Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10490    Accepted Submission(s): 2706

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#include"stdio.h"
#include"stdlib.h"
#include"string.h"
#define N 105
int a[N];
int main()
{
	int i,j,n,n1,flag;
	char ch[N];
	scanf("%d",&n);
	while(n--)
	{		
		memset(a,0,sizeof(a));      //数组初始化为零
		while(scanf("%s",ch),ch[0]!=‘0‘)     //输入零结束
		{
			n1=strlen(ch);
			for(i=n1-1,j=0;i>=0;i--)
			{
				a[j++]+=ch[i]-‘0‘;         
				a[j]+=a[j-1]/10;         //进位
				a[j-1]%=10;              //取余
			}
		}
		flag=0;
		for(i=N-1;i>0;i--)
		{
			if(flag)
				printf("%d",a[i]);
			else if(a[i])
			{
				printf("%d",a[i]);
				flag=1;
			}
		}
		printf("%d\n",a[0]);         //输入0时应输出0，错了n多次
		if(n)                    //输出块之间有空行
			printf("\n");
	}
	return 0;
}

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hdu 1047 Integer Inquiry

原文：http://blog.csdn.net/u011721440/article/details/19932453