以特定格式输出两个数的和

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

参考：http://blog.csdn.net/hanangellove/article/details/42486723

//一看就是ACM题，贴上AC代码

#include<stdio.h>
#include<string.h>
int main()
{
    char a[1000],b[1000],c[1001];
    int i,j=1,p=0,n,n1,n2;
    scanf("%d",&n);
  while(n--)
    {
        scanf("%s %s",a,b);
        printf("Case %d:\n",j);
        printf("%s + %s = ",a,b);
        n1=strlen(a)-1;
        n2=strlen(b)-1;
        for(i=0;n1>=0||n2>=0;i++,n1--,n2--)
        {
            if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]-'0'+p;}
            if(n1>=0&&n2<0){c[i]=a[n1]+p;}
            if(n1<0&&n2>=0){c[i]=b[n2]+p;}
            p=0;
            if(c[i]>'9'){c[i]=c[i]-10;p=1;}
        }
        if(p==1)  printf("%d",p);
        while(i--)
            printf("%c",c[i]);
        j++;
        if(n!=1) printf("\n\n");
        else printf("\n");
    }
}