*大数求和

Problem P
Time Limit: 1000 MS Memory Limit: 32 MB 64bit IO Format: %I64d
Submitted: 107 Accepted: 19
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Description

One of the first users of BIT’s new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.
This supercomputer is great,'' remarked Chip.I only wish Timothy were here to see these results.” (Chip moved to a new apartment,once one became available on the third floor of the Lemon Sky apartments on Third Street.)

Input

The input will consist of at most 100 lines of text,each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length,and will only contain digits (no VeryLongInteger will be negative).

The final input line will contain a single zero on a line by itself.

Output

Your program should output the sum of the VeryLongIntegers given in the input.

This problem contains multiple test cases!

The first line of a multiple input is an integer N,then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

两个字简洁有木有？
大神的代码就是不一样！膜拜膜拜~
我自己写这个程序写了很久，也没写出个什么，看到这个代码，简直觉得整个世界都亮了啊^_^

#include<stdio.h>
#include<string.h>
#define N 110
int main()
{
    char input[N];
    int n,m,sum[N],s[N],i,j,t;
    scanf("%d",&n);
    while(n--)
    {
        memset(sum,0,sizeof(sum));
        while(scanf("%s",input)&&strcmp(input,"0"))//输入0时结束输入
         {                                       //开始对输入的数进行相加
            memset(s,sizeof(s));
            m=strlen(input);
            for(i=0;i<m;i++)
                s[i]=input[m-1-i]-'0';    //将输入的数倒序存放在int s[N]中
            for(i=0;i<m;i++)
            {
                sum[i]+=s[i];
                if(sum[i]>9)
                {
                    sum[i]%=10;
                    sum[i+1]++;           //进位
                }
            }
        }
        for(i=N-1;sum[i]==0&&i>=0;i--)
                j=i;
        if(j==0)
            printf("0");
        for(i=j-1;i>=0;i--)             //倒序输出 正好输出来就是正序了
            printf("%d",sum[i]);
        printf("n");
        if(n>0)
            printf("n");
    }

}