HDU 1002 A + B Problem 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 shoshould 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 shoshould 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

I almost got abused. In this simple question, I wrote an inline function, max, which is different from the function in algorithm.
Only AC is handed in three times. I will give you some data for WA:
0001 1000
0 0
000 0000
9999 1
1 9999
99900 00999
00999 99900
If these data and examples are all used, you should be able to access

LANGUAGE: C ++
CODE:
# Include <stdio. h>
# Include <string. h>
# Include <algorithm>
# Include <stdlib. h>
 
# Define maxn1005
 
Using namespace std;
 
Char ans [maxn];
Int anslen;
 
Void plus (char s1 [], char s2 [])
{
Int len1 = strlen (s1 );
Int len2 = strlen (s2 );
 
For (int I = 0; I <len1; I ++)
S1 [I]-= '0 ';
For (int I = 0; I <len2; I ++)
S2 [I]-= '0 ';
 
Int mid1 = len1> 1;
Int mid2 = len2> 1;
 
// Printf ("% d \ n", len1, len2, mid1, mid2 );
 
For (int I = 0; I <mid1; I ++)
Swap (s1 [I], s1 [len1-1-i]);
For (int I = 0; I <mid2; I ++)
Swap (s2 [I], s2 [len2-1-i]);
 
Anslen = max (len1, len2 );

// Printf ("\ n anslen = % d \ n", anslen );
 
Int s = 0, sum;
For (int I = 0; I <= anslen; I ++)
{
Sum = (s1 [I] + s2 [I] + s );
Ans [I] = sum % 10;
S = sum/10;
}
}
 
Void printans (char ans [], int anslen)
{
// If (ans [anslen + 1] = 0)
While (anslen> 0 & (ans [anslen] = 0) anslen --;
// Else printf ("% d", ans [anslen + 1]);
If (anslen = 0)
{
Printf ("% d \ n", ans [0]);
Return;
}
For (int I = anslen; I>-1; I --)
{
Printf ("% d", ans [I]);
}
Printf ("\ n ");
}
 
Int main ()
{
Int cas;
Char s1 [maxn], s2 [maxn];
Scanf ("% d", & cas );
For (int I = 1; I <= cas; I ++)
{
Memset (s1, 0, sizeof (s1 ));
Memset (s2, 0, sizeof (s2 ));
Scanf ("% s", s1, s2 );
Printf ("Case % d: \ n", I );
Printf ("% s + % s =", s1, s2 );
Plus (s1, s2 );
Printans (ans, anslen );
If (I! = Cas) printf ("\ n ");
}
Return 0;
}