hdoj 1047 Integer Inquiry

Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15221    Accepted Submission(s): 3912

Problem 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

 

英语烂真心伤不起，愣是看不懂题目的意思，刚开始以为是首先输入一个整数，表示输入数据的组数

（我以为是每一组三个字符串）当整数为0时不作处理程序结束，结果提交上一直超时；后来看了舍友的题解才知道

此题的意思是：

    输入一个整数代表可以测试数据组数，输入多组字符串，当字符串为0时输入结束，计算所有字符串（超长整数的和）

#include<stdio.h>   
#include<string.h>
#define MAX 510
#define max(x,y)(x>y?x:y)
char s1[MAX];
int a[MAX],b[MAX];
int main()
{
    int n,m,j,i,s,t,l1,k,ok;
    int len;
    scanf("%d",&t);
    while(t--)
    {
    	ok=0;
        memset(b,0,sizeof(b));
        while(scanf("%s",s1))
        {
        	if(s1[0]=='0')
        	break;
        	ok++;
        	memset(a,0,sizeof(a));
            l1=strlen(s1);
            for(i=l1-1,j=0;i>=0;i--)
            {
                a[j]=s1[i]-'0';
                j++;
            }
            for(i=0;i<MAX;i++)
            {
                b[i]+=a[i];
                if(b[i]>=10)
                {
                    b[i]-=10;
                    b[i+1]++;
                }
            }   
        }    
		if(!ok)
		printf("0\n");
		else
		{
			for(i=MAX-1;i>=0;i--)            
            if(b[i]!=0)
            break;
            for(;i>=0;i--)
            printf("%d",b[i]);
            printf("\n");
		}            
        if(t)
        printf("\n");
    }
    return 0;
}