hdu1002 A + B Problem II

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

InputThe 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. 
OutputFor 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

#include<stdio.h>
#include<string.h>
int main()
{
	char a[1001],b[1001];
	int c[1001],d[1001],e[1001],n,i;
	scanf("%d",&n);
	int x = 0;
	while(n--){
		x++;
		scanf("%s%s",a,b);
		memset(c,0,sizeof(c));
		memset(d,0,sizeof(d));
		memset(e,0,sizeof(e));
		int a1 = strlen(a);
		int a2 = strlen(b);
               //将字符串倒序，将字符转化为数字才能进行加法
		int a3 = a1-1,a4 = a2-1;
		for(i = 0; i<a1; i++){
			c[a3] = a[i]-48;//减个'0'就可以转化了
			a3--;
		}
		for(i = 0; i<a2; i++){
			d[a4] = b[i]-48;
			a4--; 
		}
		int l;
		if(a1>a2){
			l = a1;
		}
		else{
			l = a2;
		}
		for(i = 0; i<l; i++){
			e[i] += c[i]+d[i];//对两组字符串中的数逐位相加，存于e[i]
			if(e[i]>=10){// 得到的结果进行进制位处理
				e[i+1]++;
				e[i] = e[i]%10;
			}
		}
		printf("Case %d:
",x);
        printf("%s + %s = ",a,b);
		if(e[l] != 0){
			printf("%d",e[l]);
		}
		for(i = l-1; i>=0; i--){
			printf("%d",e[i]);
		}
		printf("
");
		if(n){
			printf("
");
		}
	}
	return 0;
}