杭州电子科技大学OJ题目分类

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完整word版杭电ACM部分题目答案

杭电ACM 1001 Sum Problem1089 1090 10911092 1093 1094A+B for Input-Output Practice (I) .A+B for Input-Output Practice(II) ... A+B for Input-OutputPractice (III)A+B for Input-Output Practice (IV) ...A+B for Input-Output Practice (V)A+B for Input-Output Practice (VI)A+B for Input-Output Practice (VII)A+B for Input-Output Practice (VIII)246891113109510962000 ASCII 码排序................ 2001 计算两点间的距离........... 2002 计算球体积................. 2003 求绝对值................... 2004 成绩转换................... 2005 第几天？................... 2006 求奇数的乘积............... 2007 平方和与立方和............. 2008 数值统计................... 2009 求数列的和................. 2010 水仙花数................... 2011 多项式求和................. 2012 素数判定................... 2014 青年歌手大奖赛_评委会打分2015 偶数求和................... 2016 数据的交换输出............. 2017 字符串统计................. 2019 数列有序! ................... 2020 绝对值排序................. 2021 发工资咯：）............... 2033 人见人爱A+B ................. 2039 三角形..................... 2040 亲和数 (14)1617 19202122242627283031 3334 36 384042434546 48 5051姓名：电商 1001学号：10105041341001 Sum P roblemP roblem Descri pti onHey, welcome to HDOJ(Ha ngzhou Dianzi Uni versity On li ne Judge).In this p roblem, your task is to calculate SUM( n) = 1 + 2 + 3 + ... + n.InputThe input will con sist of a series of in tegers n, one in teger per line.Out putFor each case, out put SUM( n) in one line, followed by a bla nk line. You may assume the result will be in the range of 32-bit sig ned in teger.Sample Input1 100Sam ple Out put15050AuthorDOOM III解答:#i ncludevstdio.h>printf }}main () {int sum while { sumforn , i , sum; =0；((scanf ("%d"，& n)!二 1))=0;(i =0; i v=n; i ++)sum +=i ；("%d\n\n" , sum);1089A+B for Input-Output Practice(I)Problem DescriptionYour task is to Calculate a + b.Too easy?! Of course! I specially designed the problem for acm beginners.You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim.InputThe input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input1 510 20Sample Output630AuthorlcyRecommendJGShininga , b;(scanf ("%d%d"，& a,& b)!= EOF) ("%d\n" , a+b);解答：#i ncludevstdio.h> main () {intwhile printf}1090A+B for Input-Output Practice(II)Problem DescriptionYour task is to Calculate a + b.InputInput contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input2 1 510 20Sample Output630AuthorlcyRecommendJGShining解答： #i ncludevstdio.h> #defi ne M 1000 void mai n () { int a , b, n, j [ M], i ; //prin tf(" please input n:\n"); scanf ("%d"，& n);for (i =0; i vn; i ++) { scanf ("%d%d"，& a,& b); //prin tf("%d %d",a,b); j[i ]= a+b;} iwhile {=0；(i <n)printf i printf++；("%d" , j [ i ]);("\n");1091A+B for Input-Output Practice(III)Problem DescriptionYour task is to Calculate a + b.InputInput contains multiple test cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input1 510 200 0Sample Output6 30AuthorlcyRecommendJGShining解答：#i ncludevstdio.h> main () {int scanf while { printf scanf } }1092 A+B for Input-Output Practice(IV)P roblem Descri pti onYour task is to Calculate the sum of some in tegers.InputInput contains mult iple test cases. Each test case contains a in teger N, and the n N in tegers follow in the same line. A test case start ing with 0 term in ates the input and this test case is not to be p rocessed.Out putFor each gro up of input in tegers you should out put their sum in one line, and with one line of out put for each line in input.Sample Input4 1 2 3 45 1 2 3 4 5Sam ple Out put10,b; ("%d %d"，& a,& b); (!( a==0&&b==0))("%d\n" , a+b); ("%d %d"，& a,&b);15AuthorIcyRecomme ndJGShi ning解答：#i nclude <stdio.h> int mai n{intwhile {()n , sum, i , t ;(seanf ("%d"，& n)!= EOF&&n!= 0)sum for {scanfsum }printf =0;(i =0; i vn; i ++)("%d"，& t);=sum+t ;("%d\n" , sum);1093A+B for Input-Output Practice(V)Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.OutputFor each group of input integers you should output their sum in one line, and with one line of output for each line in input.Sample Input24 1 2 3 45 1 2 3 4 5Sample Output1015Authorlcy解答：#i ncludevstdio.h> main () {intsum while {n , a, b, i , j , sum;=0;(seanf ("%d\n" ，& n)!=- 1)for{(i =0; i vn; i ++)scanf for {scanfsum} printf sum("%d"，& b);(j =0;j <b; j++)("%d"，& a);+=a;("%d\n" , sum);=0；1094A+B for Input-Output Practice(VI)P roblem Descri pti onYour task is to calculate the sum of some in tegers.InputInput contains mult iple test cases, and one case one line. Each case starts with an in teger N, and the n N in tegers follow in the same line.Out putFor each test case you should out put the sum of N in tegers in one line, and with one line of out put for each line in input.Sample Input4 1 2 3 45 1 2 3 4 5Sam ple Out put1015AuthorIcyRecomme ndJGShi ning解答：#i ncludevstdio.h> main () {intsum while {n , a, b, i , j , sum;=0；(seanf ("%d\n" ，& n)!=- 1)for{ scanf sum} printfsum(j =0；j <n； j++)("%d"，& a);+=a;("%d\n" , sum);=0；[Copy to Clip board ] [ Save to File]1095A+B for Inp ut-Out put P ractice(VII)P roblem Descri pti onYour task is to Calculate a + b.InputThe input will con sist of a series of p airs of in tegers a and b, sep arated by a sp ace, one p air of in tegers per line.Out putFor each p air of input in tegers a and b you should out put the sum of a and b, and followed by a bla nk line.Sample Input1 510 20Sam ple Out put630AuthorIcyRecomme ndJGShi ning156a , b; (scanf ("%d%d"，& a,& b)!= EOF) ("%d\n\n" , a+b); A+B for Inp ut-Out put P ractice (VIII)P roblem Descri pti onYour task is to calculate the sum of some in tegers. InputInput contains an in teger N in the first line, and the n N lines follow. Each line starts with a in teger M, and the n M in tegers follow in the same line.Out putFor each group of input in tegers you should out put their sum in one line, and you must note that there is a bla nk line betwee n out pu ts.Sample Input3Sam pie Out put10解答：#i ncludevstdio.h>main (){intwhileprintf} 1096AuthorIcy Recomme nd JGShi ning解答： int mai n { intscanf ()getchar for l for { sca nf getchar for { a , b, i , j , l [ 1000], k; ("%d"，& i );(); (j =1; j <=i ; j ++)[j ]= 0; (j =1; j <=i ; j ++) ("%d"，& a); ()； (k=1; k<=a; k++)scanf ("%d"，& b); getchar (); l [j ]+= b; } for printf printf (j =1; j <=i-1; j ++) ("%d\n\n" , l [ j ]);("%d\n" , l [ i ]); 2000 ASCII 码排序P roblem Descri pti on 输入三个字符后，按各字符的 ASCII 码从小到大的顺序输出这三个字符。

杭电ACM试题详细分类,杭电oj详细分类,hdu详细分类,详细,ACM

数论
1164 1211 1215 1222 1286 1299
计算几何
1086 1115 1147
贪心
1009 1052 1055 1232 1272
线段树,离散化
1199 1255
图论最短路相关的问题
1142 1162 1217 1301
二分图问题
1054 1068 1150 1151 1281
杭电ACM试题分类枚举
1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201 1202 1205 1209 1212(大数取模) 1216（链表）1218 1219 1225 1228 1229 1230 1234 1235 1236 1237 1239 1250 1256 1259 1262 1263 1265 1266 1276 1279 1282 1283 1287 1296 1302 1303 1304 1305 1306 1309 1311 1314
2549、2550、2551、2552、2555、2560、2561、2562、2566、2567、 2568、2700、2710、
DP：1003、10240、1029、1069、1074、1087、1114、1159、1160、
1171、1176、1203、1231、1257、1260、1284、1421、1789、1978、 2059、2084、2159、2191、2544、2571、2602、2709、

杭电acm部分题目及答案答案

自己刷的题这是我在杭电做题的记录，希望我的分享对你有帮助！！！1001 Sum Problem***********************************************************1 1089 A+B for Input-Output Practice (I)********************************21090 A+B for Input-Output Practice (II)********************************51091A+B for Input-Output Practice (III)****************************************7 1092A+B for Input-Output Practice (IV)********************************81093 A+B for Input-Output Practice (V)********************************101094 A+B for Input-Output Practice (VI)***************************************12 1095A+B for Input-Output Practice (VII)*******************************131096 A+B for Input-Output Practice (VIII)******************************15How to Type***************************************************************161001 Sum ProblemProblem DescriptionHey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.InputThe input will consist of a series of integers n, one integer per line.OutputFor each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.Sample Input1100Sample Output15050AuthorDOOM III解答：#include<stdio.h>main(){int n,i,sum;sum=0;while((scanf("%d",&n)!=-1)){sum=0;for(i=0;i<=n;i++)sum+=i;printf("%d\n\n",sum);}}1089 A+B for Input-Output Practice(I)Problem DescriptionYour task is to Calculate a + b.Too easy?! Of course! I specially designed the problem for acm beginners.You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim.InputThe input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input1 510 20Sample Output630AuthorlcyRecommendJGShining解答：#include<stdio.h>main(){int a,b;while(scanf("%d%d",&a,&b)!=EOF)printf("%d\n",a+b);}1090 A+B for Input-Output Practice(II)Problem DescriptionYour task is to Calculate a + b.InputInput contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input21 510 20Sample Output630AuthorlcyRecommendJGShining解答：#include<stdio.h>#define M 1000void main(){int a ,b,n,j[M],i;//printf("please input n:\n"); scanf("%d",&n);for(i=0;i<n;i++){scanf("%d%d",&a,&b);//printf("%d %d",a,b);j[i]=a+b;}i=0;while(i<n){printf("%d",j[i]);i++;printf("\n");}}1091A+B for Input-Output Practice(III)Problem DescriptionYour task is to Calculate a + b.InputInput contains multiple test cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.OutputFor each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.Sample Input1 510 200 0Sample Output630AuthorlcyRecommendJGShining解答：#include<stdio.h>main(){int a,b;scanf("%d %d",&a,&b);while(!(a==0&&b==0)){printf("%d\n",a+b);scanf("%d %d",&a,&b);}}1092A+B for Input-Output Practice(IV)Problem DescriptionYour task is to Calculate the sum of some integers.InputInput contains multiple test cases. Each test case contains a integer N, and then N integers followin the same line. A test case starting with 0 terminates the input and this test case is not to be processed.OutputFor each group of input integers you should output their sum in one line, and with one line of output for each line in input.Sample Input4 1 2 3 45 1 2 3 4 5Sample Output1015AuthorlcyRecommendJGShining解答：#include <stdio.h>int main(){int n,sum,i,t;while(scanf("%d",&n)!=EOF&&n!=0) {sum=0;for(i=0;i<n;i++){scanf("%d",&t);sum=sum+t;}printf("%d\n",sum);}}1093 A+B for Input-Output Practice(V)Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.OutputFor each group of input integers you should output their sum in one line, and with one line of output for each line in input.Sample Input24 1 2 3 45 1 2 3 4 5Sample Output1015Authorlcy解答：#include<stdio.h>main(){int n,a,b,i,j,sum;sum=0;while(scanf("%d\n",&n)!=-1) {for(i=0;i<n;i++){scanf("%d",&b);for(j=0;j<b;j++){scanf("%d",&a);sum+=a;}printf("%d\n",sum); sum=0;}}}1094 A+B for Input-Output Practice (VI) Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains multiple test cases, and one case one line. Each case starts with an integer N, and then N integers follow in the same line.OutputFor each test case you should output the sum of N integers in one line, and with one line of output for each line in input.Sample Input4 1 2 3 45 1 2 3 4 5Sample Output1015AuthorlcyRecommendJGShining解答：#include<stdio.h>main(){int n,a,b,i,j,sum;sum=0;while(scanf("%d\n",&n)!=-1){for(j=0;j<n;j++){scanf("%d",&a);sum+=a;}printf("%d\n",sum);sum=0;}}[ Copy to Clipboard ][ Save to File]1095A+B for Input-Output Practice (VII) Problem DescriptionYour task is to Calculate a + b.InputThe input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line. OutputFor each pair of input integers a and b you should output the sum of a and b, and followed by a blank line.Sample Input1 510 20Sample Output630AuthorlcyRecommendJGShining解答：#include<stdio.h>main(){int a,b;while(scanf("%d%d",&a,&b)!=EOF)printf("%d\n\n",a+b);}1096 A+B for Input-Output Practice (VIII) Problem DescriptionYour task is to calculate the sum of some integers.InputInput contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.OutputFor each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.Sample Input34 1 2 3 45 1 2 3 4 53 1 2 3Sample Output10156AuthorlcyRecommendJGShining解答：int main(){int a,b,i,j,l[1000],k;scanf("%d",&i);getchar();for(j=1;j<=i;j++)l[j]=0;for(j=1;j<=i;j++){scanf("%d",&a);getchar();for(k=1;k<=a;k++){scanf("%d",&b);getchar();l[j]+=b;}}for(j=1;j<=i-1;j++)printf("%d\n\n",l[j]);printf("%d\n",l[i]);}How to TypeProblem DescriptionPirates have finished developing the typing software. He called Cathy to test his typing software. She is good at thinking. After testing for several days, she finds that if she types a string by some ways, she will type the key at least. But she has a bad habit that if the caps lock is on, she must turn off it, after she finishes typing. Now she wants to know the smallest times of typing the key to finish typing a string.InputThe first line is an integer t (t<=100), which is the number of test case in the input file. For each test case, there is only one string which consists of lowercase letter and upper case letter. The length of the string is at most 100.OutputFor each test case, you must output the smallest times of typing the key to finish typing this string.Sample Input 3Pirates HDUacm HDUACM Sample Output 888#include <stdio.h>#include <string.h>#define MAX 200int arr[MAX][4];char str[MAX];int letter(char ch){if(ch>='A'&&ch<='Z') return 1;return 0;}void proc(){int i;int tmp,min;int len=strlen(str);for(i=0;i<len;i++){if(i==0){if(letter(str[i])) { arr[i][1]=2; arr[i][2]=2; }else { arr[i][0]=1; arr[i][3]=3; }}else{if(letter(str[i])==letter(str[i-1])){if(arr[i-1][0]){ arr[i][0]=arr[i-1][0]+1;arr[i][3]=arr[i-1][0]+3;}if(arr[i-1][1]) { arr[i][1]=arr[i-1][1]+2; arr[i][2]=arr[i-1][1]+2;}if(arr[i-1][2]){if(arr[i][0]>arr[i-1][2]+1||!arr[i][0]) arr[i][0]=arr[i-1][2]+1;if(arr[i][3]>arr[i-1][2]+3||!arr[i][3]) arr[i][3]=arr[i-1][2]+3;}if(arr[i-1][3]){if(arr[i][1]>arr[i-1][3]+2||!arr[i][1]) arr[i][1]=arr[i-1][3]+2;if(arr[i][2]>arr[i-1][3]+2||!arr[i][2]) arr[i][2]=arr[i-1][3]+2;}}else{if(arr[i-1][0]){ arr[i][1]=arr[i-1][0]+2; arr[i][2]=arr[i-1][0]+2;}if(arr[i-1][1]){ arr[i][0]=arr[i-1][1]+1; arr[i][3]=arr[i-1][1]+3;}if(arr[i-1][2]){if(arr[i][1]>arr[i-1][2]+2||!arr[i][1]) arr[i][1]=arr[i-1][2]+2;if(arr[i][2]>arr[i-1][2]+2||!arr[i][2]) arr[i][2]=arr[i-1][2]+2;}if(arr[i-1][3]){if(arr[i][0]>arr[i-1][3]+1||!arr[i][0]) arr[i][0]=arr[i-1][3]+1;if(arr[i][3]>arr[i-1][3]+3||!arr[i][3]) arr[i][3]=arr[i-1][3]+3;}}}}min=3*MAX;if(letter(str[len-1])){if(arr[len-1][0]){ tmp=arr[len-1][0]+1; if(tmp<min) min=tmp;}if(arr[len-1][1]){ tmp=arr[len-1][1]; if(tmp<min) min=tmp; }if(arr[len-1][2]){ tmp=arr[len-1][2]+1; if(tmp<min) min=tmp;}if(arr[len-1][3]){ tmp=arr[len-1][3]; if(tmp<min) min=tmp; } }else{if(arr[len-1][0]) { tmp=arr[len-1][0]; if(tmp<min) min=tmp; }if(arr[len-1][1]) { tmp=arr[len-1][1]+1; if(tmp<min) min=tmp;}if(arr[len-1][2]) { tmp=arr[len-1][2]; if(tmp<min) min=tmp; }if(arr[len-1][3]) { tmp=arr[len-1][3]+1; if(tmp<min) min=tmp;} }printf("%d\n",min);}//Caps Shift:0-00;1-01;2-10;3-11int main(){int num;scanf("%d",&num);while(num--){scanf("%s",str);memset(arr,0,strlen(str)*4*sizeof(int));proc();}return 0;}。

杭州电子科技大学学生考试卷(B)卷

A.说明静态数据成员时前面要加修饰词
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.静态数据成员可以由类的所有成员函数访问
．派生类的构造函数的成员初始化列表中，不能包含。
.基类的构造函数.派生类中子对象的初始化
.基类中子对象的初始化.派生类中一般数据成员的初始化
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.友元函数只能访问该类的成员变量.友元函数只能访问该类的成员函数
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杭州电子科技大学acm习题集锦

目录1、数塔问题 (2)2、并查集类问题 (4)3、递推类问题 (9)4、动态规划系列 (10)5、概率类题型 (13)6、组合数学类题型 (15)7、贪心策略 (16)8、几何问题 (19)数塔类问题数塔Problem Description在讲述DP算法的时候，一个经典的例子就是数塔问题，它是这样描述的：有如下所示的数塔，要求从顶层走到底层，若每一步只能走到相邻的结点，则经过的结点的数字之和最大是多少？已经告诉你了，这是个DP的题目，你能AC吗?Input输入数据首先包括一个整数C,表示测试实例的个数，每个测试实例的第一行是一个整数N(1 <= N <= 100)，表示数塔的高度，接下来用N行数字表示数塔，其中第i行有个i个整数，且所有的整数均在区间[0,99]内。

Output对于每个测试实例，输出可能得到的最大和，每个实例的输出占一行。

Sample Input1573 88 1 02 7 4 44 5 2 6 5Sample Output 30#include<stdio.h>#include<string.h>#define MAX 101int arr[MAX][MAX][2];void res(){int n; int i,j;memset(arr,0,MAX*MAX*sizeof(int));scanf("%d",&n);for(i=0;i<n;i++) //输入数塔for(j=0;j<=i;j++) { scanf("%d",&arr[i][j][0]); arr[i][j][1]=arr[i][j][0]; }for(i=n-2;i>=0;i--){for(j=0;j<=i;j++){if(arr[i+1][j][1]>arr[i+1][j+1][1]) arr[i][j][1]+=arr[i+1][j][1];else arr[i][j][1]+=arr[i+1][j+1][1];}}printf("%d\n",arr[0][0][1]);}int main(){int num;scanf("%d",&num);while(num--) { res(); }return 0;}免费馅饼Problem Description都说天上不会掉馅饼，但有一天gameboy正走在回家的小径上，忽然天上掉下大把大把的馅饼。

杭电OJ：1089----1096（c++）（ACM入门第一步：所有的输入输出格式）

杭电OJ：1089----1096（c++）（ACM⼊门第⼀步：所有的输⼊输出格式）1089:输⼊输出练习的A + B（I）问题描述您的任务是计算a + b。

太容易了？！当然！我专门为ACM初学者设计了这个问题。

您⼀定已经发现某些问题与此标题具有相同的名称，是的，所有这些问题都是出于相同的⽬的⽽设计的。

输⼊项输⼊将由⼀系列由空格隔开的整数对a和b组成，每⾏⼀对整数。

输出量对于每对输⼊整数a和b，应该在⼀⾏中输出a和b的总和，并且在输⼊中每⾏输出⼀⾏。

样本输⼊1 5 10 20样本输出6 30题解：#include<cstdio>#include<iostream>using namespace std;int main(){int a, b,sum;while(cin >> a >> b){sum = a+b;cout << sum << endl;}return 0;}1090：投⼊产出练习的A + B（II）问题描述您的任务是计算a + b。

输⼊项输⼊的第⼀⾏包含⼀个整数N，然后是N⾏。

每⾏由⼀对整数a和b组成，每对之间⽤空格隔开，每⾏⼀对整数。

输出量对于每对输⼊整数a和b，应该在⼀⾏中输出a和b的总和，并且在输⼊中每⾏输出⼀⾏。

样本输⼊2 1 5 10 20样本输出6 30题解：#include<cstdio>#include<iostream>using namespace std;int a,b,n,sum;cin >> n;while (n){cin >> a >> b;sum = a + b;cout << sum << endl;n--;}return 0;}1091：投⼊产出练习的A + B（III）问题描述您的任务是计算a + b。

完整word版杭电ACM试题答案

【杭电ACM1000】A +B ProblemProblem DescriptionCalculate A + B.InputEach line will contain two integers A and B. Process to end of file.OutputFor each case, output A + B in one line.Sample Input1 1Sample Output2# include <stdio.h>int main(){int a, b;while(scanf(%d%d, &a, &b)!=EOF)printf(%d\n, a+b);return 0;}【杭电ACM1001】Sum ProblemProblem DescriptionHey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.InputThe input will consist of a series of integers n, one integer per line.OutputFor each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.Sample Input1 100Sample Output1 5050# include <stdio.h>int main(){int n, i, sum = 0;while(scanf(%d, &n)!=EOF){for(i=1; i<=n; ++i)sum = sum + i;printf(%d\n\n, sum);sum = 0;}return 0;}【杭电ACM1002】A +B Problem IIProblem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.InputThe first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integersare very large, that means you should not process them by using 32-bit integer. You may assumethe length of each integer will not exceed 1000.OutputFor each test case, you should output two lines. The first line is Case #:, # means the number ofthe test case. The second line is the an equation A + B = Sum, Sum means the result of A + B.Note there are some spaces int the equation. Output a blank line between two test cases.Sample Input2 1 2 112233445566778899 998877665544332211Sample OutputCase 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110#include<stdio.h>#include<string.h>int shu(char a){return (a-'0');}int main(){char a[1000],b[1000];int num[1001];int n,i,j=1,al,bl,k,t;scanf(%d,&n);while(n--){getchar();if(j!=1)printf(\);scanf(%s,a);al=strlen(a);scanf(%s,b);bl=strlen(b);k=(al>bl)?al:bl;for(i=0;i<=k;i++)num[i]=0;t=k;for(k;al>0&&bl>0;k--){num[k]+=shu(a[--al])+shu(b[--bl]);if(num[k]/10){num[k-1]++;num[k]%=10;}}while(al>0){num[k--]+=shu(a[--al]);if(num[k+1]/10){num[k]++;num[k+1]%=10;}}while(bl>0){num[k--]+=shu(b[--bl]);if(num[k+1]/10){num[k]++;num[k+1]%=10;}}printf(Case %d:\n,j++);printf(%s + %s = ,a,b);for(i=0;i<=t;i++){if(i==0&&num[i]==0)i++;printf(%d,num[i]);}printf(\);}return 0;}。

杭电OJ水题答案2

Hdu 2000#include<stdio.h>void main(){char a,b,c,t;while(scanf("%c%c%c",&a,&b,&c)>0){if(a>b) t=a,a=b,b=t;if(a>c) t=c,c=a,a=t;if(c>a && c<b) t=b,b=c,c=t;scanf("%*c",&a);printf("%c %c %c\n",a,b,c);}}Hdu 2001#include <stdio.h>#include <math.h>int main(){double x1;double x2;double y1;double y2;while(scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2)!=EOF){double num=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));printf("%.2lf\n",num);}return 0;}Hdu 2002#include<stdio.h>int main(){double pi=3.1415927;double r;while (scanf("%lf",&r)==1)printf("%.3lf\n",((pi*r*r*r)*4)/3);}Hdu 2003#include <stdio.h>#include <math.h>int main(){double x;while (scanf("%lf",&x)==1)printf("%.2lf\n",fabs(x));return 0;}Hdu 2004#include<stdio.h>void main(){double x;while(scanf("%lf",&x)==1)if(x>100||x<0)printf("Score is error!\n");else if(0<x&&x<60) printf("E\n");else if(60<=x&&x<70) printf("D\n");else if(70<=x&&x<80)printf("C\n");else if(80<=x&&x<90)printf("B\n");elseprintf("A\n");}Hdu 2005#include<stdio.h>void main(){int y,m,d,sum=0;while(scanf("%d/%d/%d",&y,&m,&d)!=EOF){if((y%100==0&&y%400==0)||(y%100!=0&&y%4==0)){switch(m=13-m){case 1: sum=sum+30;case 2: sum=sum+31;case 3: sum=sum+30;case 4: sum=sum+31;case 5: sum=sum+31;case 6: sum=sum+30;case 7: sum=sum+31;case 8: sum=sum+30;case 9: sum=sum+31;case 10: sum=sum+29;case 11: sum=sum+31;}}else{switch(m=13-m){case 1: sum=sum+30;case 2: sum=sum+31;case 3: sum=sum+30;case 4: sum=sum+31;case 5: sum=sum+31;case 6: sum=sum+30;case 7: sum=sum+31;case 8: sum=sum+30;case 9: sum=sum+31;case 10: sum=sum+28;case 11: sum=sum+31;}}printf("%d\n",sum+d);sum=0;}}#include<stdio.h>void main(){int n,i,a,sum=1;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++){scanf("%d",&a);if(a%2==1)sum*=a;}printf("%d\n",sum);sum=1;}}Hdu 2007#include<stdio.h>void main(){int a,b,x=0,y=0,t;while(scanf("%d%d",&a,&b)!=EOF){if(b<a) { t=a; a=b; b=t;goto m;}else{m:{ while(a<=b){if(a%2==0) { x=a*a+x;a++;}else { y=a*a*a+y; a++;}}}printf("%d %d\n",x,y);x=0; y=0;}}}#include<stdio.h>void main(){double a; int i,x=0,y=0,z=0,n;while(scanf("%d",&n)==1){if(n==0)break;else{while(n--){scanf("%lf",&a);if(a<0) {x++;}else if(a==0) {y++;}else {z++;}}}printf("%d %d %d\n",x,y,z);x=0;y=0;z=0;}}Hdu 2009#include<stdio.h>#include<math.h>void main(){double n,m;double sum=0;while (scanf("%lf%lf",&n,&m)==2){for(int i=0;i<m;i++){sum=sum+n;n=sqrt(n);}printf("%.2lf\n",sum);sum=0;}}Hdu 2010#include<stdio.h>void main(){int a,b,i,t,x,y,z,count=0;while(scanf("%d%d",&a,&b)!=EOF){if(a>b) {t=a;a=b,b=t;}if(a<100||b>999) goto m;for(i=a;i<=b;i++){x=(i%100)%10;y=(int(i/10))%10;z=int(i/100);if(x*x*x+y*y*y+z*z*z==i){count+=1;if(count==1)printf("%d",i);else printf(" %d",i);}}m: {if(count>=1) printf("\n");else { printf("no\n");x=0;y=0;z=0; }}count=0;}}Hdu 2011#include<stdio.h>#include<math.h>int main(){int n;double a,i=1.0;double sum=0,b=0;while(scanf("%d",&n)!=EOF){while(n--){scanf("%lf",&a);for(i=1;i<=a;i++){sum=sum+(1/i)*pow(-1.0,i+1);}printf("%.2lf\n",sum);sum=0;}}}Hdu 2012#include<stdio.h>#include<math.h>int fun1(int x){return x*x+x+41;}int fun2(int x){for(int i=2;i<x;i++){if(x%i==0) { return 1;}}return 0;}void main(){int a,b,m,n,t,sum=0;while(scanf("%d %d",&a,&b)!=EOF){if(a==0&&b==0)continue;else if(a>b){t=a;a=b;b=t;}{/*if(a>b) { c=a;a=b;b=c;}*/for(int i=a;i<=b;i++){m=fun1(i);n=fun2(m);sum+=n;}if(sum!=0){printf("Sorry\n");}else{printf("OK\n");}sum=0;}}}Hdu 2013#include<stdio.h>#include<math.h>int main(){int x,n;while(scanf("%d",&n)!=EOF){if(n==0) break;n=pow(2,n-1);printf("%d\n",3*n-2);}}Hdu 2014#include<stdio.h>int main(){int n,i;float a[100],sum,max,min;while(scanf("%d",&n)!=EOF){sum=0;for(i=0;i<n;i++)scanf("%f",&a[i]);max=min=a[0];for(i=0;i<n;i++){if(a[i]>max)max=a[i];if(a[i]<min)min=a[i];sum+=a[i];}printf("%.2f\n",(sum-min-max)/(n-2));}}Hdu 2016#include<stdio.h>void main(){int a[99],n,i,min,t=0;while(scanf("%d",&n)!=EOF){if(n==0) break;else{for(i=0;i<n;i++) scanf("%d",&a[i]);min=a[0];for(i=0;i<n;i++){if(a[i]<=min) {min=a[i]; t=i;}}a[t]=a[0]; a[0]=min;}printf("%d",a[0]);for(i=1;i<n;i++)printf(" %d",a[i]);t=0;printf("\n");}}Hdu 2017#include <stdio.h>void main(){int count=0,i,n;char a[1000];scanf("%d",&n);while(n--){ scanf("%s",a);getchar();for(i=0;a[i];i++) //为什么？？？？？？{if(a[i]>='0'&&a[i]<='9')count++;}printf("%d\n",count);count=0;}}Hdu 2018#include<stdio.h>int main(){int a[55]={1,2,3};int n,i;while(scanf("%d",&n)==1&&n!=0){for(i=3;i<55;i++)a[i]=a[i-1]+a[i-3];printf("%d\n",a[n-1]);}}Hdu 2019#include<stdio.h>int main(){int a[100];int n,i,k=0,m;while(scanf("%d%d",&n,&m)==2&&(n!=0&&m!=0)) {for(i=0;i<n;i++){scanf("%d",&a[i]);}if(m>=a[n-1]){for(i=0;i<n;i++){printf("%d ",a[i]);}printf("%d\n",m);}else if(m<=a[0]){printf("%d",m);for(i=0;i<n;i++){printf(" %d",a[i]);}printf("\n");}else{for(i=0;i<n;i++){if(a[i]>=m){k=i;break;}}for(i=0;i<k;i++){printf("%d ",a[i]);}printf("%d",m);for(i=k;i<n;i++){printf(" %d",a[i]);}printf("\n");}}}Hdu 2020#include<stdio.h>#include<math.h>int main (){int i,j,t,n,a[100];while(scanf("%d",&n)!=EOF&&n!=0){for(i=0;i<n;i++)scanf("%d",&a[i]);for(i=0;i<n;i++){for(j=0;j<n-i-1;j++){if(abs(a[j])<abs(a[j+1])){t=a[j];a[j]=a[j+1];a[j+1]=t;}}}printf("%d",a[0]);for(i=1;i<n;i++)printf(" %d",a[i]);printf("\n");}}Hdu 2021#include<stdio.h>int fun(int x){int a=0,b=0,c=0,d=0,e=0,f=0;a=(int)(x/100);b=(int)((x-100*a)/50);c=(int) ((x-100*a-50*b)/10);d=(int)((x-100*a-50*b-10*c)/5);e=(int) ((x-100*a-50*b-10*c-5*d)/2);f=x-100*a-50*b-10*c-5*d-2*e;return a+b+c+d+e+f;}void main(){int n,a[100],sum;while(scanf("%d",&n)!=EOF){if(n==0) break;sum=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);sum+=fun(a[i]);}printf("%d\n",sum);}}Hdu 2025#include<stdio.h>#include<string.h>int main(){char a[100],max;int i,l;while(scanf("%s",a)!=EOF){getchar();l=strlen(a);max=a[0];for(i=1;i<l;i++){if(a[i]>=max){max=a[i];}}for(i=0;i<l;i++){if(a[i]<max)printf("%c",a[i]);if(a[i]==max)printf("%c(max)",a[i]);}printf("\n");}}Hdu 2027#include <stdio.h>void main(){int n,i,a,b,c,d,e,k=0;char f[100];scanf("%d",&n);while(n--){a=b=c=d=e=k=0;scanf("%s",f);getchar();for(i=0;f[i];i++){if(f[i]=='a'||f[i]=='A')a++;if(f[i]=='e'||f[i]=='E')b++;if(f[i]=='i'||f[i]=='I')c++;if(f[i]=='o'||f[i]=='O')d++;if(f[i]=='u'||f[i]=='U')e++;}printf("a:%d\ne:%d\ni:%d\no:%d\nu:%d\n",a,b,c,d,e);if(n) printf("\n"); //重要！！！！！！！！！！！！！！！！！！}}Hdu 2028#include<stdio.h>void main(){int a[100];int i,j,min,max,n,count=0;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++){min=a[0];max=a[0];scanf("%d",&a[i]);if(a[i]>=max)max=a[i];if(a[i]<=min)min=a[i];}for(j=max;;j++){count=0;for(i=0;i<n;i++)if(j%a[i]==0) count++;if(count==n) break;}printf("%d\n",j);}}Hdu 2029#include <stdio.h>#include <stdlib.h>#include <string.h>int main(){int n,i,m=0,b,j;char a[1000];while(scanf("%d",&n)!=EOF){getchar();for(i=0;i<n;i++){gets(a);b=strlen(a);for(j=0;j<b/2;j++){if(a[j]!=a[b-j-1])m++;}if(m==0)printf("yes\n");elseprintf("no\n");m=0;}}return 0;}Hdu 2030#include<stdio.h>#include<string.h>int main(){int i,j,n,m,count,b;char a[5000];scanf("%d",&b);getchar();while(b--){gets(a); //用gets！！！！！！！count=0;j=strlen(a);for(i=0;i<j;i++){if(a[i]<0)count++;}printf("%d\n",count/2);}}Hdu 2033#include<stdio.h>void main(){int a1,b1,c1,n,i,a2,b2,c2,x,y,z,a3,b3,c3;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d%d%d%d%d%d",&a1,&b1,&c1,&a2,&b2,&c2);c3=(c1+c2)%60; x=(c1+c2-c3)/60;b3=(b1+b2+x)%60; y=(b1+b2+x-b3)/60;a3=a1+a2+y;printf("%d %d %d\n",a3,b3,c3);a1=a2=c1=b1=b2=c2=x=y=z=0;}}Hdu 2035#include <stdio.h>void main(){int n,m,t,i;while(scanf("%d%d",&n,&m)!=EOF){if(n==0&&m==0)break;t=1;for(i=0;i<m;i++){t=t*n;t=t%1000;}printf("%d\n",t);}}Hdu 2039#include<stdio.h>int main(){double a,b,c;int N;scanf("%d",&N);while (scanf("%lf %lf %lf",&a,&b,&c)==3){if(a+b>c&&a+c>b&&b+c>a&&a-b<c&&a-c<b&&b-c<a) printf("YES\n");elseprintf("NO\n");}return 0;}Hdu 2040#include<stdio.h>int fun(int x){int i,sum=0;for(i=1;i<x;i++){if(x%i==0){sum+=i;}}return sum;}int main(){int n,a,b;while(scanf("%d",&n)>0){while(n--){scanf("%d%d",&a,&b);if(fun(a)==b&&fun(b)==a)printf("YES\n");else printf("NO\n");}}}Hdu 2041#include<stdio.h>int main(){double f[40]={1,1};int n,m;for(int i=2;i<40;i++)f[i]=f[i-1]+f[i-2];while(scanf("%d",&n)!=EOF){for(int j=0;j<n;j++){scanf("%d",&m);printf("%.0lf\n",f[m-1]);}}}Hdu 2042#include<stdio.h>int main(){double f[31]={3};int n,m;for(int i=1;i<31;i++)f[i]=2*f[i-1]-2;while(scanf("%d",&n)!=EOF){for(int j=0;j<n;j++){scanf("%d",&m);printf("%.0lf\n",f[m]);}}}Hdu 2043#include<string.h>#include <stdio.h>void main(){int x=0,y=0,z=0,m=0,b,i,n;char a[50];scanf("%d",&n);while(n--){scanf("%s",a);getchar();b=strlen(a);if(strlen(a)>=8&&strlen(a)<=16){for(i=0;i<b;i++){if(a[i]>='A'&&a[i]<='Z')x=1;if(a[i]>='a'&&a[i]<='z')y=1;if(a[i]>='0'&&a[i]<='9')z=1;if(a[i]=='~'||a[i]=='!'||a[i]=='@'||a[i]=='#'||a[i]=='$'||a[i]=='%'||a[i]=='^')m=1;}if(x+y+z+m>=3)printf("YES\n");elseprintf("NO\n");x=y=z=m=0;}elseprintf("NO\n");}}Hdu 2044#include<stdio.h>int main (){__int64 a[50]={0,1,2};int i,n,c,b;while(scanf("%d",&n)!=EOF){while(n--){for(i=3;i<50;i++)a[i]=a[i-1]+a[i-2];scanf("%d%d",&c,&b);printf("%I64d\n",a[b-c]);}}}Hdu 2045#include<stdio.h>int main(){__int64 a[51]={3,6,6};int i,n;while(scanf("%d",&n)!=EOF){for(i=3;i<51;i++){ a[i]=a[i-1]+2*a[i-2]; }printf("%I64d\n",a[n-1]);}}Hdu 2046#include<stdio.h>int main(){double f[51]={1,2,3};int n;for(int i=3;i<51;i++)f[i]=f[i-1]+f[i-2];while(scanf("%d",&n)!=EOF){printf("%.0lf\n",f[n-1]);}return 0;}Hdu 2047#include<stdio.h>int main(){__int64 a[40]={3,8};int i,n;while(scanf("%d",&n)!=EOF){for(i=2;i<40;i++)a[i]=2*(a[i-2]+a[i-1]);printf("%I64d\n",a[n-1]);}}。

杭电计算机笔试题

杭电计算机笔试题
以下是十道常见的杭电计算机笔试题：
题目一：
已知一个含有n个元素的数组arr，编写一个函数，将数组中的元素全部复制到另一个数组brr中，并逆序排列。

题目二：
编写一个程序，实现字符串翻转功能。

给定一个字符串，输出其逆序字符串。

题目三：
给定一个整数n，编写程序计算n!的值（即n的阶乘）。

题目四：
已知一个含有n个元素的数组arr，编写一个函数，找到数组中的最大值和最小值，并输出。

题目五：
已知一个n阶的方阵A，编写一个函数，计算矩阵中主对角线元素之和。

题目六：
给定一个整数n，编写程序计算斐波那契数列的第n项。

题目七：
已知一个含有n个元素的数组arr和一个目标值target，编写一个函数，在数组中查找target并返回其下标。

如果target不存在，返回-1
题目八：
编写程序实现插入排序算法。

给定一个含有n个元素的数组arr，按照非降序排列数组元素。

题目九：
给定一个整数n，编写程序输出n个斐波那契数列的项。

题目十：
给定一个整数n，编写程序计算n的平方根，并保留3位小数。

杭电多校题解

杭电多校题解1. 简介杭电多校题解是指对杭州电子科技大学（Hangzhou Dianzi University，简称杭电）多校联合训练赛（简称多校赛）中的题目进行解答和分析的过程。

多校赛是中国大学生计算机竞赛中的一项重要赛事，吸引了全国各高校的优秀选手参与。

本文将从以下几个方面对杭电多校赛的题目进行解答和分析：1.多校赛的背景和意义2.题目类型和难度分析3.解题思路和算法分析4.代码实现和优化策略5.比赛经验和技巧总结2. 多校赛的背景和意义多校赛是中国大学生计算机竞赛中的一项重要赛事，由杭州电子科技大学、浙江大学、浙江工业大学、浙江理工大学和浙江农林大学等五所高校联合举办。

多校赛旨在提高大学生的算法设计和编程能力，促进各高校之间的学术交流和合作。

参加多校赛可以帮助选手提高解决问题的能力，培养团队协作精神，提升算法和数据结构的应用能力。

同时，多校赛也是选拔优秀人才的重要途径，各大高校和企业会通过多校赛的成绩来选拔和录取优秀的计算机专业学生。

3. 题目类型和难度分析多校赛的题目类型多样，涵盖了各个领域的算法和数据结构。

常见的题目类型包括动态规划、图论、字符串处理、贪心算法、搜索算法等。

题目的难度也不一，有些题目较为简单，适合初学者入门，而有些题目则较为困难，需要较高的算法和编程能力才能解答。

在解题过程中，需要仔细阅读题目描述，理解题目要求和限制条件。

对于较难的题目，可以先尝试分析问题的性质和特点，然后设计相应的算法。

对于已有的经典算法和数据结构，可以尝试将其应用到题目中，提高解题效率。

4. 解题思路和算法分析解题思路和算法分析是解决多校赛题目的关键步骤。

在解题过程中，需要根据题目的要求和限制条件，设计合适的算法来解决问题。

以下是一些常见的解题思路和算法分析方法：•动态规划：对于一些具有最优子结构性质的问题，可以使用动态规划算法来解决。

动态规划将问题分解为子问题，并通过保存子问题的解来避免重复计算，从而提高算法效率。

zoj浙江大学oj题目分类

搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight Moves1101 Gamblers1204 Additive equations1221 Risk1230 Legendary Pokemon1249 Pushing Boxes1364 Machine Schedule1368 BOAT1406 Jungle Roads1411 Anniversary1453 Surround the Trees 凸包1516 Uncle Tom's Inherited Land1525 Air Raid1586 QS Network1602 Multiplication Puzzle dp1649 Rescue1671 Walking Ant1711 Sum It Up dfs1901 A Star not a Tree? 说他搜索可能有点争议（不管了）1940 Dungeon Master2100 Seeding2110 Tempter of the Bone2140 Bal难题&经典1003 Crashing Balloon1015 Fishing Net 完美图1144 Robbery1149 Dividing up1161 Gone Fishing1197 Sorting Slides1217 Eight1228 Farewell, My Friend1237 Fans and Gems1455 Schedule Problem1456 Minimum Transport Cost 图论最短路径要保存路径1492 Maximum Clique 图论经典算法--最大团1600 Market Place1605 One-way Traffic1568 WishingBone's Room Plan1742 Gap 至今一点想法都没有，难！！！1743 Concert Hall Scheduling1827 The Game of 31 博弈1855 Maze1903 Jogging Trails 中国邮路问题1909 Square 经典的dfs.2064 Bomberman - Just Search! 经典！2094 Max Angle 计算几何+博弈2125 Rocket Mania2126 Rocket Mania Plus2127 Zuma2128 Seven Seas2129 Mummy Maze2142 Light The Square （24道）dp1499 Increasing Sequences 经典题1107 FatMouse and Cheese 很好的一题1039 Number Game 没有完美解决的题，感觉可以直接以所有剩下的数作为状态DP 1227 Free Candies SRbGa的经典题1234 Chopsticks SRbGa的经典题……1554 Folding 推荐2059 The Twin Towers 推荐2097 Walking on a Chessboard 推荐1011 NTA 可用dp，也可以不用1013 Great Equipment1024 Calendar Game1027 Human Gene Functions1037 Gridland1052 Algernon's Noxious Emissions1058 Currency Exchange1076 Gene Assembly1092 Arbitrage1093 Monkey and Banana1094 Matrix Chain Multiplication1100 Mondriaan's Dream DP可以过，不过有组合公式1103 Hike on a Graph1134 Strategic Game1147 Formatting Text1148 The Game1161 Gone Fishing1180 Self Numbers1192 It's not a Bug, It's a Feature!1196 Fast Food1136 Multiple ，BFS1276 Optimal Array Multiplication Sequence1255 The Path1250 Always On the Run1213 Lumber Cutting1206 Win the Bonus1301 The New Villa1303 Jury Compromise 其实不是很难，但是很容易错1345 Best Deal1360 Radar Installation1396 The Umbrella Problem: 20541425 Crossed Matchings1438 Asteroids!1459 String Distance and Transform Process1462 Team Them Up!1556 Heroes Of Might And Magic1520 Duty Free Shop1524 Supermarket1536 Labyrinth1479 Dweep1245 Triangles 可用dp也可搜索1011 NTA 简单题1013 Great Equipment 简单题1024 Calendar Game 简单题1027 Human Gene Functions 简单题1037 Gridland简单题1052 Algernon\'\'s Noxious Emissions 简单题1409 Communication System 简单题，但是很容易看错~~~ 1425 Crossed Matchings简单题1438 Asteroids! 简单题1459 String Distance and Transform Process 简单题1462 Team Them Up! 简单题1556 Heroes Of Might And Magic 简单题，不过背景蛮有意思的……1520 Duty Free Shop 简单题1524 Supermarket 简单题1301 The New Villa 简单题1303 Jury Compromise 其实不是很难，但是很容易错，555……1345 Best Deal 简单题，但是也很容易错……555……1360 Radar Installation 简单题1396 The Umbrella Problem: 2054 简单题1058 Currency Exchange 简单题1076 Gene Assembly 简单题1092 Arbitrage 简单题1093 Monkey and Banana 简单题1094 Matrix Chain Multiplication 简单题1536 Labyrinth 简单题1100 Mondriaan\'\'s Dream 简单题，DP可以过，不过据说有复杂的组合公式1103 Hike on a Graph 简单题1134 Strategic Game 简单题1147 Formatting Text 简单题1148 The Game 简单题1161 Gone Fishing 简单题1180 Self Numbers 简单题1192 It\'\'s not a Bug, It\'\'s a Feature! 简单题1196 Fast Food 简单题1107 FatMouse and Cheese 简单题，不过题目描述有些混乱1136 Multiple 简单题，BFS1276 Optimal Array Multiplication Sequence 简单题1255 The Path 简单题1250 Always On the Run 简单题1213 Lumber Cutting 简单题1206 Win the Bonus 简单题1479 Dweep无聊题1587 UP 100 无聊题，DP应该可以……但是太烦了……没做……1066 Square Ice 无聊题，目前已知的O(nlogn)算法要用AVL……您有没有简单点的O(nlogn)的算法？1245 Triangles 无聊题1022 Parallel Expectations 经典题，想了n久，最后发现可以DP，相当好的一道题1499 Increasing Sequences 经典题{}1039 Number Game 没有完美解决的题，感觉可以直接以所有剩下的数作为状态DP，但是缺乏证明……1227 Free Candies SRbGa的经典题，我看了oibh上的解题报告才做出来的……1234 Chopsticks SRbGa的经典题……图论:1525 Air Raid 最小路径覆盖1500 Pre-Post-erous! 简单题1501 Knockout Tournament 简单题1508 Intervals 对您来说应该是简单题，但我想了n久……，差分限制系统1333 Galactic Import 简单题1304 Tin Cutter 简单题，但是似乎有空间复杂度为O(n)的算法1310 Robot 简单题1311 Network 简单题1344 A Mazing Problem 简单题1395 Door Man 简单题，欧拉回路1372 Networking 简单题1406 Jungle Roads 简单题1053 FDNY to the Rescue! 简单题1055 Oh, Those Achin\'\' Feet 不错的简单题1059 What\'\'s In a Name 二分图完美匹配1064 Roads Scholar 简单题1082 Stockbroker Grapevine 简单题1085 Alien Security 简单题，我觉得我当时的算法好巧妙1097 Code the Tree 简单题1060 Sorting It All Out 简单题，但是规模要是大些的话……1105 FatMouse\'\'s Tour 简单题1119 SPF 简单题1127 Roman Forts 简单题1140 Courses 简单题1157 A Plug for UNIX 蛮不错的简单题1203 Swordfish 简单题1221 Risk 简单题1197 Sorting Slides 简单题，匹配1268 Is It A Tree? 不错的题，图论1273 It\'\'s Ir-Resist-Able! 简单题，图论1298 Domino Effect 简单题，最长路1260 King 简单题，差分限制系统……1291 MPI Maelstrom 不错的题，最长路1266 Gossiping 简单题1285 Shipping Routes 无聊题，最短路1313 Gears on a Board 无聊题1502 Plugged In 无聊题，匹配应该可以，但是太烦了，没做……1568 WishingBone\'\'s Room Plan 无聊题，最大最小匹配，不过容易看错题~~~~~~~~1077 Genetic Combinations 无聊题，匹配1364 Machine Schedule 匹配1137 Girls and Boys 背诵题，匹配……1023 University Entrace Examination 经典题，但是和1576重复1576 Marriage is Stable 经典题，感人的背景，经典的算法……1249 Pushing Boxes 经典题，某人的论文写过，求割点和块+BFS可以在O(面积)的时间内做出来，没时间看论文，我用双重BFS过的1141 Closest Common Ancestors 没有完美解决，最简单的算法就能过，但要是规模大了……1084 Channel Allocation 尚未完美解决，我用搜索过的，不过是不是有更好的算法呢……？1231 Mysterious Mountain SRbGa的经典题……1232 Adventure of Super Mario SRbGa的经典题……数学:1007 Numerical Summation of a Series 简单题，还是蛮有意思的1045 HangOver简单题1049 I Think I Need a Houseboat 简单题1028 Flip and Shift 简单题，可以DP/BFS/……，但是实际上有数学方法可以直接判断出来1026 Modular multiplication of polynomials 简单题，感觉有比较简单的好算法，但想不出来1307 Packets 简单题，不过也蛮经典的……1312 Prime Cuts 简单题1334 Basically Speaking 简单题1337 Pi 简单题1342 Word Index 简单题1349 Four Quarters 简单题1350 The Drunk Jailer 简单题1352 Number Base Conversion 简单题1353 Unimodal Palindromic Decompositions 规模不大，所以是简单题……1354 Extended Lights Out 简单题1362 Game Prediction 简单题1365 Mileage Bank 简单题1382 A Simple Task 简单题1383 Binary Numbers 简单题1403 Safecracker 简单题1408 The Fun Number System 简单题1486 Color the Tree 简单题1487 Playing Cards 简单题1489 2^x mod n = 1 简单题，应该有好算法，枚举过的……1503 One Person "The Price is Right" 简单题，POI Eggs的翻版1512 Water Treatment Plants 简单题，组合计数1526 Big Number 简单题，不过O(1)和O(n)还是有区别的1529 Enigmatic Travel 简单题，不过个人感觉题目描述很令人费解1530 Find The Multiple 简单题1537 Playing with a Calculator 简单题1577 GCD & LCM 简单题，分区联赛的题……1005 Jugs 简单题1543 Stripies简单题1569 Partial Sums 简单题1062 Trees Made to Order 简单题1070 Bode Plot 简单题1073 Round and Round We Go 简单题1078 Palindrom Numbers 简单题1086 Octal Fractions 简单题1199 Point of Intersection 简单题1104 Leaps Tall Buildings 简单题1110 Dick and Jane 简单题1115 Digital Roots 简单题1113 u Calculate e 简单题1152 A Mathematical Curiosity 简单题1154 Niven Numbers 简单题1160 Biorhythms 简单题1163 The Staircases 简单题1177 K-Magic Number 简单题1184 Counterfeit Dollar 简单题1182 Keeps Going and Going and ... 简单题1284 Perfection 简单题1272 Numerically Speaking 简单题1269 Coconuts, Revisited 简单题1247 There\'\'s Treasure Everywhere! 简单题1241 Geometry Made Simple 简单题1202 Divide and Count 简单题1216 Deck 简单题1218 Ratio 简单题1261 Prime Land 简单题1212 Mountain Landscape 无聊题1410 Number Sequence 无聊题1401 Hilbert Curve Intersections 无聊题1331 Perfect Cubes 无聊题1322 Random Number 无聊题1535 Lucky Ticket 无聊题1539 Lot 无聊题1363 Chocolate 经典题……1366 Cash Machine 经典题！强烈推荐！1149 Dividing up 经典题，应该可以用1366的方法做，但似乎可以利用问题的特殊性用贪心+DP在O(1)的时间内做出来1222 Just the Facts 经典题，没有完美解决,据说可能有O(logn)的做法1475 Ranklist1572 Bracelet ~~~题义不明，感觉可能是判定欧拉回路的存在性1133 Smith Numbers 没有完美解决，数学1080 Direct Subtraction1229 Gift?! SRbGa的经典题……1238 Guess the Number SRbGa的经典题……1239 Hanoi Tower Troubles Again! SRbGa的经典题……字符串处理:1050 Start Up the Startup 简单题1315 Excuses, Excuses! 简单题1151 Word Reversal 简单题1170 String Matching 简单题1174 Skip Letter Code 不错的简单题1175 Word Process Machine 简单题1181 Word Amalgamation 简单题1038 T9 无聊题，单词树1330 DNA Translation 无聊题1335 Letter Sequence Analysis 无聊题1099 HTML 无聊题1243 URLs 无聊题1540 Censored! 经典题！强烈推荐！1511 Word Puzzles 后缀树模拟:1051 A New Growth Industry 简单题1300 Border 简单题1326 M*A*S*H 简单题1494 Climbing Worm 简单题1072 Microprocessor Simulation 简单题1098 Simple Computers 简单题1056 The Worm Turns 简单题1195 Blowing Fuses 简单题1189 Numbers That Count 简单题1144 Robbery 简单题1153 Tournament Seeding 简单题，但是直接计算好像不行，得模拟……1167 Trees on the Level 简单题1200 Mining 简单题1278 Pseudo-Random Numbers 简单题1257 Parking Lot 简单题1270 Nonstop Travel 简单题1207 The Knight, the Princess, and the Dragons 无聊题1169 Square Cipher 无聊题1176 Die and Chessboard 无聊题1178 Booklet Printing 无聊题1009 Enigma 无聊题，但是很容易错……1012 Mainframe 无聊题，但是很容易错……1324 Unix ls无聊题，输出格式没说清楚~~~~~~~1336 Mark-up 无聊题1277 Transferable Voting 无聊题1279 Cowculations无聊题1281 Hi-Q 无聊题1282 Call Forwarding 无聊题1065 Robots 变态题，太复杂啦~~~~~~~~~~~1208 Roll the Die! 变态题1388 Exchanges 经典题！强烈推荐！1236 Eat or Not to Eat? SRbGa的经典题……模拟几何:1575 Koch Curve 简单题1010 Area 简单题1565 Input 简单题1081 Points Within 简单题1165 Laser Lines 简单题1248 Video Surveillance 简单题，李彭煦的论文中写到过，好像是某年CTSC的……1299 Pendulum 简单题1090 The Circumference of the Circle 无聊题1271 Doing Windows 无聊题1280 Intersecting Lines 无聊题1296 Stars 经典变态题……1030 Farmland 变态题1041 Transmitters 变态题1158 Treasure Hunt 经典题1139 Rectangles 没有完美解决其它:1006 Do the Untwist 简单题1014 Operand 简单题1016 Parencodings 简单题1042 W\'\'s Cipher 简单题1047 Image Perimeters 简单题1514 Fake Tickets 简单题1029 Moving Tables 简单题，好像是线段树的经典题目之一，但是这题规模比较小，所以不必用。

杭电OJ水题答案2

杭电OJ水题答案2 Hdu 2000#include<stdio.h>void main(){char a,b,c,t;while(scanf("%c%c%c",&a,&b,&c)>0){if(a>b) t=a,a=b,b=t;if(a>c) t=c,c=a,a=t;if(c>a && c<b) t=b,b=c,c=t;scanf("%*c",&a);printf("%c %c %c\n",a,b,c);}}Hdu 2001#include <stdio.h>#include <math.h>int main(){double x1;double x2;double y1;double y2;while(scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2)!=EOF) {double num=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); printf("%.2lf\n",num);}return 0;}Hdu 2002#include<stdio.h>int main(){- 1 -double pi=3.1415927;double r;while (scanf("%lf",&r)==1)printf("%.3lf\n",((pi*r*r*r)*4)/3);}Hdu 2003#include <stdio.h>#include <math.h>int main(){double x;while (scanf("%lf",&x)==1)printf("%.2lf\n",fabs(x));return 0;}Hdu 2004#include<stdio.h>void main(){double x;while(scanf("%lf",&x)==1)if(x>100||x<0)printf("Score is error!\n");else if(0<x&&x<60) printf("E\n"); else if(60<=x&&x<70) printf("D\n"); else if(70<=x&&x<80)printf("C\n");else if(80<=x&&x<90)printf("B\n");elseprintf("A\n");}Hdu 2005#include<stdio.h>void main()- 2 -{int y,m,d,sum=0;while(scanf("%d/%d/%d",&y,&m,&d)!=EOF){if((y%100==0&&y%400==0)||(y%100!=0&&y%4==0)) {switch(m=13-m){case 1: sum=sum+30;case 2: sum=sum+31;case 3: sum=sum+30;case 4: sum=sum+31;case 5: sum=sum+31;case 6: sum=sum+30;case 7: sum=sum+31;case 8: sum=sum+30;case 9: sum=sum+31;case 10: sum=sum+29;case 11: sum=sum+31;}}else{switch(m=13-m){case 1: sum=sum+30; case 2: sum=sum+31; case 3: sum=sum+30; case 4: sum=sum+31; case 5: sum=sum+31; case 6: sum=sum+30; case 7: sum=sum+31; case 8: sum=sum+30; case 9: sum=sum+31; case 10: sum=sum+28; case 11: sum=sum+31; }}printf("%d\n",sum+d); sum=0;}}- 3 -Hdu 2006#include<stdio.h> void main(){int n,i,a,sum=1;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++){scanf("%d",&a);if(a%2==1)sum*=a;}printf("%d\n",sum);sum=1;}}Hdu 2007#include<stdio.h>void main(){int a,b,x=0,y=0,t;while(scanf("%d%d",&a,&b)!=EOF) {if(b<a) { t=a; a=b; b=t;goto m;} else{m:{ while(a<=b){if(a%2==0) { x=a*a+x;a++;} else { y=a*a*a+y; a++;}}}printf("%d %d\n",x,y);x=0; y=0;}}}- 4 -Hdu 2008#include<stdio.h>void main(){double a; int i,x=0,y=0,z=0,n; while(scanf("%d",&n)==1){if(n==0)break;else{while(n--){scanf("%lf",&a);if(a<0) {x++;}else if(a==0) {y++;}else {z++;}}}printf("%d %d %d\n",x,y,z);x=0;y=0;z=0;}}Hdu 2009#include<stdio.h>#include<math.h>void main(){double n,m;double sum=0;while (scanf("%lf%lf",&n,&m)==2) {for(int i=0;i<m;i++){sum=sum+n;n=sqrt(n);}printf("%.2lf\n",sum);sum=0;- 5 -}}Hdu 2010#include<stdio.h>void main(){int a,b,i,t,x,y,z,count=0;while(scanf("%d%d",&a,&b)!=EOF){if(a>b) {t=a;a=b,b=t;}if(a<100||b>999) goto m;for(i=a;i<=b;i++){x=(i%100)%10;y=(int(i/10))%10;z=int(i/100); if(x*x*x+y*y*y+z*z*z==i){count+=1;if(count==1)printf("%d",i);else printf(" %d",i);}}m: {if(count>=1) printf("\n");else { printf("no\n");x=0;y=0;z=0; } }count=0;}}Hdu 2011#include<stdio.h>#include<math.h>int main(){int n;double a,i=1.0;double sum=0,b=0;- 6 -while(scanf("%d",&n)!=EOF){while(n--){scanf("%lf",&a);for(i=1;i<=a;i++){sum=sum+(1/i)*pow(-1.0,i+1);}printf("%.2lf\n",sum);sum=0;}}}Hdu 2012#include<stdio.h>#include<math.h>int fun1(int x){return x*x+x+41;} int fun2(int x) {for(int i=2;i<x;i++){if(x%i==0) { return 1;}}return 0;}void main(){int a,b,m,n,t,sum=0;while(scanf("%d %d",&a,&b)!=EOF) {if(a==0&&b==0)continue;else if(a>b){t=a;a=b;b=t;}{/*if(a>b) { c=a;a=b;b=c;}*/ for(int i=a;i<=b;i++){m=fun1(i);- 7 -n=fun2(m);sum+=n;}if(sum!=0){printf("Sorry\n");}else{printf("OK\n");}sum=0;}}}Hdu 2013#include<stdio.h> #include<math.h> int main() {int x,n;while(scanf("%d",&n)!=EOF){if(n==0) break;n=pow(2,n-1);printf("%d\n",3*n-2);}}Hdu 2014#include<stdio.h> int main(){int n,i;float a[100],sum,max,min;while(scanf("%d",&n)!=EOF){sum=0;- 8 -for(i=0;i<n;i++)scanf("%f",&a[i]);max=min=a[0];for(i=0;i<n;i++){if(a[i]>max)max=a[i];if(a[i]<min)min=a[i];sum+=a[i];}printf("%.2f\n",(sum-min-max)/(n-2));}}Hdu 2016#include<stdio.h>void main(){int a[99],n,i,min,t=0;while(scanf("%d",&n)!=EOF){if(n==0) break;else{for(i=0;i<n;i++) scanf("%d",&a[i]);min=a[0]; for(i=0;i<n;i++){if(a[i]<=min) {min=a[i]; t=i;}}a[t]=a[0]; a[0]=min;}printf("%d",a[0]);for(i=1;i<n;i++)printf(" %d",a[i]);t=0;printf("\n");}}- 9 -Hdu 2017#include <stdio.h>void main(){int count=0,i,n;char a[1000];scanf("%d",&n);while(n--){ scanf("%s",a);getchar();for(i=0;a[i];i++) //为什么,,,,,, {if(a[i]>='0'&&a[i]<='9')count++;}printf("%d\n",count);count=0;}}Hdu 2018#include<stdio.h>int main(){int a[55]={1,2,3};int n,i;while(scanf("%d",&n)==1&&n!=0) {for(i=3;i<55;i++)a[i]=a[i-1]+a[i-3];printf("%d\n",a[n-1]);}}Hdu 2019#include<stdio.h>int main(){- 10 -int a[100];int n,i,k=0,m;while(scanf("%d%d",&n,&m)==2&&(n!=0&&m!=0)) {for(i=0;i<n;i++){scanf("%d",&a[i]);}if(m>=a[n-1]){for(i=0;i<n;i++){printf("%d ",a[i]);}printf("%d\n",m);}else if(m<=a[0]){printf("%d",m);for(i=0;i<n;i++){printf(" %d",a[i]);}printf("\n");}else{for(i=0;i<n;i++) {if(a[i]>=m){k=i;break;}}for(i=0;i<k;i++) {printf("%d ",a[i]); }printf("%d",m);for(i=k;i<n;i++) {printf(" %d",a[i]); }printf("\n");}}- 11 -}Hdu 2020#include<stdio.h> #include<math.h>int main (){int i,j,t,n,a[100];while(scanf("%d",&n)!=EOF&&n!=0) {for(i=0;i<n;i++)scanf("%d",&a[i]);for(i=0;i<n;i++){for(j=0;j<n-i-1;j++){if(abs(a[j])<abs(a[j+1])){t=a[j];a[j]=a[j+1];a[j+1]=t;}}}printf("%d",a[0]);for(i=1;i<n;i++)printf(" %d",a[i]);printf("\n");}}Hdu 2021#include<stdio.h>int fun(int x){int a=0,b=0,c=0,d=0,e=0,f=0;a=(int)(x/100);b=(int)((x-100*a)/50);c=(int) ((x-100*a-50*b)/10);d=(int)((x-100*a-50*b-10*c)/5);e=(int) ((x-100*a-50*b-10*c-5*d)/2); f=x-100*a-50*b-10*c-5*d-2*e;- 12 -return a+b+c+d+e+f; }void main(){int n,a[100],sum;while(scanf("%d",&n)!=EOF){if(n==0) break;sum=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);sum+=fun(a[i]);}printf("%d\n",sum);}}Hdu 2025#include<stdio.h> #include<string.h> int main() {char a[100],max;int i,l;while(scanf("%s",a)!=EOF){getchar();l=strlen(a);max=a[0];for(i=1;i<l;i++){if(a[i]>=max){max=a[i];}}for(i=0;i<l;i++){if(a[i]<max)printf("%c",a[i]);if(a[i]==max)printf("%c(max)",a[i]); - 13 -}printf("\n");}}Hdu 2027#include <stdio.h>void main(){int n,i,a,b,c,d,e,k=0; char f[100];scanf("%d",&n);while(n--){a=b=c=d=e=k=0;scanf("%s",f);getchar();for(i=0;f[i];i++){if(f[i]=='a'||f[i]=='A') a++;if(f[i]=='e'||f[i]=='E') b++;if(f[i]=='i'||f[i]=='I')c++;if(f[i]=='o'||f[i]=='O')d++;if(f[i]=='u'||f[i]=='U')e++;}printf("a:%d\ne:%d\ni:%d\no:%d\nu:%d\n",a,b,c,d,e);if(n) printf("\n"); //重要～～～～～～～～～～～～～～～～～～}}Hdu 2028#include<stdio.h>void main(){int a[100];int i,j,min,max,n,count=0;- 14 -while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++){min=a[0];max=a[0];scanf("%d",&a[i]);if(a[i]>=max)max=a[i];if(a[i]<=min)min=a[i];}for(j=max;;j++){count=0;for(i=0;i<n;i++)if(j%a[i]==0) count++;if(count==n) break;}printf("%d\n",j);}}Hdu 2029#include <stdio.h>#include <stdlib.h> #include <string.h> int main() {int n,i,m=0,b,j;char a[1000];while(scanf("%d",&n)!=EOF){getchar();for(i=0;i<n;i++){gets(a);b=strlen(a);for(j=0;j<b/2;j++){if(a[j]!=a[b-j-1])- 15 -m++;}if(m==0)printf("yes\n");elseprintf("no\n");m=0;}}return 0;}Hdu 2030#include<stdio.h> #include<string.h> int main() {int i,j,n,m,count,b;char a[5000];scanf("%d",&b);getchar();while(b--){gets(a); //用gets～～～～～～～count=0;j=strlen(a);for(i=0;i<j;i++){if(a[i]<0)count++;}printf("%d\n",count/2);}}Hdu 2033#include<stdio.h> void main(){int a1,b1,c1,n,i,a2,b2,c2,x,y,z,a3,b3,c3; scanf("%d",&n);- 16 -for(i=0;i<n;i++){scanf("%d%d%d%d%d%d",&a1,&b1,&c1,&a2,&b2,&c2);c3=(c1+c2)%60; x=(c1+c2-c3)/60;b3=(b1+b2+x)%60; y=(b1+b2+x-b3)/60; a3=a1+a2+y;printf("%d %d %d\n",a3,b3,c3);a1=a2=c1=b1=b2=c2=x=y=z=0;}}Hdu 2035#include <stdio.h>void main(){int n,m,t,i;while(scanf("%d%d",&n,&m)!=EOF) {if(n==0&&m==0)break;t=1;for(i=0;i<m;i++){t=t*n;t=t%1000;}printf("%d\n",t);}}Hdu 2039#include<stdio.h>int main(){double a,b,c;int N;scanf("%d",&N);while (scanf("%lf %lf %lf",&a,&b,&c)==3) {if(a+b>c&&a+c>b&&b+c>a&&a-b<c&&a-c<b&&b-c<a) printf("YES\n");else- 17 -printf("NO\n");}return 0;}Hdu 2040#include<stdio.h> int fun(int x){int i,sum=0;for(i=1;i<x;i++){if(x%i==0){sum+=i;}}return sum;}int main(){int n,a,b;while(scanf("%d",&n)>0){while(n--){scanf("%d%d",&a,&b);if(fun(a)==b&&fun(b)==a) printf("YES\n");else printf("NO\n");}}}Hdu 2041#include<stdio.h> int main() {double f[40]={1,1};int n,m;- 18 -for(int i=2;i<40;i++)f[i]=f[i-1]+f[i-2];while(scanf("%d",&n)!=EOF) {for(int j=0;j<n;j++){scanf("%d",&m);printf("%.0lf\n",f[m-1]); }}}Hdu 2042#include<stdio.h>int main(){double f[31]={3};int n,m;for(int i=1;i<31;i++)f[i]=2*f[i-1]-2;while(scanf("%d",&n)!=EOF) {for(int j=0;j<n;j++){scanf("%d",&m);printf("%.0lf\n",f[m]);}}}Hdu 2043#include<string.h>#include <stdio.h>void main(){int x=0,y=0,z=0,m=0,b,i,n;char a[50];scanf("%d",&n);while(n--){scanf("%s",a);- 19 -getchar();b=strlen(a);if(strlen(a)>=8&&strlen(a)<=16) {for(i=0;i<b;i++){if(a[i]>='A'&&a[i]<='Z')x=1;if(a[i]>='a'&&a[i]<='z')y=1;if(a[i]>='0'&&a[i]<='9')z=1;if(a[i]=='~'||a[i]=='!'||a[i]=='@'||a[i]=='#'||a[i]=='$'||a[i]=='%'| |a[i]=='^')m=1;}if(x+y+z+m>=3)printf("YES\n");elseprintf("NO\n");x=y=z=m=0;}elseprintf("NO\n");}}Hdu 2044#include<stdio.h>int main (){__int64 a[50]={0,1,2};int i,n,c,b;while(scanf("%d",&n)!=EOF) {while(n--){for(i=3;i<50;i++)a[i]=a[i-1]+a[i-2];scanf("%d%d",&c,&b);printf("%I64d\n",a[b-c]); }}}- 20 -Hdu 2045#include<stdio.h>int main(){__int64 a[51]={3,6,6};int i,n;while(scanf("%d",&n)!=EOF) {for(i=3;i<51;i++){ a[i]=a[i-1]+2*a[i-2]; }printf("%I64d\n",a[n-1]); }}Hdu 2046#include<stdio.h>int main(){double f[51]={1,2,3};int n;for(int i=3;i<51;i++)f[i]=f[i-1]+f[i-2];while(scanf("%d",&n)!=EOF) {printf("%.0lf\n",f[n-1]); }return 0;}Hdu 2047#include<stdio.h>int main(){__int64 a[40]={3,8};int i,n;while(scanf("%d",&n)!=EOF){for(i=2;i<40;i++)a[i]=2*(a[i-2]+a[i-1]); - 21 -printf("%I64d\n",a[n-1]); }}- 22 -。

杭州电子科技大学大三计算机专业计算机网络试卷及答案 (1)

杭州电子科技大学2021计算机学院计算机网络期末考试〔手抄本，纯手打，求给分〕真可恶，同一个专业同一门课，用不同的教科书，竟然也有两份不同的考卷。

〔王相林版〕一，选择题〔选项大局部没有抄〕1.曼切斯特编码的特点〔〕A在0比特的前沿有电平翻转2.HDLC协议是一种〔〕A面相比特的同步链路控制协议3.因特网中的协议应该满足规定的层次关系，表示协议层次和对应关系的是〔〕4.在快速以太网物理层标准中，使用对5类元屏蔽的双绞线是〔〕A100base-tx5.编码方式属于差分曼切斯特编码的是〔〕6.在RIP协议中，可以用水平分割法解决路由环路问题，下面说法正确的选项是〔〕7.OSPF采用〔链路状态〕算法计算最正确路由8.局域网的协议结构一般不包括〔〕9.在一部通信中，每个字符包含1位起始位，7位数据位，1为奇偶检验位，1位终止位，每秒钟传送100个字符，则有效数据速率是〔70%〕10.BGP协议的作用是〔〕11.ARP协议数据单元封装在〔以太帧〕发送12.ICMP协议数据封装在〔IP数据报〕发送13.TCP是互联网中的传输层协议，TCP协议进行流量控制的方法是用〔使用可变大小的滑动窗口协议〕14.RIP是一种基于〔距离向量算法/BELLMAN-FORD〕的路由协议15.一个B类网络的子网掩码为，被分成〔2〕个子网16.某公司的网络的地址是下面属于〔B〕这个网络A20 C17 私网地址用于配置公司内部网络，(B)属于私网地址A 128.168.101B 10.128.10.1 C18 通过交换机连接的一组工作站〔〕冲突域，播送域19采用CRC校验的生成多项式为X16+X15+X2+1，产生的校验码是〔16〕位20在TCP协议中采用〔套接字〕来区分不同的应用进程二填空题1计算机网络由〔资源子网〕〔通信子网〕两个子网组成，地址由〔4〕个字节组成，包括〔网络号〕和主机号，MAC地址由〔6〕个字节组成3电信网络一般分为线路交换网络和分组交换网络，线路交换网络可以采用频分复用和〔时分复用〕技术，而分组交换网络可分为〔数据报〕和虚电路交换网络4.局域网常用的拓扑结构有总线型，星形和〔树形〕三种，其中以太网采用〔星形〕结构5.邮件效劳器发邮件时通过〔SMTP〕协议来实现，利用FOSMAIL收邮件是通过〔FTP〕协议实现的。

杭电计算机笔试题

杭电计算机笔试题
以下是十道杭电计算机笔试题：
1.编写一个程序，输出1到100之间的所有偶数。

2.编写一个函数，计算一个整数数组的平均值。

3.编写一个程序，实现斐波那契数列。

要求用户输入一个正整数，然后输出该数列的前n项。

4.编写一个程序，判断一个字符串是否为回文字符串。

回文字符串是指从前向后和从后向前读取都一样的字符串。

5.编写一个程序，求一个整数数组中最大的两个数之和。

6.实现一个简单的计算器程序，用户输入两个操作数和一个操作符（+、-、*、/），然后输出计算结果。

7.编写一个程序，判断一个数是否为质数。

质数是只能被1和自身整除的数。

8.编写一个程序，找出一个字符串中出现次数最多的字符，并输出其出现次数。

9.编写一个函数，统计一个字符串中包含的英文字母、数字和其他字符的个数。

10.编写一个程序，实现选择排序算法对一个整数数组进行排序。

请注意，这只是一小部分可能的笔试题目。

在实际笔试中，还可能包含其他类型的题目，如编程题、算法题、数据结构题等。

杭电OJ的输入输出格式题

杭电OJ的输入/输出格式题输入一、1000 A + B Problem题目名称：A + B Problem链接地址：/showproblem.php?pid=1000Time Limit: 1 Seconds Memory Limit:32768KTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Problem DescriptionCalculate A + B.InputEach line will contain two integers A and B. Process to end of file.OutputFor each case, output A + B in one line.Sample Input1 1Sample Output2参考答案#include <stdio.h>int main(void){int a,b;while(scanf("%d%d",&a,&b)!=EOF)printf("%d\n",a+b);return 0;}二、1002 A + B Problem II题目名称：A + B Problem II链接地址：/showproblem.php?pid=1002Time Limit: 1 Seconds Memory Limit:32768KTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22627 Accepted Submission(s): 4035Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.InputThe first line of the input contains an integer T(1<=T<=20) which me ans the number of test cases. Then T lines follow, each line consis ts of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.OutputFor each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B . Note there are some spaces int the equation. Output a blank line between two test cases.Sample Input21 2112233445566778899 998877665544332211Sample OutputCase 1:1 +2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110思路：本题是手工模拟两个大数相加的问题。

杭电oj部分水题 c语言源代码

{
if(n==0)
break;
a=b=c=0;
for (i = 1 ; i <= n ; i++){
scanf("%d",&m);
if(m>0)
{c=c+1;}
else if(m==0)
{ b=b+1;}
}
return 0;
}
1091 A+B for Input-Output Practice (III)
#include<stdio.h>
int main()
{
int a,b;
while(scanf("%d%d",&a,&b)){
if(a==0&&b==0)
break;
for(s = 0.0,i=0; i<n;i++ ){
s=s+x; x = sqrt(x);
}
printf("%.2lf\n", s);
}
return 0;
}
2010 水仙花数
#include<stdio.h>
for (i = m ; i <= n ; i++){
if(i%2==0)
{ a=pow(i,2);
x=x+a;}
else
{ b=pow(i,3);
y=y+b;}
}
printf("%.0lf %.0lf\n", x, y);
1000和1089 A+B for Input-Output Practice (I)

杭州电子科技大学acm习题集锦

Output对于每个测试实例，输出可能得到的最大和，每个实例的输出占一行。

杭电OJ水题答案

Hdu1000,1090#include<stdio.h>int main(){int a,b;while (scanf("%d %d",&a,&b)==2)printf("%d\n",a+b);return 0;}Hdu1008#include<stdio.h>int main(){int n,i,j;int t;while(scanf("%d",&n)!=EOF){t=5*n;j=0;if(n==0)break;while(n--){scanf("%d",&i);if(i>j)t+=(i-j)*6;if(i<j)t+=(j-i)*4;j=i;}printf("%d\n",t);}}Hdu1021#include <stdio.h>int main(){int n;while(scanf("%d",&n)!=EOF){n=n%8;if(n==2||n==6)printf("yes\n");elseprintf("no\n");}return 0;}Hdu1040#include<stdio.h>int main(){int i,j,n,m,a[2000],t;scanf("%d",&n);while(n--){scanf("%d",&m);for(i=0;i<m;i++)scanf("%d",&a[i]);for(i=0;i<m;i++)for(j=0;j<m-1-i;j++)if(a[j]>a[j+1]){t=a[j];a[j]=a[j+1];a[j+1]=t;}printf("%d",a[0]);for(i=1;i<m;i++)printf(" %d",a[i]);printf("\n");}}Hdu1076#include<stdio.h>int main(){int count=0,i,n,j,a,b;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++){scanf("%d %d",&a,&b);for(j=a;;j++){if((j%100==0&&j%400==0)||(j%100!=0&&j%4==0)){ count++; }if(count==b) break;}printf("%d\n",j);count=0;}}}Hdu1091#include<stdio.h>int main(){double a;double b;while(scanf("%lf %lf",&a,&b))if(a==0 && b==0)break;elseprintf("%.0lf\n",a+b);return 0;}Hdu1092#include<stdio.h>void main(){int n,i,a,sum=0;while(scanf("%d",&n)!=EOF){if(n==0)continue;elsefor(i=1;i<=n;i++){scanf("%d",&a);sum+=a;}printf("%d\n",sum);sum=0;}}}Hdu1093#include<stdio.h>int main(){int n,i,j,a,sum=0;scanf("%d",&j);for(int b=0;b<j;b++){while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++){scanf("%d",&a);sum+=a;}printf("%d\n",sum);sum=0;}}}Hdu1094#include<stdio.h>int main(){int n,i,a,sum=0;while(scanf("%d",&n)!=EOF) {for(i=1;i<=n;i++)scanf("%d",&a);sum+=a;}printf("%d\n",sum);sum=0;}}Hdu1095#include<stdio.h>int main(){int a,b;while (scanf("%d %d",&a,&b)==2)printf("%d\n\n",a+b);return 0;}Hdu1096#include<stdio.h>int main(){int N;scanf("%d",&N);while(N--){int M,i,a,sum=0;scanf("%d",&M);for(i=0;i<M;i++){scanf("%d",&a);sum+=a;}if(N!=0)printf("%d\n\n",sum);elseprintf("%d\n",sum);}}Hdu1108#include<stdio.h>void main (){unsigned int b, a, i=1;while(scanf("%d %d",&a,&b)!=EOF){for(i=1;;i++){if(i%a==0&&i%b==0){printf("%d\n",i);break;}}}}Hdu1201#include<stdio.h>void main(){int a,b,c,i,n,count1,count2,sum;while(scanf("%d",&n)!=EOF){while(n--){scanf("%d-%d-%d",&a,&b,&c);if(b==2&&c==29){ printf("-1\n"); }else{count1=0;sum=0;for(i=a;i<a+18;i++){if((((i%100!=0&&i%4==0)||(i%400==0))&&b<=2)||((((i+1)%100!=0&&(i+1)%4==0)||((i+1) %400==0))&&b>2))sum+=366;elsesum+=365;}printf("%d\n",sum);}}}}Hdu1235#include<stdio.h>void main(){int a,b,c,i,n,f[1000],count=0;while(scanf("%d",&n)>0){if(n==0) break;for(i=0;i<n;i++)scanf("%d",&f[i]);scanf("%d",&c);for(i=0;i<n;i++){if(f[i]==c)count++;}printf("%d\n",count);count=0;}}Hdu1046#include<stdio.h>int f(int x){int j,sum=0;for(j=1;j<x;j++){if(x%j==0)sum+=j;}if(x==sum)return 1;else return 0;sum=0;}void main(){unsigned int i,n,a,b,t,x,count=0;while(scanf("%d",&n)>0){while(n--){scanf("%d%d",&a,&b);if(a>b) {t=a;a=b;b=t; }for(i=a;i<=b;i++){x=f(i);if(x==1) //1是完数count++;}printf("%d\n",count);count=0;}}}。

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杭州电子科技大学OJ题目分类1001 整数求和水题1002 C语言实验题——两个数比较水题1003 1、2、3、4、5... 简单题1004 渊子赛马排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想数论：容斥定理1007 童年生活二三事递推题1008 University 简单hash1009 目标柏林简单模拟题1010 Rails 模拟题（堆栈）1011 Box of Bricks 简单题1012 u Calculate e 简单数学计算1013 STAMPS 搜索or动态规划1014 Border 模拟题1015 Simple Arithmetics 高精度计算1016 Shoot-out 博弈+状态压缩DP1017 Tour Guide1018 Card Trick 简单题1019 Necklace Decomposition 贪心1020 Crashing Robots 模拟题1021 Electrical Outlets 简单题1022 Watchdog 简单题1023 Taxi Cab Scheme 图论：最小路径覆盖--->最大二分匹配1024 Pseudo-random Numbers 数论1025 Card Game Cheater 简单题1026 Investment 动态规划1027 Pipes1028 SETI 数学:高斯消元法1029 Minimax Triangulation 计算几何1030 Unequalled Consumption 母函数1031 Declaration of Content1032 Laserbox 搜索：DFS1033 Bowlstack1034 Pesky Heroes1035 Reduced ID Numbers 暴力1036 Tantrix1037 Guardian of Decency 图论：匈牙利算法求二分图的最大匹配1038 Up the Stairs 简单数学题1039 Sudoku 搜索：DFS1040 The SetStack Computer1041 Pie 二分法1042 Ticket to Ride 动态规划1043 The Bookcase 动态规划1044 Printer Queue 模拟题1045 Prime Path 搜索：BFS1046 Lineland's Airport1047 Leonardo's Notebook 数学题：群置换1048 简易版最长序列简单题1049 Jesse's problem 搜索：DFS1050 Error Correction 模拟题1051 A ×B problem 高精度计算1052 Redraiment的走法动态规划1053 Word Encoding 动态规划1054 Jesse's Code 组合数学：排列1055 简单密码破解水题1056 英文金曲大赛水题1057 有假币水题1058 寄居蟹与海葵水题1059 天仙配水题1060 鹊桥相会水题1061 杨辉三角水题1062 蟠桃记水题1063 养兔子水题1064 字符统计水题1065 完美数水题1066 亲和数水题1067 成绩评估水题1068 找零钱水题1069 漂亮菱形水题1070 Least Common Multiple 水题1071 第几天水题1072 编辑距离水题1073 支配值数目水题1074 等值数目水题1075 两数组最短距离水题1076 输入入门(1) 水题1077 输入入门(2) 水题1078 输入入门(3) 水题1079 输出入门水题1080 Counterfeit Dollar 组合数学1081 Dividing 动态规划1082 Sorting It All Out 图论：拓扑排序1083 False coin 暴力法1084 File Mapping1085 Color Me Less 简单题1086 Round and Round We Go 简单题1087 Microprocessor Simulation 简单题1088 求奇数的乘积水题1089 平方和与立方和水题1090 绝对值排序水题1091 JudgeOnline 水题1092 More Beautiful 水题1093 猴子分桃水题1094 C语言实验题——一元二次方程水题1095 C语言实验题——保留字母水题1096 C语言实验题——排列水题1097 C语言实验题——矩阵转置水题1098 C语言实验题——素数水题1099 Ambiguous permutations 简单题1100 Home Work 贪心法1101 Redraiment的遭遇数学题：找规律1102 Decorate the wall 搜索or动态规划1103 Economic phone calls 动态规划or贪心1104 Any fool can do it 记忆化搜索1105 Wine trading in Gergovia 贪心法1106 Homogeneous squares 随机算法1107 Automatic Correction of Misspellings 字符串处理：字典序1108 Black and white painting 简单数学题1109 Cylinder 计算几何：公式推导1110 Deli Deli 水题1111 Expressions 数据结构：树的遍历1112 Flavius Josephus Reloaded 数论：Pollard's Rho算法1113 Annoying painting tool 贪心法1114 Frequent values RMQ区间最值问题OR 线段树1115 Anagram Groups 字符串匹配1116 Let it Bead 组合数学->Polya定理1117 Simple Computers 简单题1118 Mondriaan's Dream 动态规划1119 Equidistance 计算几何1120 How many Fibs? 高精度计算1121 Hike on a Graph 搜索：BFS1122 ASCII Art1123 Billing Tables1124 Cellular Automaton 矩阵计算1125 Exchange1126 Fool's Game1127 Java vs C++ 字符串处理1128 Kickdown 字符串处理1129 Copying Books 贪心+二分法1130 Adding Reversed Numbers 简单题1131 Glass Beads 字符串的最小表示1132 The Circumference of the Circle 计算几何题1133 Knight Moves 搜索：BFS1134 Eeny Meeny Moo 变形的约瑟夫问题1135 Lotto 组合数学1136 Humble Numbers 动态规划1137 Average is not Fast Enough! 简单题1138 Etaoin Shrdlu 简单题1139 Hard to Believe, but True! 简单题1140 Code the Tree 简单题1141 Fiber Network 图论：全源最短路径，Floyd-Warshall算法1142 Global Roaming 3D几何题1143 All in All 字符串处理1144 The Sierpinski Fractal 递归1145 Assistance Required 简单题：预处理1146 Drink, on Ice 模拟题1147 All Discs Considered 搜索：BFS1148 In Danger 模拟题1149 Run Length Encoding 字符串处理1150 Bee Maja 模拟题1151 Friends 表达式求值1152 John 博弈论1153 Double Queue 最大堆与最小堆1154 ‘JBC’1155 Loan Scheduling 贪心+堆1156 Showstopper1157 Highway 贪心法1158 Computers 动态规划1159 The Stable Marriage Problem 组合数学1160 Arne Saknussemm 模拟题1161 Sum Problem 水题1162 Fire Net 搜索题1163 统计1到N之间数字1的个数推理题1164 最大公因子水题1165 C语言实验题——三个整数水题1166 C语言实验题——大小写转换水题1167 C语言实验题——分数序列水题1168 C语言实验题——最值水题1169 C语言实验题——保留整数水题1170 C语言实验题——矩阵下三角元素之和水题1171 C语言实验题——字符逆序水题1172 C语言实验题——打印菱形水题1173 C语言实验题——分割整数水题1174 C语言实验题——删除指定字符水题1175 C语言实验题——时间间隔水题1176 C语言实验题——数组逆序水题1177 C语言实验题——打印数字图形水题1178 C语言实验题——单词统计水题1179 C语言实验题——最小公倍数和最大公约数水题1180 Crashing Balloon 搜索题1181 念数字模拟题1182 A+B for Input-Output Practice(1) 水题1183 Anagrams by Stack 搜索：回溯1184 Elevator 数学：找规律1185 Substrings 字符串处理1186 Calling Extraterrestrial Intelligence Again 搜索：枚举法1187 Do the Untwist 简单数学题1188 数字对水题1189 A+B for Input-Output Practice (2) 水题1190 火星A+B 简单题1191 三齿轮问题：三个齿轮啮合简单数学题1192 A + B Problem II 高精度计算1193 The ones to remain 数学题1194 Chinese Chess 博弈论1195 Page Replacement 数据结构：队列or hash1196 RSA Signing 数论：Pollard's Rho算法1197 Number Guessing 搜索：穷举1198 求n的阶乘高精度计算1199 Area 计算几何1200 求两直线的夹角水题1201 三角形面积水题1202 Max Sum 动态规划1203 Number Sequence 大数问题1204 u Calculate e 水题1205 斐波那契数列高精度计算1206 Fibonacci Again 大数问题1207 Let the Balloon Rise 字符串处理1208 还是A+B 水题1209 A + B 水题1210 The area 简单计算几何1211 Ignatius's puzzle 简单数学问题1212 Computer Transformation 高精度计算1213 N! 高精度计算1217 Text Reverse 水题1220 填数字游戏搜索：DFS1221 Tempter of the Bone 搜索：DFS or BFS+剪枝1226 Last non-zero Digit in N! 数论1227 三角形递推求解1228 回文数猜想简单题1229 Factorial 简单题1230 Specialized Four-Digit Numbers 简单数学题1231 Lowest Bit 简单题1232 To and Fro 简单题1233 AC Me 简单题1234 Wolf and Rabbit 数论1235 最大连续子序列动态规划1236 开门人和关门人字符串处理1237 排名排序1238 统计难题字符串处理：字典树1239 Tick and Tick 数学题1240 Quoit Design 分治法1241 钱币兑换问题递推求解1242 求出前m大的数简单题1243 角谷猜想简单题1244 Reverse Number 简单题1245 寻找素数对简单题1246 ZJUTACM 简单题1247 Hat's Fibonacci 高精度计算1248 Encoding 简单题1249 四数相加高精度计算1250 两数相减高精度计算1251 Square Coins 母函数1252 Counting Triangles 递推求解1253 2^x mod n = 1 数论：费尔马小定理1254 Minimum Inversion Number 简单题1255 Surround the Trees 计算几何：凸包1256 Number Steps 简单题1257 Binary Numbers 简单题1258 Knight Moves 搜索：BFS1259 Lotto 组合数学1260 A Simple Task 简单题1261 The Drunk Jailer 数论1262 Hanoi Tower Troubles Again! 递推求解1263 IBM Minus One 水题1264 Definite Values 简单题1265 Box of Bricks 水题1266 Perfection 简单题1267 Reverse Text 水题1268 Inversion 模拟题1269 Prime Cuts 简单题1270 How Many Fibs? 高精度计算1271 Round and Round We Go 简单题1272 Red and Black 搜索：DFS1273 What Day Is It? 简单题1274 String Matching 字符串匹配1275 A Contesting Decision 简单题1276 Doubles 简单题1277 The Snail 简单题1278 Jungle Roads 图论：最小生成树1279 Prime Ring Problem 搜索：DFS1280 Big Number 大数问题1281 Least Common Multiple 简单题1283 简单排序水题1284 Gridland 简单题1285 An Easy Task 简单题1286 Calendar Game 模拟题1287 Human Gene Functions 动态规划1288 计算几何练习题——线段相交计算几何1289 计算几何练习题——线段相交II 计算几何1290 计算几何练习题——直线交点计算几何1291 Trees Made to Order 递归求解1292 排序简单题1293 18岁生日简单题1294 吃糖果递推求解1295 变种汉诺塔递推求解1296 洗牌递推求解1297 大数求余数论1298 圆桌会议递推求解1299 畅通工程并查集1300 还是畅通工程最小生成树1301 统计同成绩学生人数水题1302 简单计算器表达式求值：栈的应用1303 改进版计算器表达式求值：栈的应用1304 FatMouse' Trade 贪心法1305 Digital Roots 大数问题1306 Uniform Generator 数论1307 A Mathematical Curiosity 穷举法1308 Safecracker 穷举法1309 The 3n + 1 problem 简单题1310 分享糖果模拟题1311 宝物收集搜索：BFS1312 Climbing Worm 简单题1313 搬桌子贪心法1314 Humble Numbers 动态规划1315 Dividing 动态规划1316 Rightmost Digit 数学问题1317 Leftmost Digit 数学问题1318 Hangover 简单数学问题1319 Exponentiation 高精度计算1320 I Think I Need a Houseboat 简单题1321 Girls and Boys DFS+二分图1322 Monkey and Banana 动态规划1323 买牛奶简单题1324 Matrix Chain Multiplication 数据结构：栈的应用1325 计算成绩简单题1326 Holding Bin-Laden Captive! 母函数1327 You can Solve a Geometry Problem too 计算几何1328 Super Jumping! Jumping! Jumping! 动态规划1329 a^b 数论1330 计算GPA 水题1331 Give me an offer! 动态规划：0-1背包1332 田忌赛马贪心法1333 Asteroids! 搜索：BFS1334 Oil Deposits 搜索：DFS1335 营救天使搜索：BFS1336 小数化分数高精度计算1337 I Hate It 线段树1338 Strange Billboard 位运算+枚举1339 Frobenius 递推求解1340 奇怪的公式数学题1341 Fibonacci again and again 博弈论1342 A New Tetris Game 博弈论1343 Sum It Up 搜索：DFS1344 速算24点搜索1345 推箱子搜索：BFS1346 Pushing Boxes 搜索：BFS1347 The Worm Turns 搜索1348 Alfredo's Pizza Restaurant 简单题1349 Broken Keyboard 字符串处理1350 Convert Kilometers to Miles 简单题1351 单词数水题1352 仙人球的残影简单题1353 Family planning 简单题1354 Rout 66 简单题1355 LC-Display 模拟题1356 A == B ? 高精度计算1357 不容易系列之一递推求解1358 折线分割平面递推求解1359 find the nth digit 二分查找1360 奇数阶魔方(II) 简单题1361 Keep on Truckin' 简单题1362 Factstone Benchmark 简单题1363 Destroy the Well of Life 模拟题1365 Brave Game 博弈论1366 ASCII码排序水题1367 计算两点间的距离水题1368 计算球体积水题1369 求绝对值水题1370 数值统计水题1371 求数列的和水题1372 水仙花数水题1373 多项式求和水题1374 素数判定水题1375 偶数求和水题1376 母牛的故事水题1377 数列有序! 水题1378 发工资咯：）水题1379 C语言合法标识符水题1380 海选女主角水题1381 查找最大元素水题1382 首字母变大写水题1383 统计元音水题1384 Palindromes _easy version 水题1385 汉字统计水题1386 进制转换水题1387 人见人爱A+B 水题1388 人见人爱A-B 水题1389 人见人爱A^B 水题1390 改革春风吹满地计算几何1391 今年暑假不AC 动态规划1392 三角形水题1393 求平均成绩水题1394 不容易系列之二递推求解1395 密码水题1396 一只小蜜蜂... 递推求解1397 不容易系列之(3)——LELE的RPG难题递推求解1398 骨牌铺方格递推求解1399 阿牛的EOF牛肉串递推求解1400 神、上帝以及老天爷递推求解1401 不容易系列之(4)——考新郎递推求解1402 Bitset 简单题1403 Picture 简单模拟题1404 Switch Game 找规律1405 An easy problem 简单模拟题1406 A + B Again 简单题1407 The sum problem 简单数学题1408 龟兔赛跑动态规划1409 Snooker 简单数学题1410 Subset sequence 简单题1411 汉诺塔III 递推求解1412 "红色病毒"问题递推求解1413 小兔的棋盘递推求解1414 RPG的错排错排+排列组合1415 无限的路简单题1416 夹角有多大数学题1417 汉诺塔IV 递推求解1418 复习时间简单题1419 选课时间暴力求解1420 手机短号字符串处理1421 找单词母函数1422 简易版之最短距离数学题1423 数塔动态规划1424 核反应堆简单题1425 A1 = ? 公式推导1426 剪花布条字符串处理1427 不要62 数学题1428 空心三角形字符串处理1429 小明A+B 简单题1430 Sky数进制转换1431 整除的尾数简单题1432 分拆素数和数论1433 正整数解数学题1434 挂盐水模拟题1435 {A} + {B} 简单题1436 小数A+B 高精度计算1437 Zigzag 简单题1438 螺旋形简单题1439 行李寄存简单题1440 判断多边形凹凸计算几何1441 The centre of polygon 计算几何1442 最小正整数简单题1443 Elevator Stopping Plan 二分+贪心法1444 TOYS 计算几何1445 The Doors 计算几何1446 Polygon And Segment 计算几何1447 Fence 计算几何1448 两圆相交面积计算几何1449 Area of Circles 计算几何1450 Pipe 计算几何1451 zero sum 搜索：DFS1452 C语言实验题——Hello World 水题1453 C语言实验题——数日子水题1454 C语言实验题——三个数排序水题1455 C语言实验题——数字串求和水题1456 C语言实验题——拍皮球水题1457 C语言实验题——求一个3*3矩阵对角线元素之和水题1458 C语言实验题——数组逆序水题1459 C实验题——求最大值水题1460 C实验题——求绝对值最大值水题1461 C语言实验题——求平均值水题1462 C语言实验题——打印直角三角形水题1463 C语言实验题——相加和最大值水题1464 C语言实验题——简单编码水题1465 C语言实验题——某年某月的天数水题1466 C语言实验题——各位数字之和排序水题1467 C语言实验题——两个数最大水题1468 C语言实验题——求级数值水题1469 Pipe II 计算几何1470 Transmitters 计算几何1471 Wall 计算几何1472 C语言实验题——逆置正整数水题1473 C语言实验题——找中间数水题1474 C语言实验题——整数位水题1475 C语言实验题——一元二次方程II 水题1476 C语言实验题——圆周率水题1477 C语言实验题——余弦水题1478 C语言实验题——打印金字塔水题1479 C语言实验题——排序水题1480 C语言实验题——约瑟夫问题水题1481 C语言实验题——鞍点水题1482 C语言实验题——计算表达式水题1483 C语言实验题——汉诺塔水题1484 C语言实验题——字符串排序水题1485 C语言实验题——整除水题1486 Solitaire 搜索：（双向）BFS1487 Abbreviation 水题1488 C语言实验题——买糖果水题1489 C语言实验题——字符编码水题1490 C语言实验题——合法的C标识符水题1491 C语言实验题——三角形面积水题1492 C语言实验题——大小写转换水题1493 C语言实验题——圆柱体计算水题1494 C语言实验题——温度转换水题1495 C语言实验题——统计字串水题1496 C语言实验题——字符过滤水题1497 Coin Change 暴力求解1498 Beautiful Meadow 搜索题1499 C语言实验题——鸡兔同笼水题1500 Coins of Luck 数学题：数学期望1501 Friends 搜索：DFS1502 Find All M^N Please 数学题1503 Incredible Cows 搜索：二分+DFS1504 计算直线的交点数递推求解1505 Number Game 动态规划1506 Sort ZOJ7 字符串处理1507 Find 7 Faster Than John Von Neumann 高精度计算1508 免费馅饼动态规划1509 Worm 动态规划1510 Common Subsequence 动态规划1511 搬寝室动态规划1512 Daydream 字符串处理1513 Ballroom Lights1514 Drop the Triples1515 Finding Seats1516 He is offside!1517 Justice League1518 星星点点搜索1519 逆波兰表达式表达式求解：栈的应用1520 十六进制高精度计算1521 Palindromic sequence1522 Hotel 模拟题1523 Intersecting Lines 计算几何1524 Heap Construction 最短路径1525 Pizza Anyone?1526 Adam's Genes1527 Risk1528 Just the Facts 数论1529 Horse Shoe Scoring 计算几何1530 哥德巴赫猜想数论1531 爱的伟大意义简单题1532 校门外的树模拟题1533 最多约数问题数论1534 Quicksum 数学题1535 找规律填数字数学题1536 Accepted Necklace 搜索：DFS1537 除法表达式数论1538 A Walk Through the Forest 图论：最短路径1539 Accurately Say "CocaCola"! 简单题1540 Build The Electric System 图论：最小生成树1541 Colorful Rainbows 计算几何1542 Easy Task 数学题1543 Faster, Higher, Stronger 简单题1544 Give Me the Number 模拟题1545 Hurdles of 110m 动态规划1546 Just Pour the Water 矩阵计算1547 Kinds of Fuwas 穷举法1548 复数运算简单题1549 元素个数排序简单题1550 Fiber Communications1551 Power Hungry Cows 搜索：BFS1552 Cow Cycling 动态规划1553 Rebuilding Roads 树型DP1554 Triangular Pastures 动态规划1555 Chores 动态规划1556 Extra Krunch1557 BUY LOW, BUY LOWER 动态规划1558 Hypnotic Milk Improvement1559 Happy Cows1560 Unary Cow Counting1561 Dairy Route1562 Calf Numbers1563 Hide and Seek1564 Mountain Majesties1565 Secret Milk Pipes1566 Circus Tickets1567 Life Cycle1568 Wiggle Numbers1569 Superwords1570 Cow Brainiacs1571 Pasture Fences1572 New Years Party1573 Strolling Cows1574 Grazing Sets1575 Factorial Power1576 Friday the Thirteenth1577 Beef McNuggets1578 Calf Flac1579 Light Bulbs1580 Cow Math 图论1581 Cow Imposters 动态规划1582 Traffic Lights 递推求解1583 Farm Tour 图论：最短路径1584 Vertical Histogram 简单题1585 Cowties 动态规划1586 Travel Games 搜索：DFS1587 Best Cow Fences 二分法1588 Cornfields RMQ问题1589 Six Degrees of Cowvin Bacon 简单题1590 Herd Sums 简单题1591 Message Decoding 简单题1592 Mountain Walking 二分+flood fill1593 Millenium Leapcow 动态规划1594 Optimal Milking 最大流+二分法1595 Bale Figures 模拟+二分法1596 Jumping Cows 动态规划1597 Lost Cows SBT树1598 Bovine Math Geniuses 简单题1599 Dividing the Path 动态规划1600 Fence Obstacle Course 动态规划1601 Cow Ski Area 图论：flood fill1602 Cleaning Shifts 贪心法1603 Bad Cowtractors 最大生成树1604 Tree Cutting 树状动态规划1605 Navigation Nightmare 并查集1606 Cow Marathon 树状动态规划1607 Distance Queries LCA，tarjan算法1608 Distance Statistics 楼天成大牛“男人八题”中的一道1609 Moo University - Team Tryouts 排序+穷举法1610 Moo University - Emergency Pizza Order1611 Moo University - Financial Aid 最大堆、最小堆1612 Cube Stacking 并查集1613 The Cow Lineup 穷举法1614 MooFest 线段树1615 Turning in Homework 动态规划1616 Alignment of the Planets1617 Finding Bovine Roots1618 Cow Bowling1619 Cow Patterns 字符串匹配的扩展1620 Barn Expansion 二分查找1621 Layout 差分约束系统1622 Knights of Ni 搜索：BFS1623 Cleaning Shifts DP＋Heap1624 Scales 搜索+剪枝1625 Secret Milking Machine 二分+网络流1626 Aggressive cows 二分法1627 Rigging the Bovine Election 穷举法1628 Feed Accounting 简单模拟题1629 Muddy Fields 穷举法1630 The Wedding Juicer 堆+flood fill1631 Naptime 动态规划1632 Sumsets 动态规划1633 Moo Volume 简单题1634 Ombrophobic Bovines Floyd-Warshall 1635 Space Elevator 动态规划1636 Yogurt factory 动态规划1637 Checking an Alibi 最短路径1638 Out of Hay1639 Satellite Photographs 搜索：BFS or DFS 1640 Asteroids 最大网络流1641 Grazing on the Run 动态规划1642 Walk the Talk 动态规划1643 City Skyline 栈的应用1644 Cow Acrobats 贪心法1645 Ant Counting 动态规划1646 Hopscotch 搜索：DFS1647 Securing the Barn 穷举法1648 Bovine Birthday 递推求解1649 Max Factor 简单题1650 Flying Right1651 Close Encounter1652 Allowance1653 Lazy Cows1654 Expedition1655 Around the world1656 Landscaping1657 Waves1658 Navigating the City1659 Disease Management1660 Muddy roads1661 Wormholes 最短路径1662 The Fewest Coins 动态规划1663 Milk Patterns 二分法or后缀树1664 Cow Picnic 搜索：BFS or DFS1665 Cow Roller Coaster 动态规划1666 River Hopscotch 二分法+贪心1667 The Moronic Cowmpouter 进制转换1668 DNA Assembly 穷举法1669 Cow Phrasebook 二分法1670 Cellphones 穷举法1671 Steady Cow Assignment 网络流1672 Treats for the Cows 动态规划1673 Backward Digit Sums 穷举法1674 Stump Removal 简单题1675 Finicky Grazers 动态规划1676 The Water Bowls 枚举二进制位1677 Redundant Paths 图论1678 Roping the Field 动态规划1679 Corral the Cows 二分法1680 The Cow Prom 图论1681 Dollar Dayz 动态规划1682 The Grove 最短路径1683 Fence Repair Huffman编码1684 Corn Fields 状态压缩DP1685 Roadblocks 图论：最短路径1686 Bad Hair Day 搜索1687 Big Square 穷举法1688 Round Numbers 枚举二进制位1689 Building A New Barn1690 Cow Sorting 置换群1691 Lilypad Pond 最短路径1692 The Cow Lexicon 动态规划1693 Silver Cow Party 最短路径1694 Problem Solving 动态规划1695 Cow School1696 Protecting the Flowers 贪心法1697 Tallest Cow 区间统计1698 Balanced Lineup RMQ问题1699 Gold Balanced Lineup RMQ问题1700 Ranking the Cows 搜索：DFS1701 Face The Right Way 穷举法1702 Cow Traffic 动态规划1703 Monthly Expense 贪心法1704 Cheapest Palindrome 动态规划1705 Dining 贪心+网络流1706 City Horizon 离散化+ 扫描1707 Catch That Cow 最短路径1708 Fliptile 枚举+位压缩1709 2-Dimensional Rubik's Cube 搜索：BFS 1710 Ball 计算几何1711 3D Camera 三维计算几何1712 Cipher 模拟题1713 Five in a Row 简单题1714 Pinhole Imaging 简单计算几何1715 URL 模拟题1716 Battle of Submarines 集合DP1717 WOJ 动态规划1718 钥匙计数之二递推求解1719 BrokenLED 模拟题1722 A+B again and again! 模拟题1723 Just calculate it! 数论1724 Guess how much I love you? 简单题1725 NBA Finals1726 Find Out an “E”1727 Judging ACM/ICPC1728 Cryptography of Alex1729 Rings of square grid1730 Fermat's Theorem1731 Cup 二分法1732 Find the Path DP+二分法1733 Five in a Row, Again 动态规划1734 Minimum Heap 递推求解1735 Name PK 模拟题1736 Pendant 动态规划1737 Radar 计算几何+搜索1738 Ring 多串模式匹配1739 Run 计算几何1740 Toxophily 简单题1741 通讯录编排简单题1742 超缘分ACM队伍简单题1743 集合运算简单题1744 矩阵计算简单题1745 Arbitrage 动态规划1746 The Tower of Babylon 动态规划1747 Binomial Showdown 组合数学1748 Dungeon Master 搜索：BFS1749 Equation Solver 表达式求值应用1750 Frogger 最短路径1751 Globetrotter 计算几何1752 Tree Recovery 数据结构：二叉树1753 Artificial Intelligence?1754 The Settlers of Catan 搜索1755 France '98 概率问题1756 Goldbach's Conjecture 数论1757 Heavy Cargo 最小生成树1758 Quadtree1759 From Dusk till Dawn or: Vladimir the Vampire 最短路径1760 Euro Cup 20001761 Quadtree II or: Florida Jones strikes back1762 HTML 简单题1763 Paths on a Grid 组合数学：T路问题1764 Balanced Food 动态规划1765 California Jones and the Gate to Freedom 组合数学1766 Diplomatic License 简单计算几何题1767 Polygon Programming with Ease 数学题1768 Hall of Fountains 搜索：BFS or DP1769 The Bottom of a Graph 图论：强连通分量1770 Edge1771 Fold1772 Largest Rectangle in a Histogram 动态规划1773 Boolean Logic1774 Code1775 In Danger 模拟题1776 Fractran1777 Huffman's Greed1778 Bullshit Bingo 字符串处理1779 A Song contest1780 Message1781 The skatepark's new ramps1782 Road1783 Warfare1784 Blackjack1785 Robintron1786 Diamond Dealer 计算几何：凸包1787 Best Compression Ever1788 Code Theft1789 Dinner1790 Event Planning1791 Getting Gold1792 Introspective Caching1793 Just A Few More Triangles!1794 Knights of the Round Table 图论：无向图的块判断奇圈1795 The Cow Doctor 穷举法1796 Wild West 线段树1797 Find the Clones1798 The Warehouse1799 Widget Factory 数论：同余方程组1800 Martian Mining 动态规划3301 字符串；AC自动机，动态规划；状态压缩3302 计算几何3303 数学；代数运算；高斯消元3304 图论；强连通分量；2-SAT3305 动态规划；凸单调性优化3306 枚举3307 贪心3308 数学；代数运算3309 最短路；佛洛伊德3310 动态规划3311 贪心3312 计数问题；递推，数状数组，二分查找3313 数论；欧拉定理，快速幂取模3314 计数问题，数状数组3315 博弈；Surreal数；Farey数列；3316 计数问题；递推，高精度3317 计数问题；容斥原理3318 递推；矩阵乘法3319 数学；概率3320 背包3321 动态规划3322 字符串；AC自动机3323 动态规划3324 博弈3325 搜索3326 贪心3327 最短路3328 数据结构（实现一种数据结构，支持要求的操作），数状数组3329 图论；二分图最大权匹配3330 数学；数论3331 递推；矩阵乘法3332 数学；数论，二分查找3333 计算几何3334 动态规划3335 字符串，后缀数组或拉宾卡普；动态规划3336 数据结构；并查集3337 计数问题，递推3338 二分查找，贪心3339 数学3340 计算几何；凸包，图论；佛洛伊德；最小环3341 动态规划3342 广搜3343 动态规划3344 计算几何3345 二分图最大匹配3346 树型DP3347 动态规划3348 数学；数论；进制3349 计数问题3350 贪心3351 数学；数论；进制3352 动态规划，数论，组合数学3353 数学；数论3354 计数；递推3355 图论；佛洛伊德3356 博弈3357 动态规划3358 数据结构；线段树，数状数组3359 计算几何，动态规划3360 博弈；SG函数3361 图论；最近公共祖先3362 图论；强连通分量；2－SAT 3363 计算几何3364 字符串；AC自动机，动态规划3365 搜索，舞蹈链3366 数学；数论3367 数学；代数运算；高斯消元3368 动态规划3369 计数问题；递推3370 网络流（错题）3371 树型DP3372 数学；高精度3373 数学；3374 RMQ3376 数学；进制3377 字符串；后缀数组3378 动态规划3379 计算几何3380 线段树3381 图论；欧拉路3382 简单题3383 字符串；AC自动机3384 广搜3385 计算几何，矩阵3386 语言处理3387 动态规划；状态压缩3388 图论；全局最小割3389 简单题3390 广搜3391 数学；Pell方程3392 背包3393 计算几何3394 广搜3395 搜索；迭代加深3396 数学；计数问题3397 数学；解方程3398 分析3399 模拟3400 数学；计数问题，数论6 热度。