求助——A + B Problem II

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本文详细解析了一个解决大数加法问题的算法，通过字符串处理和逐位相加的方法，实现了两个大数的加法运算。代码示例清晰展示了如何处理超过32位整数的加法，适用于如HDU ACM 1002题等竞赛场景。

http://acm.hdu.edu.cn/showproblem.php?pid=1002

感觉是对的但是一直wrong，哪位好心人帮我看看呀。代码在下面！

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 458574 Accepted Submission(s): 88799

Problem Description

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input

The 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.

Output

For 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 Input

2 1 2 112233445566778899 998877665544332211

Sample Output

Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

Author

Ignatius.L

#include<stdio.h>
#include<string.h>

int main(){
	char a[1000],b[1000];
	int  p[1000]={0},q[1000]={0};
	int i,j,n,m,x,y=0;
	scanf("%d",&x);
	while(y++<x)
	{	
		memset(a, 0, sizeof(a));
		memset(b, 0, sizeof(b));
		memset(p, 0, sizeof(p));
		memset(q, 0, sizeof(q));
		scanf("%s",&a);
		scanf("%s",&b);
		printf("case %d:\n",y);
		printf("%s ",a);
		printf("+ ");
		printf("%s ",b);
		printf("= ");
		n=strlen(a);
		m=strlen(b);
		j=0;
		for(i=n-1;i>=0;i--)
		{
			p[j++]=a[i]-'0';
		}
		j=0;
		for(i=m-1;i>=0;i--)
		{
			q[j++]=b[i]-'0';
		}
		for(i=0;i<205;i++)
		{
			p[i]=p[i]+q[i];
			if(p[i]>=10)
			{
				p[i]-=10;
				p[i+1]+=1;
			}
		}
	    for(i=204;i>=0;i--)
		{  
			if(p[i]!=0)
			{
				for(j=i;j>=0;j--)
				{
				  printf("%d",p[j]);
				}
	    	break;
			}	
		}
		if(i==-1) printf("0");
		if(y<x) printf("\n");
		printf("\n");
	}
	return 0;
}