str_key while p -1 substring str_input tmp p mprintf "s\n" tmpp search pn1 str_input str_keyprintf "\n"return 0void table char keyint k nn strlen keyfor k 0 k 255 kskip[k] nfor k 0 k n - 1 kskip[key[k]] n - k - 1int search int p char input char keyint i m nchar tmp[80] \0m strlen inputn strlen key while p msubstring input tmp p-n1 pif strcmp tmp key 比较两字串是否相同return p-n1p skip[input[p]]return -1void substring char text char tmp int s int e int i jfor i s j 0 i e i jmp[j] text[i]tmp[j] \012Algorithm Gossip 双色三色河内塔说明双色河内塔与三色河内塔是由之前所介绍过的河内塔规则衍生而来双色河内塔的目的是将下图左上的圆环位置经移动成为右下的圆环位置而三色河内塔则是将下图左上的圆环经移动成为右上的圆环解法无论是双色河内塔或是三色河内塔其解法观念与之前介绍过的河内塔是类似的同样也是使用递回来解不过这次递回解法的目的不同我们先来看只有两个盘的情况这很简单只要将第一柱的黄色移动至第二柱而接下来第一柱的蓝色移动至第三柱再来是四个盘的情况首先必须用递回完成下图左上至右下的移动接下来最底层的就不用管它们了因为它们已经就定位只要再处理第一柱的上面两个盘子就可以了那么六个盘的情况呢一样首先必须用递回完成下图左上至右下的移动接下来最底层的就不用管它们了因为它们已经就定位只要再处理第一柱上面的四个盘子就可以了这又与之前只有四盘的情况相同接下来您就知道该如何进行解题了无论是八个盘十个盘以上等都是用这个观念来解题那么三色河内塔呢一样直接来看九个盘的情况首先必须完成下图的移动结果接下来最底两层的就不用管它们了因为它们已经就定位只要再处理第一柱上面的三个盘子就可以了双色河内塔 C 实作include void hanoi int disks char source char temp char targetif disks 1printf "move disk from c to c\n" source targetprintf "move disk from c to c\n" source target elsehanoi disks-1 source target temphanoi 1 source temp targethanoi disks-1 temp source targetvoid hanoi2colors int diskschar source Achar temp Bchar target Cint ifor i disks 2 i 1 i--hanoi i-1 source temp targetprintf "move disk from c to c\n" source temp printf "move disk from c to c\n" source temp hanoi i-1 target temp sourceprintf "move disk from c to c\n" temp targetprintf "move disk from c to c\n" source tempprintf "move disk from c to c\n" source targetint mainint nprintf "请输入盘数"scanf "d" n hanoi2colors n return 0C 实作include void hanoi int disks char source char temp char targetif disks 1printf "move disk from c to c\n" source target printf "move disk from c to c\n" source target printf "move disk from c to c\n" source target elsehanoi disks-1 source target temphanoi 1 source temp targethanoi disks-1 temp source targetvoid hanoi3colors int diskschar source Achar temp Bchar target Cint iif disks 3printf "move disk from c to c\n" source tempprintf "move disk from c to c\n" source tempprintf "move disk from c to c\n" source targetprintf "move disk from c to c\n" temp targetprintf "move disk from c to c\n" temp sourceprintf "move disk from c to c\n" target tempelsehanoi disks3-1 source temp targetprintf "move disk from c to c\n" source tempprintf "move disk from c to c\n" source tempprintf "move disk from c to c\n" source temp hanoi disks3-1 target temp sourceprintf "move disk from c to c\n" temp targetprintf "move disk from c to c\n" temp targetprintf "move disk from c to c\n" temp target hanoi disks3-1 source target tempprintf "move disk from c to c\n" target sourceprintf "move disk from c to c\n" target source hanoi disks3-1 temp source targetprintf "move disk from c to c\n" source temp for i disks 3 - 1 i 0 i--if i 1hanoi i-1 target source tempprintf "move disk from c to c\n"target source printf "move disk from c to c\n"target source if i 1hanoi i-1 temp source targetprintf "move disk from c to c\n" source tempint mainint nprintf "请输入盘数"scanf "d" n hanoi3colors nreturn 013Algorithm Gossip 背包问题Knapsack Problem说明假设有一个背包的负重最多可达8公斤而希望在背包中装入负重范围内可得之总价物品假设是水果好了水果的编号单价与重量如下所示0 李子 4KG NT4500 1 苹果 5KG NT57002 橘子 2KG NT22503 草莓 1KG NT1100 4甜瓜 6KG NT6700解法背包问题是关于最佳化的问题要解最佳化问题可以使用「动态规划」Dynamic programming从空集合开始每增加一个元素就先求出该阶段的最佳解直到所有的元素加入至集合中最后得到的就是最佳解以背包问题为例我们使用两个阵列value与itemvalue表示目前的最佳解所得之总价item表示最后一个放至背包的水果假设有负重量 1~8的背包8个并对每个背包求其最佳解逐步将水果放入背包中并求该阶段的最佳解放入李子背包负重 1 2 3 4 5 6 7 8 value 0004500 4500 4500 4500 9000 item ---00000放入苹果背包负重 1 2 3 4 5 6 7 8 value 0004500 5700 5700 5700 9000 item ---0 1 1 1 0放入橘子背包负重 1 2 3 4 5 6 7 8 value 02250 2250 4500 5700 6750 7950 9000 item -2 2 0 1 2 2 0放入草莓背包负重 1 2 3 4 5 6 7 8 value 11002250 3350 4500 5700 6800 7950 9050 item 3 23 0 1 3 2 3放入甜瓜背包负重 1 2 3 4 5 6 7 8 value 11002250 3350 4500 5700 6800 7950 9050 item 3 23 0 1 3 2 3由最后一个表格可以得知在背包负重8公斤时最多可以装入9050元的水果而最后一个装入的水果是3号也就是草莓装入了草莓背包只能再放入7公斤8-1的水果所以必须看背包负重7公斤时的最佳解最后一个放入的是2号也就是橘子现在背包剩下负重量5公斤7-2所以看负重5公斤的最佳解最后放入的是1号也就是苹果此时背包负重量剩下0公斤5-5无法再放入水果所以求出最佳解为放入草莓橘子与苹果而总价为9050元实作 Cincludeinclude define LIMIT 8 重量限制define N 5 物品种类define MIN 1 最小重量 struct bodychar name[20]int sizeint pricetypedef struct body object int main voidint item[LIMIT1] 0int value[LIMIT1] 0int newvalue i s p object a[] "李子" 4 4500 "苹果" 5 5700"橘子" 2 2250"草莓" 1 1100"甜瓜" 6 6700 for i 0 i N i for s a[i]size s LIMIT sp s - a[i]sizenewvalue value[p] a[i]priceif newvalue value[s] 找到阶段最佳解value[s] newvalueitem[s] iprintf "物品\t价格\n"for i LIMIT i MIN i i - a[item[i]]sizeprintf "s\td\n"a[item[i]]name a[item[i]]priceprintf "合计\td\n" value[LIMIT] returnJavaclass Fruitprivate String nameprivate int sizeprivate int price public Fruit String name int size int pricethisname namethissize sizethisprice pricepublic String getNamereturn namepublic int getPricereturn pricepublic int getSizereturn sizepublic class Knapsackpublic static void main String[] argsfinal int 8final int MIN 1int[] item new int[1]int[] value new int[1] Fruit fruits[]new Fruit "李子" 4 4500new Fruit "苹果" 5 5700new Fruit "橘子" 2 2250new Fruit "草莓" 1 1100new Fruit "甜瓜" 6 6700 fo。
数值分析算法C语言程序
数值分析算法C语言程序数值分析是研究数学问题的近似解法的一门学科,其中包括了各种数值方法和算法。
本文将介绍数值分析中的常见算法,并给出相应的C语言程序。
1.二分法(Bisection Method)二分法是一种求函数零点的简单且常用的方法。
该方法的基本思想是通过不断将区间进行二分,并比较中点处函数值的正负来找到零点所在的区间。
```c#include <stdio.h>double f(double x)return x * x - 2;double bisection(double a, double b, double eps)double c;while ((b - a) > eps)c=(a+b)/2;if (f(c) == 0)break;}else if (f(a) * f(c) < 0)b=c;}elsea=c;}}return c;int maidouble a = 0.0;double b = 2.0;double result = bisection(a, b, eps);printf("The root is: %lf\n", result);return 0;```2.牛顿迭代法(Newton's Method)牛顿迭代法是一种高效的求函数零点的方法。
该方法的基本思想是通过对函数进行线性逼近,不断逼近函数的零点。
```c#include <stdio.h>#include <math.h>double f(double x)return x * x - 2;double df(double x)return 2 * x;double newton(double x0, double eps) double x = x0;double deltaX = f(x) / df(x);while (fabs(deltaX) > eps)deltaX = f(x) / df(x);x = x - deltaX;}return x;int maidouble x0 = 2.0;double result = newton(x0, eps); printf("The root is: %lf\n", result); return 0;```3.高斯消元法(Gaussian Elimination)高斯消元法是一种用于求解线性方程组的方法。
C程序代码大全
C程序代码大全Prepared on 21 November 2021<<endl;cout<<"------------------"<<endl;<<endl;cout<<"pi="<<pi<<" r="<<r<<" s="<<s<<endl;}#include <> 1 c<<endl;;int a[10];double xy[10];cout<<"sizeof(str)="<<sizeof(str)<<endl;cout<<"sizeof(a)="<<sizeof(a)<<endl;cout<<"sizeof(xy)="<<sizeof(xy)<<endl;<<endl;else if (score>=80)cout<<"Your grade is a B."<<endl;else if (score>=70)cout<<"Your grade is a C."<<endl;else if (score>=60)cout<<"Your grade is a D."<<endl;elsecout<<"Your grade is a E."<<endl;}#include <>main(){int n;cout<<"n=";cin>>n;if (n>=0 && n<=100 &&n%2==0)cout<<"n="<<n<<endl;elsecout<<"The "<<n<<" is out of range!"<<endl; }#include <>main(){int a,b,Max;.10for(int i=1;i<=10;i++)cout<<i<<" ";cout<<endl;.1for(int j=10;j>=1;j--)cout<<j<<" ";cout<<endl;.9for(int k=1;k<=10;k=k+2)cout<<k<<" ";cout<<endl;.Zfor(char c='A';c<='Z';c++)cout<<c;cout<<endl;.for(float x=0;x<=;x=x+cout<<x<<" ";cout<<endl;.for(float x1=0;x1<=+2;x1=x1+cout<<x1<<" ";cout<<endl;.+100int s=0;for(int n=1;n<=100;n++)s=s+n;cout<<"s="<<s<<endl;}#include<>main(){.+100int s=0,n=1;while(n<=100) {s=s+n;n++;}cout<<"s="<<s<<endl;.+100int s=0,n=0;do {n++;s+=n;}while(n<100);cout<<"s="<<s<<endl;um;cin>>st[i].name;cin>>st[i].maths;cin>>st[i].physics;cin>>st[i].chemistry;}otal=st[i].maths+st[i].physics+st[i].chemistry; um<<'\t';cout<<st[i].name<<'\t';cout<<st[i].maths<<'\t';cout<<st[i].physics<<'\t';cout<<st[i].chemistry<<'\t';cout<<st[i].total<<endl;}}#include<>main(){ame="<<(*p).name<<endl;cout<<"(*p).sex="<<(*p).sex<<endl;cout<<"(*p).age="<<(*p).age<<endl;cout<<"p->name="<<p->name<<endl;cout<<"p->sex="<<p->sex<<endl;cout<<"p->age="<<p->age<<endl;ame;cout<<"sex:";cin>>(*p).sex;cout<<"age:";cin>>(*p).age;ame="<<(*p).name<<endl;cout<<"(*p).sex="<<(*p).sex<<endl;cout<<"(*p).age="<<(*p).age<<endl;cout<<"-------------------------"<<endl;ame<<'\t';cout<<x[i].sex<<'\t';cout<<x[i].age<<endl;}cout<<"----------------"<<endl;;int int_values[] = {51, 23, 2, 44, 45,0,11};float float_values[] = {, , , , };student st_arr[]={101,"WangLin",92,102,"LiPing",85,103,"ZhaoMin",88};um<<" ";cout<<st_arr[i].name<<" ";cout<<st_arr[i].grade<<endl;}}#include<>otal<a[j].total)swap_Rec(&a[i],&a[j]); <<"\t"<<str_len("This is a test.")<<endl; }#include<>void disp(void); <<endl;}#include<><<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); um;cin>>st[i].name;cin>>st[i].grade;fprintf(fp1,"%d %s %f\n",st[i].num,st[i].name,st[i].grade);}fclose(fp1); <<endl; cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); <<endl;cout<<"Exiting program."<<endl;exit(1); */putc( 'A', stdin );if( ferror( stdin ) ){perror( "Write error" );clearerr( stdin );}/* See if read causes an error. */printf( "Will input cause an error " );c = getc( stdin );if( ferror( stdin ) ){perror( "Read error" );clearerr( stdin );}}#include<>#include<><<endl;for (i=1; i<=MAX; i++) {cin>>x;(x);}<<endl;}void push(float x) <<endl;for (i=1; i<=MAX; i++) {cin>>x;(x);}<<endl;}void push(float x) <<endl;}~stack(void) << endl; << endl;}<< endl; isp_count();delete p;();}#include<><< endl;}~ex_class() {cout << "The Object destroyed." <<endl;}void set_value(int n);void show_val(char *name);} ;<<endl;}#include<>um :";cout<<num<<endl;}};um :";cout<<<<endl;}um="<<num<<endl;}public: um=";cout<<<<endl;}how_value("(*p)obj2");al="<<val<<endl;DispFirst();}};<<endl;}virtual void aFn2(void) {cout<<"aFn2 is in Base class."<<endl;}<<endl;}};<<endl;}<<endl;<<endl;}<<endl;<<endl;}void aFn2(void) {cout<<"aFn2 is in First derived class."<<endl; }};<<endl;}void aFn2(void){cout<<"aFn2 is in Second derived class."<<endl; }};;cout<<"s1: "<<s1<<endl;cout<<"s2: "<<s2<<endl;10.2f10.5f;cout<<"String: "<<string<<endl;cout<<"p : "<<p<<endl;}#include<>#include <>;int n;1.7320534f << endl;}~stack(void) {cout << "Stack Destroyed." << endl;}void push(T);T pop(void);};<< endl;return;}stck[tos++] = i;}template <class T> T stack<T>::pop(void){if(tos==0){cout << "Stack underflow." << endl;return 0;}return stck[--tos];}irst;cout<<","<<(*theIterator).second<<" ";}cout<<endl;irst;cout<<","<<(*theIterator).second<<" ";}cout<<endl;irst;cout<<","<<(*it).second<<" ";}cout<<endl;econd << " ";elsecout << "[err] ";}cout << endl;}}#include <iostream>#include <string>#include <map>using namespace std;econd;cout<<"\t"<<(*theIterator).first<<endl; }econd;cout<<"\t"<<(*i).first<<endl;}econd;cout<<"\t"<<(*it).first<<endl;}econd;cout<<"\t"<<(*i).first<<endl;}}#include <iostream>#include <valarray>#include <>using namespace std;#define ARRAY_SIZE 3 .入栈for (i=0;i<10;i=i+2)(i);if (!()) {cout << "()=20;" << endl;()=20;}//弹出栈中所有的数据并显示cout<<"stack1: ";while (!()) {cout<<()<<" ";();}cout<<endl;}#include <iostream>#include <list>#include <numeric>using namespace std;//创建一个list容器的实例LISTINT,其存放int型数据typedef list<int> LISTINT;void main(void){//用LISTINT创建一个名为listOne的list对象 LISTINT listOne;//指定i为迭代器变量LISTINT::iterator i;LISTINT::reverse_iterator ir;//从前面向listOne容器中添加数据(2);(1);//从后面向listOne容器中添加数据(3);(4);//从前向后显示listOne中的数据for (i = (); i != (); ++i)cout << *i << " ";cout << endl;//从后向后显示listOne中的数据for (ir =();ir!=(); ++ir)cout << *ir << " ";cout << endl;//从键盘上输入数据for (i = (); i != (); ++i) {cout<<"listOne :";cin>>(*i);}//从前向后显示listOne中的数据for (i = (); i != (); ++i)cout << *i << " ";cout << endl;//bidirectional迭代器不允许加减运算// i=()+1;}#include <iostream>#include <iostream>#include <numeric>#include <vector>#include <list>#include <set>using namespace std;//利用类模板生成类实例typedef vector < int > IntArray;typedef list <int> LISTINT;typedef set<int> SET_INT;int add(int a, int b) {return a+b;}//在main()函数中测试accumulate算法void main (){//--------------------------------------------// accumulate算法对于普通数组的计算//--------------------------------------------- int x[]={1,3,5,7,9};cout<<"x[]:";for (int i=0;i<5;i++)cout<<x[i]<<" ";cout<<endl;cout<<"accumulate(x,x+5,0)=";cout<<accumulate(x,x+5,0)<<endl;int val=100;cout<<"val="<<val<<endl;cout<<"accumulate(x,x+5,val)=";cout<<accumulate(x,x+5,val)<<endl;//--------------------------------------------// accumulate算法对于vector容器的计算//--------------------------------------------- //声明intvector容器和迭代器iiIntArray intvector;IntArray::iterator ii;//向intvector容器中插入元素for (i=1; i<=5; i++) {(i);};//显示intvector容器中的元素值和累加结果cout << "intvector: "<<endl;for (ii=();ii !=();++ii)cout<<(*ii)<<" ";cout<<endl;cout<<"accumulate(),(),0)=";cout<<accumulate(),(),0)<<endl;//--------------------------------------------// accumulate算法对于list容器的计算//--------------------------------------------- //声明list容器对象和迭代器LISTINT::iterator iL;LISTINT list1;//向list1容器对象中插入元素并显示(1);(3);(5);(2);(6);//显示list1容器的元素值和累加结果cout << "list1: "<<endl;for (iL=();iL !=();++iL)cout<<(*iL)<<" ";cout<<endl;cout<<"accumulate(),(),0)=";cout<<accumulate(),(),0)<<endl;//--------------------------------------------// accumulate算法对于set容器的计算//--------------------------------------------- //声明set容器对象和迭代器SET_INT set1;SET_INT::iterator si;//向set1容器中插入元素(5);(20);(10);(15);(25);//显示set1容器的元素值和累加结果cout <<"set1: "<<endl;for (si=();si !=();++si)cout<<(*si)<<" ";cout<<endl;cout<<"accumulate(),(),0)=";cout<<accumulate(),(),0)<<endl;cout<<"accumulate(),(),100)=";cout<<accumulate(),(),100)<<endl;}#include <iostream>#include <algorithm>#include <vector>#include <list>#include <set>#define size 10using namespace std;//产生指定范围的整数随机数int getrand(int min,int max) {int m;m=(max-min);m=min+double(rand())/RAND_MAX*m ;return m;}//利用类模板生成实例typedef vector < int > IntArray;typedef list <int> LISTINT;typedef set<int> SET_INT;//在main()函数中测试accumulate算法void main (){//--------------------------------------------// count算法对于普通数组的计算//--------------------------------------------- int x[size];cout<<"x[]:";for (int i=0;i<size;i++) {x[i]=getrand(1,3);cout<<x[i]<<" ";}cout<<endl;cout<<"count(x,x+size,2)=";cout<<count(x,x+size,2)<<endl;cout<<"count(x+2,x+8,2)=";cout<<count(x+2,x+8,2)<<endl;//--------------------------------------------// count算法对于vector容器的计算//声明intvector容器和迭代器iiIntArray intvector;IntArray::iterator ii;//向intvector容器中插入元素for (i=1; i<size; i++) {(getrand(2,6));};//显示intvector容器中的元素值和统计结果cout << "intvector: ";for (ii=();ii !=();++ii)cout<<(*ii)<<" ";cout<<endl;cout<<"count(),(),4)=";cout<<count(),(),4)<<endl;//--------------------------------------------// count算法对于list容器的计算//--------------------------------------------- //声明list容器对象和迭代器LISTINT::iterator iL;LISTINT list1;//向list1容器对象中插入元素并显示for (i=1; i<size; i++) {(getrand(3,5));};//显示list1容器的元素值和统计结果cout << "list1: ";for (iL=();iL !=();++iL)cout<<(*iL)<<" ";cout<<endl;cout<<"count(),(),3)=";cout<<count(),(),3)<<endl;//--------------------------------------------// count算法对于set容器的计算//--------------------------------------------- //声明set容器对象和迭代器SET_INT set1;SET_INT::iterator si;//向set1容器中插入元素for (i=1; i<size; i++) {(getrand(1,10));};//显示set1容器的元素值和统计结果cout <<"set1: ";for (si=();si !=();++si)cout<<(*si)<<" ";cout<<endl;cout<<"count(),(),5)=";cout<<count(),(),5)<<endl;}#include <iostream>#include <algorithm>#include <string>#include <vector>//如果字符串以'S'开头,则返回trueint MatchFirstChar( const string& str){string s("S") ;return s == (0,1) ;}//测试count_if算法void main(){const int VECTOR_SIZE = 8 ;//生成成员类型为strings的vector容器类typedef vector<string > StringVector ;//定义迭代器类型typedef StringVector::iterator StringVectorIt ;//声明vector容器的对象StringVector NamesVect(VECTOR_SIZE) ;//声明迭代器StringVectorIt start, end, it ;int result = 0 ; // 存放统计数据//初始化vector容器NamesVectNamesVect[0] = "She" ;NamesVect[1] = "Sells" ;NamesVect[2] = "Sea" ;NamesVect[3] = "Shells" ;NamesVect[4] = "by" ;NamesVect[5] = "the" ;NamesVect[6] = "Sea" ;NamesVect[7] = "Shore" ;//设置容器的起始位置和终止位置start = () ;end = () ;//显示NamesVect容器的元素cout << "NamesVect: " ;for(it = start; it != end; it++)cout << *it << " " ;cout <<endl ;//统计并显示NamesVect容器的所有元素中以'S'字符开头的字符串 result = count_if(start, end, MatchFirstChar) ;cout << "Number of elements that start with letter \"S\" = " << result << endl ;//显示NamesVect容器[1,6]之间的元素cout <<"NamesVect[1]--NamesVect[6]: " ;for(it =&NamesVect[1]; it != &NamesVect[7]; it++)cout << *it << " " ;cout <<endl ;//统计并显示NamesVect容器的所有元素中以'S'字符开头的字符串cout << "Number of elements that start with letter \"S\" = " << result << endl ;}#include <iostream>#include <algorithm>#include <vector>using namespace std;//利用类模板生成实例typedef vector < int > IntArray;//显示数组void put_array(int x[],int size) {for(int i=0;i<size;i++)cout<<x[i]<<" ";cout<<endl;}//显示vector容器中的元素void put_vector(IntArray v){IntArray::iterator theIterator;for (theIterator=();theIterator!=();++theIterator){cout<<(*theIterator)<<" ";}cout<<endl;}//在main()函数中测试fill和fill_n算法void main (){//--------------------------------------------// fill和fill_n算法对普通数组的计算//---------------------------------------------int x[]={1,3,5,7,9};cout << "x[]: ";put_array(x,5);//填数处理fill(x+1,x+3,2);cout << "fill(x+1,x+3,2): "<<endl;put_array(x,5);fill_n(x,3,8);cout << "fill_n(x,3,8): "<<endl;put_array(x,5);//--------------------------------------------// fill和fill_n算法对于vector容器的计算//---------------------------------------------//声明intvector容器和迭代器iiIntArray intvector;//向intvector容器中插入元素for (int i=1; i<=10; i++) {(i);};//显示intvector容器中的元素值和统计结果cout << "intvector: "<<endl;put_vector(intvector);//填数处理fill(),()+3,2);put_vector(intvector);fill_n(&intvector[5],3,8);cout << "fill_n(&intvector[5],3,8): "<<endl;put_vector(intvector);}#include <iostream>#include <algorithm>#include <vector>#define ARRAY_SIZE 10using namespace std;//利用类模板生成实例typedef vector < int > IntArray;//显示数组void put_array(int x[],int size) {for(int i=0;i<size;i++)cout<<x[i]<<" ";cout<<endl;}//显示vector容器中的元素void put_vector(IntArray v){IntArray::iterator theIterator;for (theIterator=();theIterator!=();++theIterator){cout<<(*theIterator)<<" ";}cout<<endl;}//在main()函数中测试find()算法void main (){int i,value,*p;//--------------------------------------------// find()算法对于普通数组的处理//---------------------------------------------int x[ARRAY_SIZE]={1,3,5,7,9,2,4,6,8,10};cout << "x[]: ";put_array(x,ARRAY_SIZE);//find()算法查找,并显示查找结果for(i=0;i<=2;i++) {cout<<"value=";cin>>value;p=find(x,x+ARRAY_SIZE,value);if (p != x + ARRAY_SIZE) { //查到cout << "First element that matches " << value;cout<< " is at location " << p - x<< endl;}else { //未查到cout << "The sequence does not contain any elements"; cout<< " with value " << value << endl ;}}//声明intvector容器对象IntArray intvector;//向intvector容器中插入元素for (i=1; i<=10; i++) {(i);};//显示intvector容器中的元素值cout << "intvector: ";put_vector(intvector);//find()算法查找,并显示查找结果IntArray::iterator pos;for (i=0;i<=2;i++) {cout<<"value=";cin>>value;pos=find(),(),value);if (pos != ()) { //查到cout << "First element that matches " << value;cout<< " is at location " <<pos - ()<< endl;}else { //未查到cout << "The sequence does not contain any elements"; cout<< " with value " << value << endl ;}}}#include <iostream>#include <algorithm>#include <vector>#define ARRAY_SIZE 10using namespace std;//利用类模板生成实例typedef vector < int > IntArray;//显示数组void put_array(int x[],int size) {for(int i=0;i<size;i++)cout<<x[i]<<" ";}//显示vector容器中的元素void put_vector(IntArray v){IntArray::iterator theIterator;for (theIterator=();theIterator!=();++theIterator){cout<<(*theIterator)<<" ";}}//在main()函数中测试find()_end()算法void main (){int x[ARRAY_SIZE]={1,3,5,7,9,2,4,6,8,10};cout << "x[]: ";put_array(x,ARRAY_SIZE);cout<<endl;int y[]={5,7,9};cout << "y[]: ";put_array(y,3);cout<<endl;// find_end()算法查找,并显示查找结果int *p=find_end(x,x+ARRAY_SIZE,&y[0],&y[2]);if (p != x + ARRAY_SIZE) { //查到cout << "The first element that matches :" ;put_array(y,3);cout<< " is at location in x" << p - x<< endl;}else { //未查到cout << "The sequence does not contain any elements";cout<< " with value " ;put_array(&x[3],3);}//--------------------------------------------// find_end()算法对vector容器的处理//---------------------------------------------//声明intvector容器对象IntArray intvector;//向intvector容器中插入元素for (int i=1; i<=10; i++) {(i);};//显示intvector容器中的元素值cout << "intvector: ";put_vector(intvector);cout<<endl;IntArray temp;(5);(6);(7);cout << "temp: ";put_vector(temp);cout<<endl;// find_end()算法查找,并显示查找结果IntArray::iterator pos;pos=find_end(),(),(),());if (pos != ()) { //查到cout << "The first element that matches ";put_vector(temp);cout<< " is at location in intvector " <<pos - ()<< endl; }else { //未查到cout << "The sequence does not contain any elements";put_vector(temp);cout<< endl ;}}#include <iostream>#include <vector>#include <algorithm>using namespace std;//返回一个Fibonacci数,其由generate_n()算法调用int Fibonacci1(void){static int r;static int f1 = 0;static int f2 = 1;r = f1 + f2 ;f1 = f2 ;f2 = r ;return f1 ;}//返回一个Fibonacci数,其由generate()算法调用int Fibonacci2(void){static int r;static int f1 = 0;static int f2 = 1;r = f1 + f2 ;f1 = f2 ;f2 = r ;return f1 ;}//定义整型数的vector容器类typedef vector<int > IntVector ;//显示vector容器中的元素void put_vector(IntVector v,char *name){IntVector::iterator theIterator;cout<<name<<":"<<endl;for (theIterator=();theIterator!=();++theIterator){cout<<(*theIterator)<<" ";}cout<<endl;}//测试generate()和generate_n()算法void main(){const int VECTOR_SIZE = 15 ;//定义迭代器类typedef IntVector::iterator IntVectorIt ;//声明vector容器对象IntVector Numbers1(VECTOR_SIZE),Numbers2(VECTOR_SIZE); int i ;//初始化vector容器对象Numbers1[i] = i ;//显示vector容器对象的元素cout << "Before calling generate_n:" << endl ;put_vector(Numbers1,"Numbers1");//利用generate_n算法用Fibonacci 数填充vector容器 generate_n(), VECTOR_SIZE, Fibonacci1) ;//显示vector容器对象的元素cout << "After calling generate_n:" << endl ;put_vector(Numbers1,"Numbers1");//利用generate算法用Fibonacci 数填充vector容器generate(),(), Fibonacci2) ;//显示vector容器对象的元素cout << "After calling generate:" << endl ;put_vector(Numbers2,"Numbers2");}#include <iostream>#include <algorithm>#include <vector>using namespace std;//利用类模板生成实例typedef vector < int > IntArray;//显示数组void put_array(int x[],int size) {for(int i=0;i<size;i++)cout<<x[i]<<" ";cout<<endl;}//显示vector容器中的元素void put_vector(IntArray v){IntArray::iterator theIterator;for (theIterator=();theIterator!=();++theIterator){ cout<<(*theIterator)<<" ";}cout<<endl;}//在main()函数中测试reverse()和reverse_copy()算法void main (){//--------------------------------------------// reverse()和reverse_copy()算法对普通数组处理//---------------------------------------------int x[]={1,3,5,7,9};cout<<"x[]:";put_array(x,5);//reverse()反转x数组并显示reverse(x,x+5);put_array(x,5);int y[]={2,4,6,8,10};cout<<"y[]:";put_array(y,5);//reverse_copy()反转y数组的部分元素并拷贝到x数组第2个元素位置 reverse_copy(y+1,y+3,x+1);cout<<"x[]:";put_array(x,5);cout<<"y[]:";put_array(y,5);//--------------------------------------------// reverse()和reverse_copy()算法对vector容器的处理//---------------------------------------------//声明intvector容器和迭代器iiIntArray intvector;//向intvector容器中插入元素for (int i=1; i<=10; i++) {(i);};//显示intvector容器中的元素值cout << "intvector: "<<endl;put_vector(intvector);//reverse()对于vector容器的处理reverse(),());cout << "intvector: "<<endl;put_vector(intvector);// reverse_copy对于vector容器的处理IntArray temp(5);reverse_copy()+2,()+7,());cout << "temp: "<<endl;put_vector(temp);}#include <iostream>#include <algorithm>#include <vector>#include <>#define ARRAY_SIZE 15using namespace std;//定义整型数的vector容器类typedef vector<int > IntVector ;//显示数组void put_array(int x[],int size) {for(int i=0;i<size;i++)cout<<x[i]<<" ";cout<<endl;}//显示vector容器中的元素void put_vector(IntVector v,char *name){IntVector::iterator theIterator;for (theIterator=();theIterator!=();++theIterator){ cout<<(*theIterator)<<" ";}cout<<endl;}//产生指定范围的整数随机数int getrand(int min,int max) {int m;m=(max-min);m=min+double(rand())/RAND_MAX*m ;return m;}//在main()函数中测试sort()和partial_sort()算法void main (){int i;//--------------------------------------------// sort()和partial_sort()算法对普通数组处理//---------------------------------------------//sort()算法处理数组,并显示int x[ARRAY_SIZE];for (i=0;i<ARRAY_SIZE;i++) {x[i]=getrand(1,20);}cout<<"x[]:";put_array(x,ARRAY_SIZE);sort(x,x+ARRAY_SIZE);cout<<"sort(x,x+ARRAY_SIZE):"<<endl;put_array(x,ARRAY_SIZE);//partial_sort()算法对于数组进行处理int y[ARRAY_SIZE];for (i=0;i<ARRAY_SIZE;i++) {y[i]=getrand(1,30) ;}cout<<"y[]:";put_array(y,ARRAY_SIZE);partial_sort(y+2,y+7,y+ARRAY_SIZE);cout<<"partial_sort(y+2,y+7,y+ARRAY_SIZE):"<<endl; put_array(y,ARRAY_SIZE);//--------------------------------------------// sort()和partial_sort()算法对vector容器的处理//---------------------------------------------IntVector Numbers1,Numbers2;for(i=0;i<15;i++) {(getrand(1,30));(getrand(1,30));}put_vector(Numbers1,"Numbers1");put_vector(Numbers2,"Numbers2");//sort()算法处理并显示sort(),());cout<<"After call sort():"<<endl;put_vector(Numbers1,"Numbers1");//partial_sort()算法处理并显示partial_sort()+2,()+7,());。
c数值算法程序大全c6-8
252Chapter 6.Special FunctionsSample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).CITED REFERENCES AND FURTHER READING:Barnett,A.R.,Feng,D.H.,Steed,J.W.,and Goldfarb,L.J.B.1974,Computer Physics Commu-nications ,vol.8,pp.377–395.[1]T emme,N.M.1976,Journal of Computational Physics ,vol.21,pp.343–350[2];1975,op.cit.,vol.19,pp.324–337.[3]Thompson,I.J.,and Barnett,A.R.1987,Computer Physics Communications ,vol.47,pp.245–257.[4]Barnett,A.R.1981,Computer Physics Communications ,vol.21,pp.297–314.Thompson,I.J.,and Barnett,A.R.1986,Journal of Computational Physics ,vol.64,pp.490–509.Abramowitz,M.,and Stegun,I.A.1964,Handbook of Mathematical Functions ,Applied Mathe-matics Series,Volume 55(Washington:National Bureau of Standards;reprinted 1968by Dover Publications,New York),Chapter 10.6.8Spherical HarmonicsSpherical harmonics occur in a large variety of physical problems,for ex-ample,whenever a wave equation,or Laplace’s equation,is solved by separa-tion of variables in spherical coordinates.The spherical harmonic Y lm (θ,φ),−l ≤m ≤l,is a function of the two coordinates θ,φon the surface of a sphere.The spherical harmonics are orthogonal for different l and m ,and they are normalized so that their integrated square over the sphere is unity:2π0dφ1−1d (cos θ)Y l m *(θ,φ)Y lm (θ,φ)=δl l δm m(6.8.1)Here asterisk denotes complex conjugation.Mathematically,the spherical harmonics are related to associated Legendre polynomials by the equationY lm (θ,φ)=2l +14π(l −m )!(l +m )!P ml (cos θ)e imφ(6.8.2)By using the relationY l,−m (θ,φ)=(−1)m Y lm *(θ,φ)(6.8.3)we can always relate a spherical harmonic to an associated Legendre polynomial with m ≥0.With x ≡cos θ,these are defined in terms of the ordinary Legendre polynomials (cf.§4.5and §5.5)byP m l (x )=(−1)m(1−x 2)m/2d mdx mP l (x )(6.8.4)6.8Spherical Harmonics 253Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).The first few associated Legendre polynomials,and their corresponding nor-malized spherical harmonics,areP 00(x )=1Y 00= 14πP 11(x )=−(1−x 2)1/2Y 11=−38πsin θe iφP 01(x )=x Y 10=34πcos θP 22(x )=3(1−x 2)Y 22=14152πsin 2θe 2iφP 12(x )=−3(1−x 2)1/2x Y 21=−158πsin θcos θe iφP 02(x )=12(3x 2−1)Y 20=54π(32cos 2θ−12)(6.8.5)There are many bad ways to evaluate associated Legendre polynomials numer-ically.For example,there are explicit expressions,such asP m l (x )=(−1)m (l +m )!2m m !(l −m )!(1−x 2)m/21−(l −m )(m +l +1)1!(m +1) 1−x 2+(l −m )(l −m −1)(m +l +1)(m +l +2)2!(m +1)(m +2) 1−x22−··· (6.8.6)where the polynomial continues up through the term in (1−x )l −m .(See [1]forthis and related formulas.)This is not a satisfactory method because evaluation of the polynomial involves delicate cancellations between successive terms,which alternate in sign.For large l ,the individual terms in the polynomial become very much larger than their sum,and all accuracy is lost.In practice,(6.8.6)can be used only in single precision (32-bit)for l up to 6or 8,and in double precision (64-bit)for l up to 15or 18,depending on the precision required for the answer.A more robust computational procedure is therefore desirable,as follows:The associated Legendre functions satisfy numerous recurrence relations,tab-ulated in [1-2].These are recurrences on l alone,on m alone,and on both l and m simultaneously.Most of the recurrences involving m are unstable,and so dangerous for numerical work.The following recurrence on l is,however,stable (compare 5.5.1):(l −m )P m l =x (2l −1)P m l −1−(l +m −1)P ml −2(6.8.7)It is useful because there is a closed-form expression for the starting value,P m m =(−1)m (2m −1)!!(1−x 2)m/2(6.8.8)(The notation n !!denotes the product of all odd integers less than or equal to n .)Using (6.8.7)with l =m +1,and setting P mm −1=0,we findP m m +1=x (2m +1)P mm(6.8.9)Equations (6.8.8)and (6.8.9)provide the two starting values required for (6.8.7)for general l .The function that implements this is254Chapter 6.Special FunctionsSample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).#include <math.h>float plgndr(int l,int m,float x)Computes the associated Legendre polynomial P m l (x ).Here m and l are integers satisfying 0≤m ≤l ,while x lies in the range −1≤x ≤1.{void nrerror(char error_text[]);float fact,pll,pmm,pmmp1,somx2;int i,ll;if (m <0||m >l ||fabs(x)>1.0)nrerror("Bad arguments in routine plgndr");pmm=1.0;Compute P m m.if (m >0){somx2=sqrt((1.0-x)*(1.0+x));fact=1.0;for (i=1;i<=m;i++){pmm *=-fact*somx2;fact +=2.0;}}if (l ==m)return pmm;else {Compute P m m +1.pmmp1=x*(2*m+1)*pmm;if (l ==(m+1))return pmmp1;else {Compute P m l ,l >m +1.for (ll=m+2;ll<=l;ll++){pll=(x*(2*ll-1)*pmmp1-(ll+m-1)*pmm)/(ll-m);pmm=pmmp1;pmmp1=pll;}return pll;}}}CITED REFERENCES AND FURTHER READING:Magnus,W.,and Oberhettinger,F .1949,Formulas and Theorems for the Functions of Mathe-matical Physics (New York:Chelsea),pp.54ff.[1]Abramowitz,M.,and Stegun,I.A.1964,Handbook of Mathematical Functions ,Applied Mathe-matics Series,Volume 55(Washington:National Bureau of Standards;reprinted 1968by Dover Publications,New York),Chapter 8.[2]6.9Fresnel Integrals,Cosine and Sine Integrals 255Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).6.9Fresnel Integrals,Cosine and Sine IntegralsFresnel IntegralsThe two Fresnel integrals are defined byC (x )=x0cos π2t 2 dt,S (x )=xsin π2t 2 dt (6.9.1)The most convenient way of evaluating these functions to arbitrary precision is to use power series for small x and a continued fraction for large x .The series areC (x )=x −π2 2x 55·2!+ π2 4x 99·4!−···S (x )=π2 x 33·1!− π2 3x 77·3!+ π2 5x 1111·5!−···(6.9.2)There is a complex continued fraction that yields both S (x )and C (x )si-multaneously:C (x )+iS (x )=1+i2erf z,z =√π2(1−i )x(6.9.3)wheree z2erfc z =1√π 1z +1/2z +1z +3/2z +2z +··· =2z √π12z 2+1−1·22z 2+5−3·42z 2+9−··· (6.9.4)In the last line we have converted the “standard”form of the continued fraction to its “even”form (see §5.2),which converges twice as fast.We must be careful not to evaluate the alternating series (6.9.2)at too large a value of x ;inspection of the terms shows that x =1.5is a good point to switch over to the continued fraction.Note that for large xC (x )∼12+1πx sin π2x 2 ,S (x )∼12−1πx cos π2x 2 (6.9.5)Thus the precision of the routine frenel may be limited by the precision of thelibrary routines for sine and cosine for large x .。
c语言15的二进制
c语言15的二进制15的二进制表示为1111。
C语言是一种广泛应用于系统软件开发和嵌入式系统的高级编程语言。
它具有代码简洁、执行效率高、跨平台等特点,被广泛应用于各种领域。
本文将以C语言15的二进制表示为标题,介绍一些与二进制相关的知识和应用。
二进制是一种计数系统,只包含0和1两个数字。
在计算机中,所有的数据都以二进制形式存储和处理。
C语言中,可以使用位运算符来处理二进制数据。
我们来看一下15的二进制表示。
15的二进制为1111。
这意味着在一个4位的二进制数中,最高位是1,其余位都是1。
在C语言中,可以使用二进制字面量0b1111来表示15。
在C语言中,我们可以使用位运算符来对二进制数据进行操作。
例如,我们可以使用位与运算符(&)来将某些位设置为0,使用位或运算符(|)来将某些位设置为1,使用位异或运算符(^)来对某些位进行翻转。
除了位运算符,C语言还提供了一些其他的操作二进制数据的函数和库。
例如,可以使用标准库中的sprintf函数将一个整数转换为二进制字符串,使用sscanf函数将一个二进制字符串转换为整数。
除了基本的二进制操作,C语言还提供了一些高级的二进制操作函数和库。
例如,可以使用位字段来定义一个结构体,使用位操作来对结构体中的位进行操作。
此外,C语言还提供了一些位操作的宏定义,如位测试、位设置等。
在实际应用中,二进制操作经常用于位字段、编码解码、位图等领域。
例如,在网络通信中,可以使用二进制编码来表示数据,使用位操作来进行数据的解析和处理。
在图像处理中,可以使用位图来表示图像数据,使用位操作来对图像数据进行处理和操作。
总结一下,C语言可以方便地处理二进制数据,提供了丰富的位运算符、函数和库来操作二进制数据。
在实际应用中,二进制操作被广泛应用于各种领域,如网络通信、图像处理等。
通过对二进制的学习和掌握,可以更好地理解和应用C语言。
C语言经典算法C语言代码大全
C语言经典算法C语言代码大全一、排序算法1、冒泡排序它重复地走访过要排序的数列,一次比较两个元素,如果他们的顺序错误就把他们交换过来。
设数组为a[0…n-1]C语言实现如下://冒泡排序void bubbleSort(int arr[], int n)int i, j, temp;bool flag;//表示n次排序过程for(i = 0; i < n - 1; i++)//每次排序将最大的数放到最右边flag = false;for(j= 0; j< n-1-i; j++)if(arr[j] > arr[j+1])temp = arr[j];arr[j] = arr[j+1];arr[j+1] = temp;flag = true;}}//如果趟排序没有进行数据交换,说明数据已经有序if (flag == false)break;}}2、快速排序它采用了分治法的思想,基于快速排序的思想,可以对数组进行非常快速的排序。
设数组为a[0…n-1]C语言实现如下://快速排序// arr[left] 为起始值,arr[right] 为末尾值void quickSort(int arr[], int left, int right)int i, j, base;if (left > right)return;}i = left;j = right;base = arr[left];//定义基准值,可以是数组的第一个值while (i != j)// 因为基准值是 arr[left],所以左边右移,直到找到小于基准值的值while (arr[j] >= base && i < j)j--;}// 因为基准值是 arr[left],所以右边左移while (arr[i] <= base && i < j)i++;}//如果i<j,表示找到了,交换位置if (i < j)int temp = arr[i];arr[i] = arr[j];arr[j] = temp;}}//将基准值放到i位置arr[left] = arr[i];。
C语言数值算法程序大全-第二版
17.5 网格点的自动分配 663 17.6 内部边界条件或奇异点的处理 664
3.0 引言 88
3.1 多项式内插和外推 89
3.2 有理函数内插法和外推法 92
3.3 三次样条插值 94
第四章 函数积分 108 4.0 引言 108 4.1 坐标等距划分的经典公式 109 4.2 基本算法 113 4.3 龙贝格积分 116 4.4 广义积分 117 4.5 高斯求积法与正交多项式 122
14.6 非参数相关或秩相关 541 14.7 二维分布不同吗 ? 546 14.8 萨维兹凯 -戈雷平滑滤波器 550 第十五章 数据的模型建立 556
15.3 两个坐标数据都有误差的直线拟合 563 15.4 一般的线性最小二乘方 568 15.5 非线性模型 576 15.6 被估模型参数的置信界限 583 15.7 稳健估计 592 第十六章 常微分方程组的积分 599 16.0 引言 599
5.1 级数和其收敛性 138 5.2 连分式求值 141 5.3 多项式和有理函数 144 5.4 复数运算 147 5.5 递推关系及克伦肖递推公式 148 5.6 二次方程和三次方程 151 5.7 数值求导 153
5.11 幂级数化简 163 5.12 Padé 逼近 164 5.13 有理切比雪夫逼近 167 5.14 线积分求函数值 171 第六章 特殊函数 174 6.0 引言 174
2.3 LU 分解和它的应用 35 2.4 三对角及带状对角系统方程 41 2.5 线性方程组解的迭代改进 45
数值计算方法实验程序源代码
上机实验1. 秦九韶算法:编程求多项式P(x)=(((0.0625x+0.0425) x+1.912) x+2.1296在x=1.0处的值。
#include "stdio.h"main(){static float a[]={2.1296,1.912,0.0425,0.0625};float y;int i;float x=1.0;clrscr();y=a[3];for (i=2;i>=0;i--)y=y*x+a[i];printf("x=%4.2f,y=%6.4f",x,y);}2. 二分法:方程f(x)=x3-x-1=0,利用逐步搜索法确定一个有根区间。
#include "stdio.h"#include "math.h"#define f(x) (x*(x*x-1)-1)#define e 0.005main(){int i=0;float x,a=1,b=1.5,y=f(a);if(y*f(b)>=0){printf("\nThe range is error!"); return;}elsedo{ x=(a+b)/2;printf("\nx%d=%6.4f",i,x);i++;if (f(x)==0) break;if(y*f(x)<0)b=x;elsea=x;}while(fabs(b-a)>e);printf("\nx=%4.2f",x);} #include "stdio.h"#include "math.h"#define f(x) ((x*x-1)*x-1)main(){int i;float x,a=1,b=1.5,y=f(a);if(y*f(b)>=0){printf("\nThe range is error!"); return;}elsefor(i=0;i<=6;i++)/*次数限制结束*/ {x=(a+b)/2;printf("\nx%d=%6.4f",i,x);if (f(x)==0) break;if(y*f(x)<0)b=x;elsea=x;}printf("\nx=%4.2f",x);}3. 迭代法:(1) 求方程f(x)=x-10x+2=0的一个根#include "stdio.h"#include "math.h"main(){float x0,x1=1;int i=1;do{x0=x1;x1=log10(x0+2);printf("\nx%d=%6.4f",i,x1);i++;}while(fabs(x1-x0)>=0.00005);printf("\nx=%6.4f",x1);printf("\nf(x)=%6.4f",fabs(x1-pow(10,x1)+2)) ;} (2) 求方程x=e-x在0.5附近的根。
数值计算C语言常用小程序
1、秦九韶算法2、二分法3、拉格朗日插值4、埃特金算法5、复化梯形法6、复化辛甫生算法7、二阶龙格库塔方法8、四阶龙格库塔方法9、改进的欧拉方法10、迭代法11、埃特金加速方法:12、牛顿迭代法13、追赶法14、雅克比迭代15、蛋白质设计:17高斯消去法:1、秦九韶算法利用秦九韶算法求多项式,在x=3时的值。
程序:#include<>#include<>void main(){float a[100],v,x;int n,i,k;scanf("%d%f",&n,&x);for(i=0;i<=n;i++)scanf("%f",&a[i]);v=a[n];k=1;do{v=x*v+a[n-k];k=k+1;}while(k<=n);printf("v=%f",v);}运行结果:2、二分法用二分法求方程法x*x*x-x-1=0在[1,2]内的近似根,要求误差不超过#include<>#include<>float fun(float);void main(){float a,b,c,x,y,y1;scanf("%f%f%f",&a,&b,&c);y1=fun(a);do{x=(a+b)/2;y=fun(x);{if(y*y1>0)a=x;elseb=x;}}while((b-a)>=c);printf("%f,%f\n",x,y);}float fun(float m){float n;n=m*m*m-m-1;return(n);}运行结果:3、拉格朗日插值程序:#include<>main(){float a,b,t,x[100],y[100]; int n,i,j,k;scanf("%f%d",&a,&n);for(i=0;i<=n;i++)scanf("%f%f",&x[i],&y[i]); k=0;b=0;for(k=0;k<=n;k++){t=1;for(j=0;j<=n;j++){if(j!=k)t=t*(a-x[j])/(x[k]-x[j]);} b=b+t*y[k];}printf("%f\n",b);}4、埃特金算法程序:#include<>#include<>main(){float a,b,c,x[100],y[100]; int i,j,n,k;scanf("%d%f",&n,&a);for(i=0;i<=n;i++)scanf("%f%f",&x[i],&y[i]);for(k=1;k<=n;k++){for(i=k;i<=n;i++)y[i]=y[k-1]+(y[i]-y[k-1])*(a-x[k-1])/(x[i]-x[k-1]);} printf("%f\n",y[n]);}5、复化梯形法设 ,用复化梯形法求积分的近似值程序:#include<>#include<>double fun(double);void main(){double a,b,h,s,x,y;int n,k;scanf("%lf%lf%d",&a,&b,&n);h=(b-a)/n;x=a;y=fun(x);for(k=1;k<=n;k++){s=s+fun(x);x=x+h;s=s+fun(x);}s=(h/2)*s;printf("s=%lf\n",s);}double fun(double m){double n;n=exp(-m)*sin(4*m)+1;return(n);}运行结果:6、复化辛甫生算法设 ,用复化辛甫生法求积分的近似值#include<>#include<>double fun(double);void main(){double a,b,h,s,x,y;int n,k;scanf("%lf%lf%d",&a,&b,&n); h=(b-a)/n;s=0;x=a;y=fun(x);for(k=1;k<=n;k++){s=s+fun(x);x=x+h/2;s=s+4*fun(x);x=x+h/2;s=s+fun(x);}s=(h/6)*s;printf("s=%lf\n",s);}double fun(double m) {double n;n=exp(-m)*sin(4*m)+1;return(n);}运行结果:7、二阶龙格库塔方法求解初值问题:取h=程序:#include<>#include<>float fun(float,float); void main(){float h,x0,y0,x1,y1,k1,k2;int n,N;scanf("%f%f%f%d",&x0,&y0,&h,&N); n=1;for(n=1;n<=N;n++){x1=x0+h;k1=fun(x0,y0);k2=fun(x0+h/2,y0+h/2*k1);y1=y0+h*k2;printf("%f,%f\n",x1,y1);x0=x1;y0=y1;}}float fun(float a,float b) {float m;m=b-2*a/b;return(m);}运行结果:8、四阶龙格库塔方法求解初值问题:取h=程序:#include<>#include<>float fun(float,float);void main(){float h,x0,y0,x1,y1,k1,k2,k3,k4; int n,N;scanf("%f%f%f%d",&x0,&y0,&h,&N); n=1;for(n=1;n<=N;n++){x1=x0+h;k1=fun(x0,y0);k2=fun(x0+h/2,y0+h/2*k1);k3=fun(x0+h/2,y0+h/2*k2);k4=fun(x1,y0+h*k3);y1=y0+h/6*(k1+2*k2+2*k3+k4); printf("%f,%f",x1,y1);x0=x1;y0=y1;}}float fun(float a,float b) {float m;m=b-2*a/b;return(m);}运行结果:9、改进的欧拉方法求解初值问题:程序:#include<>#include<>float fun(float,float);void main(){float x0,y0,h,x1,y1,yp,yc;int n,N;scanf("%f%f%f%d",&x0,&y0,&h,&N); for(n=1;n<=N;n++){x1=x0+h;yp=y0+h*fun(x0,y0);yc=y0+h*fun(x1,yp);y1=(yp+yc)/2;printf("%f,%f",x1,y1);x0=x1;y0=y1;}}float fun(float a,float b){float m;m=b-2*a/b;return(m);}运行结果:10、迭代法P131 例2 用迭代法求方程在附近的一个根,要求精度为程序:#include<>#include<>float fun(float);void main(){float x0,x1,c;int k,N;scanf("%f%f%d",&x0,&c,&N); for(k=1;k<=N;k++){x1=fun(x0);printf("%f\n",x1);if(fabs(x1-x0)<c) break; x0=x1;}if(k-1==N)printf("Failure!\n");}float fun(float m){float n;n=exp(-m);return(n);}运行结果:11、埃特金加速方法:程序:#include<>#include<>float fun(float);void main(){float x0,x1,x2,c;int k,N;scanf("%f%f%d",&x0,&c,&N);for(k=1;k<=N;k++){x1=fun(x0);x2=fun(x1);x2=x2-(x2-x1)*(x2-x1)/(x2-2*x1+x0); if(fabs(x2-x0)<c){printf("%f\n",x2);break;}x0=x2;}if(k-1==N)printf("Failure!\n"); }float fun(float m) {float n;n=exp(-m);return(n);}运行结果:12、牛顿迭代法:例5、用牛顿法解方程牛顿公式为:,取x=程序:#include<>#include<>float fun(float); float ff(float);void main(){float x0,x1,c;int k,N;scanf("%f%f%d",&x0,&c,&N); for(k=1;k<=N;k++){if(ff(x0)!=0){x1=x0-fun(x0)/ff(x0);printf("%f\n",x1);if(fabs(x1-x0)<c) break;x0=x1; }elseprintf("****"); }if(k==N)printf("Failure!\n");}float fun(float m){float n;n=m-exp(-m);return(n);}float ff(float m){float n;n=1+m;return(n);}运行结果:13、追赶法:.用追赶法求解下列方程组:程序:#include<>#include<>void main(){float a[100],b[100],c[100],d[100],t; int i,n;scanf("%d",&n);for(i=2;i<=n;i++)scanf("%f",&a[i]);for(i=1;i<=n;i++)scanf("%f",&b[i]);for(i=1;i<=n-1;i++)scanf("%f",&c[i]);for(i=1;i<=n;i++)scanf("%f",&d[i]);c[1]=c[1]/b[1];d[1]=d[1]/b[1];for(i=2;i<n;i++){t=b[i]-c[i-1]*a[i];c[i]=c[i]/t;d[i]=(d[i]-d[i-1]*a[i])/t;}d[n]=(d[n]-d[n-1]*a[n])/(b[n]-c[n-1]*a[n]); printf("%f\n",d[n]);for(i=n-1;i>=1;i--){d[i]=d[i]-c[i]*d[i+1];printf("%f\n",d[i]);}}运行结果:14、雅克比迭代程序:#include<>#include<>#define N 50#define M 4void main(){double x[M],y[M],a[M][M],b[M],d[M],c,t; double ff(double [],int);int k,i,j,n;n=M-1;scanf("%lf",&c);for(i=0;i<=n;i++){scanf("%lf",&x[i]);scanf("%lf",&b[i]);}for(i=0;i<=n;i++)for(j=0;j<=n;j++)scanf("%lf",&a[0][i*M+j]);for(k=1;k<=N;k++){for(i=1;i<=n;i++){for(j=1,t=0;j<=n;j++){if(j==i)continue;elset=t+a[i][j]*x[j];}y[i]=(b[i]-t)/a[i][i];}for(i=1;i<=n;i++)d[i]=fabs(x[i]-y[i]);t=ff(d,n+1);if(t<c) break;elsefor(i=1;i<=n;i++)x[i]=y[i];}if(k==N)printf("Failure!\n");if(k<N){printf("k=%d\n",k);for(i=1;i<=n;i++)printf("y[i]=%f\n",y[i]);} }double ff(double a[],int n){double p;int t;p=a[1];for(t=2;t<n;t++){if(p<a[t])p=a[t];}return(p);}运行结果:15、蛋白质设计:程序:#include <>#include <>#include <>void main(){printf("横坐标纵坐标竖坐标\n");FILE *fp;char c[100],x[3000][7],y[3000][7],z[3000][7];float a[3000],b[3000],d[3000],r[3000];int i,k=0,j,n=0,m,p;float X[3000],Y[3000],Z[3000],s1=0,s2=0,s3=0,av_x,av_y,av_z; fp=fopen("1a1c.pdb","r");for(i=0;i<=10000;i++){fgets(c,81,fp);if(c[0]=='A'&&c[1]=='T'&&c[2]=='O'&&c[3]=='M'){for(j=0;j<6;j++){x[k][j]=c[32+j];printf("%c",x[k][j]);}printf(" ");for(j=0;j<6;j++){y[k][j]=c[40+j];printf("%c",y[k][j]);}printf(" ");for(j=0;j<6;j++){z[k][j]=c[48+j]; printf("%c",z[k][j]); }printf(" ");X[k]=atof(x[k]);s1+=X[k];Y[k]=atof(y[k]);s2+=Y[k];Z[k]=atof(z[k]);s3+=Z[k];printf("\n");k++;if(c[77]=='C') n++; }}av_x=s1/k;av_y=s2/k;av_z=s3/k;printf("共有:%d\n",k+1);printf("av_x=%f,av_y=%f,av_z=%f\n",av_x,av_y,av_z);printf("C原子个数为:%d\n",n);printf("平移原点后的坐标:\n");for(m=0;m<k;m++){a[m]=X[m]-av_x;printf("%f ",a[m]);b[m]=Y[m]-av_y;printf("%f ",b[m]);d[m]=Z[m]-av_z;printf("%f ",d[m]);r[m]=sqrt(a[m]*a[m]+b[m]*b[m]+d[m]*d[m]);printf("%f ",r[m]) ;printf("\n");}}17高斯消去法:#include ""#include""main(){double a[3][3]={1,1,1,0,4,-1,2,-2,1},b[3]={6,5,1},x[10]={0};int i,j,k,n=3;for(k=0;k<n-1;k++){ for(i=k+1;i<n;i++){ for(j=k+1;j<n;j++){ a[i][j]=a[i][j]-a[k][j]*a[i][k]/a[k][k];}b[i]=b[i]-b[k]*a[i][k]/a[k][k];}}x[n-1]=b[n-1]/a[n-1][n-1];for(i=2;i<=n;i++){ k=n-i;for(j=k+1;j<n;j++){ x[k]+=a[k][j]*x[j];}x[k]=(b[k]-x[k])/a[k][k];}for(k=0;k<n;k++)printf("x[%d]=%f",k,x[k]);}。
C数值算法程序代码大全
1.4.1.4 扩散的物理
(1.4.1.1-7)式实际上描述了热扩散的运动,因此随机行走和扩散之间有着 必然的联系。扩散是由于粒子浓度梯度的存在 ∇ρ 形成粒子往低密度区域迁移的 趋势,单位时间内通过某一方向垂直截面的粒子数即为粒子流密度,J = −D∇ρ , 由 粒 子 数 守 恒 的 Liouvill 连 续 性 方 程 ∂ρ ∂t + ∇ ⋅ J = 0 可 得 扩 散 方 程 ,
第一章 Monte Carlo 方法基础
§1.4 随机行走与生长问题
§1.4随机行走与生长问题
1.4.1 随机行走
将随机性引入到物理实在的模型中的想法可以追溯到古希腊时代,Epicurean (伊壁鸠鲁)就认为原子的无规运动源于单独的原子没有什么理由就可以偏离直 线而游荡,这种观点理所当然地受到其他持有决定论或因果性观念的学者的反 对。按现代的观点,我们对物理世界随机性的认识是这样的:在建立物理实在的 模型时,可以暂时只考虑一个小的物理子系统,但这个子系统的边界可以受系统 外部的力、场和碰撞的影响。为了避免在描述中包含大的物理系统和多自由度, 我们用一个适当选择的“偶然”力、场和碰撞等物理量来代替外部自由度对子系 统施加的影响,这就是物理中引入随机性思想的基础。
扩散是由于粒子浓度梯度的存在形成粒子往低密度区域迁移的趋势单位时间内通过某一方向垂直截面的粒子数即为粒子流密度由粒子数守恒的liouvill连续性方程dx是粒子在t时刻存在于x14141第一章montecarlo方法基础14随机行走与生长问题127图14142一维扩散时粒子的数密度分布峰随行走时间的展宽注意纵轴的尺度是不一样的
2πσ 2
式中的方差表示峰宽,
σ = x2 − x 2 = 2Dt 。 (1.4.1.4-7)
C语言求解一元二次方程的解源码(迭代法,二分法,牛顿迭代法)
#include <stdio.h>#include <stdlib.h>#include <math.h>#define maxint 32767.0#define minint -32768.0#define accuracy 0.0000001//精确度,值越小计算结果越精确float a,b,c;//系数float dt;//b^2-4acfloat x1=0.0,x2=0.0;//方程的解void read();void setDt();int assertX();void binarySolution();void interation();void newtonInteration();double f(double x);double f1(double x);double absolute(double x);void accurate();int main(void){int end=1;while(end!=0)//继续运算{accurate();printf("按任意键继续(输入0退出):\n");scanf("%d",&end);}}//读取a,b,cvoid read(){printf("请输入方程ax^2+bx+c=0的系数a,b,c:\n");printf("请输入二次项系数a:");while(0==scanf("%f",&a)||a==0){while('\n' != getchar()){}printf("输入无效!请重新输入二次项系数a:");}printf("请输入一次项系数b:");while(0==scanf("%f",&b)){while('\n' != getchar()){}printf("输入无效!请重新输入一次项系数b:");}printf("请输入常数项c:");while(0==scanf("%f",&c)){while('\n' != getchar()){}printf("输入无效!请重新输入常数项c:");}}//计算dtvoid setDt(){dt=b*b-4*a*c;}//判断是否有解int assertX(){if(dt>=0) return 1;return 0;}//循环计算控制void accurate(){read();setDt();int method=0;printf("请选择求解方法:\n\t1.二分法\n\t2.迭代法\n\t3.牛顿迭代法\n请选择:");while((0==scanf("%d",&method))||(method!=1&&method!=2&&method!=3)){while('\n' != getchar()){}printf("输入无效!请重新选择:");}if(!assertX()){printf("该方程无解!\n");}else{switch(method){case 1:binarySolution();break;case 2:interation();break;case 3:newtonInteration();break;}printf("方程%fx^2+%fx+%f=0的解为:x1=%.10f x2=%.10f\n",a,b,c,x1,x2); }}//二分法void binarySolution(){double min=minint,temp=(-1.0*b)/(2*a),max=maxint,middle=0.0;//求解X1while((max-temp)>=accuracy){middle=(max+temp)/2;if(a>0)//开口向上{if(f(middle)>0)max=middle;elsetemp=middle;}else//开口向下{if(f(middle)>0)temp=middle;elsemax=middle;}}x2=temp;//求解X2temp=(-1.0*b)/(2*a);while((temp-min)>=accuracy){middle=(min+temp)/2;if(a>0)//开口向上{if(f(middle)>0)min=middle;elsetemp=middle;}else//开口向下{if(f(middle)>0)temp=middle;elsemin=middle;}}x1=temp;}//迭代法void interation(){//求解X1,在曲线对称轴处选择初始点double index=(-1.0*b)/(2*a),temp;if(b!=0)//b不等于0时进行迭代{temp=index;index=-1.0*(a*temp*temp+c)/b;while((absolute(index-temp))>accuracy) {temp=index;index=-1.0*(a*temp*temp+c)/b;}x1=index;x2=(-1.0*b)/a-x1;}else//b=0时ax^2+c=0直接求解{x1=sqrt(-1.0*c/a);x2=-x1;}}//牛顿迭代法void newtonInteration(){//求解X1,在曲线对称轴右侧选取初始点double index=(-1.0*b)/(2*a)+10,temp; temp=index;index=temp-f(temp)/f1(temp);while((absolute(index-temp))>accuracy) {temp=index;index=temp-f(temp)/f1(temp);}x1=index;//求解X2,在曲线对称轴左侧选取初始点index=(-1.0*b)/(2*a)-10,temp;temp=index;index=temp-f(temp)/f1(temp);;while((absolute(index-temp))>accuracy) {temp=index;index=temp-f(temp)/f1(temp);}x2=index;}//函数f(x)double f(double x){return a*x*x+b*x+c;}//函数f(x)的一次导函数double f1(double x){return 2.0*a*x+b;}//求解绝对值double absolute(double x) {if(x<=0) return (-1.0*x); return x*1.0;}。
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Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).as a distribution can be.Almost always,the cause of too good a chi-square fit is that the experimenter,in a “fit”of conservativism,has overestimated his or her measurement errors.Very rarely,too good a chi-square signals actual fraud,data that has been “fudged”to fit the model.A rule of thumb is that a “typical”value of χ2for a “moderately”good fit isχ2≈ν.More precise is the statement that the χ2statistic has a mean νand a standard deviation √2ν,and,asymptotically for large ν,becomes normally distributed.In some cases the uncertainties associated with a set of measurements are not known in advance,and considerations related to χ2fitting are used to derive a value for σ.If we assume that all measurements have the same standard deviation,σi =σ,and that the model does fit well,then we can proceed by first assigning an arbitrary constant σto all points,next fitting for the model parameters by minimizing χ2,and finally recomputingσ2=N i =1[y i −y (x i )]2/(N −M )(15.1.6)Obviously,this approach prohibits an independent assessment of goodness-of-fit,a fact occasionally missed by its adherents.When,however,the measurement error is not known,this approach at least allows some kind of error bar to be assigned to the points.If we take the derivative of equation (15.1.5)with respect to the parameters a k ,we obtain equations that must hold at the chi-square minimum,0=N i =1y i −y (x i )σ2i∂y (x i ;...a k ...)∂a kk =1,...,M (15.1.7)Equation (15.1.7)is,in general,a set of M nonlinear equations for the M unknowna k .Various of the procedures described subsequently in this chapter derive from (15.1.7)and its specializations.CITED REFERENCES AND FURTHER READING:Bevington,P .R.1969,Data Reduction and Error Analysis for the Physical Sciences (New York:McGraw-Hill),Chapters 1–4.von Mises,R.1964,Mathematical Theory of Probability and Statistics (New York:AcademicPress),§VI.C.[1]15.2Fitting Data to a Straight LineA concrete example will make the considerations of the previous section more meaningful.We consider the problem of fitting a set of N data points (x i ,y i )to a straight-line modely (x )=y (x ;a,b )=a +bx(15.2.1)This problem is often called linear regression,a terminology that originated,long ago,in the social sciences.We assume that the uncertaintyσi associated with each measurement y i is known,and that the x i’s(values of the dependent variable) are known exactly.To measure how well the model agrees with the data,we use the chi-square merit function(15.1.5),which in this case isχ2(a,b)=Ni=1y i−a−bx iσi2(15.2.2)If the measurement errors are normally distributed,then this merit function will give maximum likelihood parameter estimations of a and b;if the errors are not normally distributed,then the estimations are not maximum likelihood,but may still be useful in a practical sense.In§15.7,we will treat the case where outlier points are so numerous as to render theχ2merit function useless.Equation(15.2.2)is minimized to determine a and b.At its minimum, derivatives ofχ2(a,b)with respect to a,b vanish.0=∂χ2∂a=−2Ni=1y i−a−bx iσ2i0=∂χ2∂b=−2Ni=1x i(y i−a−bx i)σ2i(15.2.3)These conditions can be rewritten in a convenient form if we define the following sums:S≡Ni=11σ2iS x≡Ni=1x iσ2iS y≡Ni=1y iσ2iS xx≡Ni=1x2iσ2iS xy≡Ni=1x i y iσ2i(15.2.4)With these definitions(15.2.3)becomesaS+bS x=S yaS x+bS xx=S xy(15.2.5)The solution of these two equations in two unknowns is calculated as∆≡SS xx−(S x)2a=S xx S y−S x S xy∆b=SS xy−S x S y∆(15.2.6)Equation(15.2.6)gives the solution for the best-fit model parameters a and b.Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).We are not done,however.We must estimate the probable uncertainties in the estimates of a and b,since obviously the measurement errors in the data must introduce some uncertainty in the determination of those parameters.If the data are independent,then each contributes its own bit of uncertainty to the parameters.Consideration of propagation of errors shows that the varianceσ2f in the value ofany function will beσ2f=Ni=1σ2i∂f∂y i2(15.2.7)For the straight line,the derivatives of a and b with respect to y i can be directly evaluated from the solution:∂a ∂y i =S xx−S x x iσ2i∆∂b ∂y i =Sx i−S xσ2i∆(15.2.8)Summing over the points as in(15.2.7),we getσ2a=S xx/∆σ2b=S/∆(15.2.9)which are the variances in the estimates of a and b,respectively.We will see in §15.6that an additional number is also needed to characterize properly the probable uncertainty of the parameter estimation.That number is the covariance of a and b, and(as we will see below)is given byCov(a,b)=−S x/∆(15.2.10)The coefficient of correlation between the uncertainty in a and the uncertainty in b,which is a number between−1and1,follows from(15.2.10)(compare equation14.5.1),r ab=−S x√SS xx(15.2.11)A positive value of r ab indicates that the errors in a and b are likely to have the same sign,while a negative value indicates the errors are anticorrelated,likely to have opposite signs.We are still not done.We must estimate the goodness-of-fit of the data to the model.Absent this estimate,we have not the slightest indication that the parameters a and b in the model have any meaning at all!The probability Q that a value of chi-square as poor as the value(15.2.2)should occur by chance isQ=gammqN−22,χ22(15.2.12)Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).Here gammq is our routine for the incomplete gamma function Q (a,x ),§6.2.If Q is larger than,say,0.1,then the goodness-of-fit is believable.If it is larger than,say,0.001,then the fit may be acceptable if the errors are nonnormal or have been moderately underestimated.If Q is less than 0.001then the model and/or estimation procedure can rightly be called into question.In this latter case,turn to §15.7to proceed further.If you do not know the individual measurement errors of the points σi ,and are proceeding (dangerously)to use equation (15.1.6)for estimating these errors,then here is the procedure for estimating the probable uncertainties of the parameters a and b :Set σi ≡1in all equations through (15.2.6),and multiply σa and σb ,as obtained from equation (15.2.9),by the additional factor χ2/(N −2),where χ2is computed by (15.2.2)using the fitted parameters a and b .As discussed above,this procedure is equivalent to assuming a good fit,so you get no independent goodness-of-fit probability Q .In §14.5we promised a relation between the linear correlation coefficient r (equation 14.5.1)and a goodness-of-fit measure,χ2(equation 15.2.2).For unweighted data (all σi =1),that relation isχ2=(1−r 2)NVar (y 1...y N )(15.2.13)whereNVar (y 1...y N )≡N i =1(y i −y )2(15.2.14)For data with varying weights σi ,the above equations remain valid if the sums inequation (14.5.1)are weighted by 1/σ2i .The following function,fit ,carries out exactly the operations that we have discussed.When the weights σare known in advance,the calculations exactly correspond to the formulas above.However,when weights σare unavailable,the routine assumes equal values of σfor each point and assumes a good fit,as discussed in §15.1.The formulas (15.2.6)are susceptible to roundoff error.Accordingly,we rewrite them as follows:Definet i =1σix i −S xS,i =1,2,...,N (15.2.15)andS tt =N i =1t 2i(15.2.16)Then,as you can verify by direct substitution,b =1S tt Ni =1t i y iσi(15.2.17)a =S y −S x bS(15.2.18)σ2a=1S1+S2xSS tt(15.2.19)σ2b=1S tt(15.2.20)Cov(a,b)=−S xSS tt(15.2.21)r ab=Cov(a,b)σaσb(15.2.22)#include<math.h>#include"nrutil.h"void fit(float x[],float y[],int ndata,float sig[],int mwt,float*a, float*b,float*siga,float*sigb,float*chi2,float*q)Given a set of data points x[1..ndata],y[1..ndata]with individual standard deviations sig[1..ndata],fit them to a straight line y=a+bx by minimizingχ2.Returned are a,b and their respective probable uncertainties siga and sigb,the chi-square chi2,and the goodness-of-fit probability q(that thefit would haveχ2this large or larger).If mwt=0on input,then the standard deviations are assumed to be unavailable:q is returned as1.0and the normalization of chi2is to unit standard deviation on all points.{float gammq(float a,float x);int i;float wt,t,sxoss,sx=0.0,sy=0.0,st2=0.0,ss,sigdat;*b=0.0;if(mwt){Accumulate sums...ss=0.0;for(i=1;i<=ndata;i++){...with weightswt=1.0/SQR(sig[i]);ss+=wt;sx+=x[i]*wt;sy+=y[i]*wt;}}else{for(i=1;i<=ndata;i++){...or without weights.sx+=x[i];sy+=y[i];}ss=ndata;}sxoss=sx/ss;if(mwt){for(i=1;i<=ndata;i++){t=(x[i]-sxoss)/sig[i];st2+=t*t;*b+=t*y[i]/sig[i];}}else{for(i=1;i<=ndata;i++){t=x[i]-sxoss;st2+=t*t;*b+=t*y[i];}}*b/=st2;Solve for a,b,σa,andσb.*a=(sy-sx*(*b))/ss;*siga=sqrt((1.0+sx*sx/(ss*st2))/ss);*sigb=sqrt(1.0/st2);Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5)Copyright (C) 1988-1992 by Cambridge University Press. P rograms Copyright (C) 1988-1992 by Numerical Recipes Software. Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copying of machine-readable files (including this one) to any server c omputer, is strictly prohibited. To order Numerical Recipes books, d iskettes, or CDROMs visit website or call 1-800-872-7423 (North America only), o r send email to trade@ (outside North America).*chi2=0.0;Calculate χ2.*q=1.0;if (mwt ==0){for (i=1;i<=ndata;i++)*chi2+=SQR(y[i]-(*a)-(*b)*x[i]);sigdat=sqrt((*chi2)/(ndata-2));For unweighted data evaluate typ-ical sig using chi2,and ad-just the standard deviations.*siga *=sigdat;*sigb *=sigdat;}else {for (i=1;i<=ndata;i++)*chi2+=SQR((y[i]-(*a)-(*b)*x[i])/sig[i]);if (ndata>2)*q=gammq(0.5*(ndata-2),0.5*(*chi2));Equation (15.2.12).}}CITED REFERENCES AND FURTHER READING:Bevington,P .R.1969,Data Reduction and Error Analysis for the Physical Sciences (New York:McGraw-Hill),Chapter 6.15.3Straight-Line Data with Errors in BothCoordinatesIf experimental data are subject to measurement error not only in the y i ’s,but also in the x i ’s,then the task of fitting a straight-line modely (x )=a +bx(15.3.1)is considerably harder.It is straightforward to write down the χ2merit function for this case,χ2(a,b )=N i =1(y i −a −bx i )2σ2y i +b 2σ2x i(15.3.2)where σx i and σy i are,respectively,the x and y standard deviations for the i th point.Theweighted sum of variances in the denominator of equation (15.3.2)can be understood both as the variance in the direction of the smallest χ2between each data point and the line with slope b ,and also as the variance of the linear combination y i −a −bx i of two random variables x i and y i ,Var (y i −a −bx i )=Var (y i )+b 2Var (x i )=σ2y i +b 2σ2x i ≡1/w i(15.3.3)The sum of the square of N random variables,each normalized by its variance,is thusχ2-distributed.We want to minimize equation (15.3.2)with respect to a and b .Unfortunately,the occurrence of b in the denominator of equation (15.3.2)makes the resulting equation for the slope ∂χ2/∂b =0nonlinear.However,the corresponding condition for the intercept,∂χ2/∂a =0,is still linear and yieldsa = iw i (y i −bx i ) iw i (15.3.4)where the w i ’s are defined by equation (15.3.3).A reasonable strategy,now,is to use themachinery of Chapter 10(e.g.,the routine brent )for minimizing a general one-dimensional function to minimize with respect to b ,while using equation (15.3.4)at each stage to ensure that the minimum with respect to b is also minimized with respect to a .。