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#include<stdio.h>void main(){int a,b;while(scanf("%d %d",&a,&b)!=EOF){printf("%d\n",a+b);}}Problem 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 Output15050#include<stdio.h>void main(){int n,sum,i;while(scanf("%d",&n)!=EOF){sum=0;for( i=0;i<=n;i++)sum+=i;printf("%d\n\n",sum);}}Problem 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 = 1111111111111111110 #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;}Problem DescriptionGiven a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.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 starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 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 contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.Sample Input25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5Sample OutputCase 1:14 1 4Case 2:7 1 6注：最大子序列是要找出由数组成的一维数组中和最大的连续子序列。

比如{5,-3,4,2}的最大子序列就是{5,-3,4,2}，它的和是8,达到最大；而{5,-6,4,2}的最大子序列是{4,2}，它的和是6。

你已经看出来了，找最大子序列的方法很简单，只要前i项的和还没有小于0那么子序列就一直向后扩展，否则丢弃之前的子序列开始新的子序列，同时我们要记下各个子序列的和，最后找到和最大的子序列#include<stdio.h>int a[100005],str[100005],start[100005];int main(){int t,n,i,num=1,end,max,k;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);}str[1]=a[1];start[1]=1;for(i=2;i<=n;i++){if(str[i-1]>=0){str[i]=str[i-1]+a[i];start[i]=start[i-1];}else{str[i]=a[i];start[i]=i;}}max=str[1];end=1;for(k=2;k<=n;k++){if(str[k]>max){max=str[k];end=k;}}printf("Case %d:\n",num);num++;printf("%d %d %d\n",max,start[end],end);if(t)printf("\n");}return 0;}Problem DescriptionContest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.This year, they decide to leave this lovely job to you.InputInput contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.A test case with N = 0 terminates the input and this test case is not to be processed.OutputFor each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.Sample Input5greenredblueredred3pinkorangepinkSample Outputredpink#include<stdio.h>#include<string.h>char a[1000][15];int b[1000];int main(){int n,i,j,max,k;while(scanf("%d",&n)!=EOF){if(n==0)break;for(i=1;i<=n;i++){b[i]=0;scanf("%s",a[i]);//二维数组的神奇用法}for(i=1;i<=n;i++){for(j=1;j<=n;j++)if(strcmp(a[i],a[j])==0){b[i]++;}}max=b[1];k=1;for(i=2;i<=n;i++){if(b[i]>max){max=b[i];k=i;}}printf("%s\n",a[k]);}return 0;}Problem 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 Output25方法1：注函数介绍void *memset(void *s, int ch, size_t n);函数解释：将s中前n个字节替换为ch并返回s；memset:作用是在一段内存块中填充某个给定的值，它是对较大的结构体或数组进行清零操作的一种最快方法。