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

套用模版做的。。。

#include<stdio.h>
#include<string.h>
char str1 [ 10000 ],str2 [ 10000 ];
char sum [ 100001 ];
void jiafa ()
{
? ? ? memset (sum , 0 , sizeof (sum ));
??? int i ,j ,p ;
??? int len1 =strlen (str1 ),len2 =strlen (str2 );
??? for (i = 0 ;i <(len1 >len2 ?len1 :len2 );i ++)
??? {
? ? ? ? ? ? ? str1 [i ]-= '0' ;
? ? ? ? ? ? ? str2 [i ]-= '0' ;
??? }
??? for (i = 0 ,j =len1 - 1 ;i <j ;i ++,j --)
??? {
??????? char c =str1 [j ];
? ? ? ? ? ? ? str1 [j ]=str1 [i ];
? ? ? ? ? ? ? str1 [i ]=c ;
??? }
??? for (i = 0 ,j =len2 - 1 ;i <j ;i ++,j --)
??? {
??????? char c =str2 [j ];
? ? ? ? ? ? ? str2 [j ]=str2 [i ];
? ? ? ? ? ? ? str2 [i ]=c ;
??? }
??? char c ;
??? for (i = 0 ,c = 0 ;i <(len1 >len2 ?len1 :len2 )||c ;i ++)
??? {
??????? if (i <len1 )
? ? ? ? ? ? ? ? ? ? ? c +=str1 [i ];
??????? if (i <len2 )
? ? ? ? ? ? ? ? ? ? ? c +=str2 [i ];
? ? ? ? ? ? ? sum [i ]=c % 10 ;
? ? ? ? ? ? ? c /= 10 ;
??? }
??? for (i =(len1 >len2 ?len1 :len2 )- 1 ;i >= 0 ;i --)
? ? ? ? ? ? ? printf ( "%d" ,sum [i ]);
? ? ? printf ( "n" );
}
int main ()
{
??? int n ;
? ? ? scanf ( "%d" ,&n );
??? int count = 1 ;
??? while (n --)
??? {
? ? ? ? ? ? ? scanf ( "%s %s" ,str1 ,str2 );
? ? ? ? ? ? ? printf ( "Case %d:n" ,count ++);
? ? ? ? ? ? ? printf ( "%s + %s = " ,str2 );
? ? ? ? ? ? ? jiafa ();
??????? if (n )
? ? ? ? ? ? ? ? ? ? ? printf ( "n" );
??? }
??? return 0 ; }