A+B II

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

#include<stdio.h>
#include<string.h>
int main()
{
    char a[1000],b[1000],c[1001];
    int T,T1;
    scanf("%d",&T);
    T1=T;
    getchar();
    while(T--)
    {
        int alen,blen,clen,i,j,k=0,p=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        scanf("%s%s",a,b);
        alen=strlen(a),blen=strlen(b);
        for(i=alen-1,j=blen-1;;)
        {
            c[k]=(a[i]+b[j]-'0'-'0'+p)%10+'0';
            p=(a[i]+b[j]-'0'-'0'+p)/10;
            if(i==0&&j==0)
            {
                if(p>0)
                {    
                    c[++k]='1';
                    break;
                }
                break;
            }
            if(i==0)
            {
                while(j>0)
                {
                    c[++k]=(p+b[--j]-'0')%10+'0';
                    p=(p+b[j]-'0')/10;
                }
                break;
            }
            if(j==0)
            {
                while(i>0)
                {
                    c[++k]=(p+a[--i]-'0')%10+'0';
                    p=(p+a[i]-'0')/10;
                }
                break;
            }
            i--,j--,k++;
        }
        printf("Case %d:
",T1-T);
        for(i=0;i<alen;i++)
            printf("%c",a[i]);
        printf(" + ");
        for(i=0;i<blen;i++)
            printf("%c",b[i]);
        printf(" = "); 
        for(i=k;i>=0;i--)
            printf("%c",c[i]);
        if(T>0)
            printf("

");
        else
        printf("
");
    }
}