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

 1 #include<stdio.h>
 2 #include<string.h>
 3 int main()
 4 {
 5     char a[1000],b[1000],c[1001];
 6     int T,T1;
 7     scanf("%d",&T);
 8     T1=T;
 9     getchar();
10     while(T--)
11     {
12         int alen,blen,clen,i,j,k=0,p=0;
13         memset(a,0,sizeof(a));
14         memset(b,0,sizeof(b));
15         memset(c,0,sizeof(c));
16         scanf("%s%s",a,b);
17         alen=strlen(a),blen=strlen(b);
18         for(i=alen-1,j=blen-1;;)
19         {
20             c[k]=(a[i]+b[j]-'0'-'0'+p)%10+'0';
21             p=(a[i]+b[j]-'0'-'0'+p)/10;
22             if(i==0&&j==0)
23             {
24                 if(p>0)
25                 {    
26                     c[++k]='1';
27                     break;
28                 }
29                 break;
30             }
31             if(i==0)
32             {
33                 while(j>0)
34                 {
35                     c[++k]=(p+b[--j]-'0')%10+'0';
36                     p=(p+b[j]-'0')/10;
37                 }
38                 break;
39             }
40             if(j==0)
41             {
42                 while(i>0)
43                 {
44                     c[++k]=(p+a[--i]-'0')%10+'0';
45                     p=(p+a[i]-'0')/10;
46                 }
47                 break;
48             }
49             i--,j--,k++;
50         }
51         printf("Case %d:\n",T1-T);
52         for(i=0;i<alen;i++)
53             printf("%c",a[i]);
54         printf(" + ");
55         for(i=0;i<blen;i++)
56             printf("%c",b[i]);
57         printf(" = "); 
58         for(i=k;i>=0;i--)
59             printf("%c",c[i]);
60         if(T>0)
61             printf("\n\n");
62         else
63         printf("\n");
64     }
65 }