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[],b[],c[];
     int T,T1;
     scanf("%d",&T);
     T1=T;
     getchar();
     while(T--)
     {
         int alen,blen,clen,i,j,k=,p=;
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         memset(c,,sizeof(c));
         scanf("%s%s",a,b);
         alen=strlen(a),blen=strlen(b);
         for(i=alen-,j=blen-;;)
         {
             c[k]=(a[i]+b[j]-''-''+p)%+'';
             p=(a[i]+b[j]-''-''+p)/;
             if(i==&&j==)
             {
                 if(p>)
                 {
                     c[++k]='';
                     break;
                 }
                 break;
             }
             if(i==)
             {
                 while(j>)
                 {
                     c[++k]=(p+b[--j]-'')%+'';
                     p=(p+b[j]-'')/;
                 }
                 break;
             }
             if(j==)
             {
                 while(i>)
                 {
                     c[++k]=(p+a[--i]-'')%+'';
                     p=(p+a[i]-'')/;
                 }
                 break;
             }
             i--,j--,k++;
         }
         printf("Case %d:\n",T1-T);
         for(i=;i<alen;i++)
             printf("%c",a[i]);
         printf(" + ");
         for(i=;i<blen;i++)
             printf("%c",b[i]);
         printf(" = ");
         for(i=k;i>=;i--)
             printf("%c",c[i]);
         if(T>)
             printf("\n\n");
         else
         printf("\n");
     }
 }