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

 

Sample Output

 

Author

Ignatius.L

 

 #include<stdio.h>

 #include<string.h>

 int main(){

     char a[],b[];

     int c[],d[],ans[];

     int n,i,k,j,h,t,l;

     int len1,len2;

     scanf("%d",&n);

         h=;

         while(n--){

             h+=;

             for(i=;i<;i++){

                 c[i]=d[i]=;

                 ans[i]=;

             }

             scanf("%s %s",&a,&b);

             len1=strlen(a);

             len2=strlen(b);

             for(i=;i<len1;i++){

                c[i]=a[i]-'';

             }

             for(i=;i<len2;i++){

                d[i]=b[i]-'';

             }

             printf("Case %d:\n",h);

             for(i=;i<len1;i++)

                 printf("%d",c[i]);

                 printf(" + ");

             for(i=;i<len2;i++)

                 printf("%d",d[i]);

                 printf(" = ");

             for(i=,j=len1-;i<len1/;i++,j--){

                 t=c[j];

                 c[j]=c[i];

                 c[i]=t;

             }

             for(i=,j=len2-;i<len2/;i++,j--){

                 t=d[j];

                 d[j]=d[i];

                 d[i]=t;

             }

             l=len1>len2?len1:len2;

             for(i=,k=;i<=l;i++,k++){

                 if(c[i]+d[i]<)

                     ans[k]=c[i]+d[i];

                 else{

                     ans[k]=(c[i]+d[i])%;

                     c[i+]+=(c[i]+d[i])/;

                 }

             }

             for(j=k-;j>=;j--){

                 if(j==k-&&ans[j]==)

                     continue;

                 printf("%d",ans[j]);

             }

                 if(n==)

                     printf("\n");

                 else

                     printf("\n\n");

         }

     return ;

 }