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 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

我勒个去，差点被虐死，这么简单的题，开始我写的一个内联函数max和algorithm里面的函数写得不一样
交3次才AC，WA的我给你们几个数据：
0001 1000
0 0
000 0000
9999 1
1 9999
99900 00999
00999 99900
这几个数据和样例全过的话，应该可以AC了

LANGUAGE:C++
CODE:
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>

#define maxn 1005

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 %d %d %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%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;
}

补充：软件开发 , C++ ,