1111111杭电：1000 A + B Problem (4)1001 Sum Problem (5)1002 A + B Problem II (6)1005 Number Sequence (8)1008 Elevator (9)1009 FatMouse' Trade (11)1021 Fibonacci Again (13)1089 A+B for Input-Output Practice (I) (14)1090 A+B for Input-Output Practice (II) (15)1091 A+B for Input-Output Practice (III) (16)1092 A+B for Input-Output Practice (IV) (17)1093 A+B for Input-Output Practice (V) (18)1094 A+B for Input-Output Practice (VI) (20)1095 A+B for Input-Output Practice (VII) (21)1096 A+B for Input-Output Practice (VIII) (22)1176 免费馅饼 (23)1204 糖果大战 (25)1213 How Many Tables (26)2000 ASCII码排序 (32)2001 计算两点间的距离 (34)2002 计算球体积 (35)2003 求绝对值 (36)2004 成绩转换 (37)2005 第几天？ (38)2006 求奇数的乘积 (40)2007 平方和与立方和 (41)2008 数值统计 (42)2009 求数列的和 (43)2010 水仙花数 (44)2011 多项式求和 (46)2012 素数判定 (47)2014 青年歌手大奖赛_评委会打分 (49)2015 偶数求和 (50)2016 数据的交换输出 (52)2017 字符串统计 (54)2019 数列有序! (55)2020 绝对值排序 (56)2021 发工资咯：） (58)2033 人见人爱A+B (59)2037 今年暑假不AC (61)2039 三角形 (63)2040 亲和数 (64)2045 不容易系列之(3)—— LELE的RPG难题 (65)2049 不容易系列之(4)——考新郎 (66)2056 Rectangles (68)2073 无限的路 (69)2084 数塔 (71)2201 熊猫阿波的故事 (72)2212 DFS (73)2304 Electrical Outlets (74)2309 ICPC Score Totalizer Software (75)2317 Nasty Hacks (77)2401 Baskets of Gold Coins (78)2500 做一个正气的杭电人 (79)2501 Tiling_easy version (80)2502 月之数 (81)2503 a/b + c/d (82)2504 又见GCD (83)2519 新生晚会 (84)2520 我是菜鸟，我怕谁 (85)2521 反素数 (86)2522 A simple problem (88)2523 SORT AGAIN (89)2524 矩形A + B (90)2535 Vote (91)2537 8球胜负 (93)2539 点球大战 (95)2547 无剑无我 (98)2548 两军交锋 (99)2549 壮志难酬 (100)2550 百步穿杨 (101)2551 竹青遍野 (103)2552 三足鼎立 (104)2553 N皇后问题 (105)2554 N对数的排列问题 (106)2555 人人都能参加第30届校田径运动会了 (107)2560 Buildings (110)2561 第二小整数 (112)2562 奇偶位互换 (113)2563 统计问题 (114)2564 词组缩写 (115)2565 放大的X (117)2566 统计硬币 (118)2567 寻梦 (119)2568 前进 (121)2569 彼岸 (123)2700 Parity (124)2577 How to Type (126)北京大学：1035 Spell checker (129)1061 青蛙的约会 (133)1142 Smith Numbers (136)1200 Crazy Search (139)1811 Prime Test (141)2262 Goldbach's Conjecture (146)2407 Relatives (150)2447 RSA (152)2503 Babelfish (156)2513 Colored Sticks (159)ACM算法：kurXX最小生成树 (163)Prim (164)堆实现最短路 (166)最短路DIJ普通版 (167)floyd (168)BELL_MAN (168)拓扑排序 (169)DFS强连通分支 (170)最大匹配 (172)还有两个最大匹配模板 (173)最大权匹配,KM算法 (175)两种欧拉路 (177)无向图： (177)有向图： (178)【最大流】Edmonds Karp (178)dinic (179)【最小费用最大流】Edmonds Karp对偶算法 (181)ACM题目：【题目】排球队员站位问题 (182)【题目】把自然数Ｎ分解为若干个自然数之和。
(184)【题目】把自然数Ｎ分解为若干个自然数之积。
(185)【题目】马的遍历问题。
(185)【题目】加法分式分解。
(186)【题目】地图着色问题 (189)【题目】放置方案 (191)【题目】找迷宫的最短路径。
(194)【题目】火车调度问题 (195)【题目】农夫过河。
(197)【题目】七段数码管问题。
(199)【题目】求相邻的格的数不连续 (200)【题目】棋盘上放棋子 (202)【题目】迷宫问题. (204)【题目】一笔画问题 (205)【题目】城市遍历问题. (207)【题目】棋子移动问题 (208)【题目】求集合元素问题（1,2x+1,3X+1类) (209)杭电：1000 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 Output2AuthorHDOJ代码：#include<stdio.h>int main(){int a,b;while(scanf("%d %d",&a,&b)!=EOF)printf("%d\n",a+b);}1001 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);}}1002 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 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 = 1111111111111111110AuthorIgnatius.L代码：#include <stdio.h>#include <string.h>int main(){char str1[1001], str2[1001];int t, i, len_str1, len_str2, len_max, num = 1, k; scanf("%d", &t);getchar();while(t--){int a[1001] = {0}, b[1001] = {0}, c[1001] = {0}; scanf("%s", str1);len_str1 = strlen(str1);for(i = 0; i <= len_str1 - 1; ++i)a[i] = str1[len_str1 - 1 - i] - '0';scanf("%s",str2);len_str2 = strlen(str2);for(i = 0; i <= len_str2 - 1; ++i)b[i] = str2[len_str2 - 1 - i] - '0';if(len_str1 > len_str2)len_max = len_str1;elselen_max = len_str2;k = 0;for(i = 0; i <= len_max - 1; ++i){c[i] = (a[i] + b[i] + k) % 10;k = (a[i] + b[i] + k) / 10;}if(k != 0)c[len_max] = 1;printf("Case %d:\n", num);num++;printf("%s + %s = ", str1, str2);if(c[len_max] == 1)printf("1");for(i = len_max - 1; i >= 0; --i){printf("%d", c[i]);}printf("\n");if(t >= 1)printf("\n");}return 0;}1005 Number SequenceProblem DescriptionA number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.Given A, B, and n, you are to calculate the value of f(n).InputThe input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.OutputFor each test case, print the value of f(n) on a single line.Sample Input1 1 31 2 100 0 0Sample Output25AuthorCHEN, ShunbaoSourceZJCPC2004RecommendJGShining代码：#include<stdio.h>int f[200];int main(){int a,b,n,i;while(scanf("%d%d%d",&a,&b,&n)&&a&&b&&n){if(n>=3){f[1]=1;f[2]=1;for(i=3;i<=200;i++){f[i]=(a*f[i-1]+b*f[i-2])%7;if(f[i-1]==1&&f[i]==1)break;}i-=2;n=n%i;if(n==0)printf("%d\n",f[i]);elseprintf("%d\n",f[n]);}elseprintf("1\n");}return 0;}1008 ElevatorProblem DescriptionThe highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.InputThere are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.OutputPrint the total time on a single line for each test case.Sample Input1 23 2 3 1Sample Output1741AuthorZHENG, JianqiangSourceZJCPC2004RecommendJGShining代码：#include<stdio.h>int a[110];int main(){int sum,i,n;while(scanf("%d",&n)&&n!=0){for(i=1;i<=n;i++)scanf("%d",&a[i]);sum=0;a[0]=0;for(i=1;i<=n;i++){if(a[i]>a[i-1])sum+=6*(a[i]-a[i-1]);elsesum+=4*(a[i-1]-a[i]);sum+=5;}printf("%d\n",sum);}return 0;}1009 FatMouse' TradeProblem DescriptionFatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.InputThe input consists of multiple test cases. Each test case begins with a line containing twonon-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.OutputFor each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.Sample Input5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1Sample Output13.333 31.500AuthorCHEN, YueSourceZJCPC2004RecommendJGShining代码：#include<stdio.h>#include<string.h>#define MAX 1000int main(){int i,j,m,n,temp;int J[MAX],F[MAX];double P[MAX];double sum,temp1;scanf("%d%d",&m,&n);while(m!=-1&&n!=-1){sum=0;memset(J,0,MAX*sizeof(int));memset(F,0,MAX*sizeof(int));memset(P,0,MAX*sizeof(double));for(i=0;i<n;i++){ scanf("%d%d",&J[i],&F[i]); P[i]=J[i]*1.0/((double)F[i]); }for(i=0;i<n;i++){for(j=i+1;j<n;j++){if(P[i]<P[j]){temp1=P[i]; P[i]=P[j]; P[j]=temp1;temp=J[i]; J[i]=J[j]; J[j]=temp;temp=F[i]; F[i]=F[j]; F[j]=temp;}}}for(i=0;i<n;i++){if(m<F[i]){ sum+=m/((double)F[i])*J[i]; break; }else { sum+=J[i]; m-=F[i]; }}printf("%.3lf\n",sum); scanf("%d%d",&m,&n);}return 0;}1021 Fibonacci AgainProblem DescriptionThere are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2). InputInput consists of a sequence of lines, each containing an integer n. (n < 1,000,000).OutputPrint the word "yes" if 3 divide evenly into F(n).Print the word "no" if not.Sample Input12345Sample OutputnonoyesnononoAuthorLeojayRecommendJGShining#include<stdio.h>int main(){long n;while(scanf("%ld",&n) != EOF)if (n%8==2 || n%8==6)printf("yes\n");elseprintf("no\n");}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(){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");}}1091 A+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 Output6AuthorlcyRecommendJGShining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);}}1092 A+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]1095 A+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]);}1176 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。