A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 242959    Accepted Submission(s):
46863

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

注意给数组清零

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
    int m,j,n,i,l1,l2,t,k=1;
    char s1[1100];
    char s2[1100];
    int s3[1100];
    int s4[1100];
	scanf("%d",&n);
	while(n--)
	{
		scanf("%s %s",s1,s2);
		memset(s3,0,sizeof(s3));  //给数组清0
		memset(s4,0,sizeof(s4));
		l1=strlen(s1);
		l2=strlen(s2);
		t=0;
		for(j=l1-1,i=0;j>=0;j--,i++)
		s3[i]=s1[j]-‘0‘;                  //将字符串转换为数字
		for(j=l2-1,i=0;j>=0;i++,j--)
		s4[i]=s2[j]-‘0‘;
		for(i=0;i<1100;i++)
		{
			s3[i]+=s4[i];
			if(s3[i]>=10)                //对各位求和
			{
				s3[i]-=10;
				s3[i+1]++;
			}
		}
		printf("Case %d:\n",k++);
		printf("%s + %s = ",s1,s2);
		for(i=1099;i>=0;i--)
		if(s3[i]!=0)
		break;
		for(;i>=0;i--)
		printf("%d",s3[i]);
		printf("\n");
		if(n>0)
		printf("\n");
	}
	return 0;
}