hdoj1002

A - A + B Problem II

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. 

InputThe 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. 
OutputFor 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
这道题一开始的思路是直接用字符数组去处理，做到一半发现实在太麻烦了，百度到他们把他放在整形数组里去处理好理解且方便.
且此题有BUG，我的程序输入1 9和1 99999999999这类数时候答案总是0，但是依然AC了- -！（玄学改变命运），不过最后还是改了一下，使程序更完美，
把进位的变量进行判断TM=1输出TM，else continue;
下面是我的程序：

#include<iostream>
using namespace std;
#include<string.h>
#include<stdio.h>
#include<stack>
#include<string.h>
#include<string>
#include<algorithm>
#include<queue>
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    char a1[1010],b1[1010];
    int a[1010],b[1010],c[1010];
    int t,i,j;
    cin>>t;
    for(int k=1;k<=t;k++)
    {
        memset(a1,0,sizeof(a1));
        memset(b1,0,sizeof(b1));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        cin>>a1>>b1;
        int lena,lenb;
        lena=strlen(a1);
        lenb=strlen(b1);
        for(i=0,j=lena-1;i<lena;i++,j--)
        {
            a[j]=a1[i]-'0';
        }
        for(i=0,j=lenb-1;i<lenb;i++,j--)
        {
            b[j]=b1[i]-'0';
        }
        int TM=0;
        if(lena>=lenb)
        {
            for(i=0;i<lena;i++)
            {
                c[i]=(a[i]+b[i]+TM)%10;
                TM=(a[i]+b[i]+TM)/10;
            }
        }
        else
        {
            for(i=0;i<lenb;i++)
            {
                c[i]=(a[i]+b[i]+TM)%10;
                TM=(a[i]+b[i]+TM)/10;
            }
        }
        int lenc;
        lenc=max(lena,lenb);
        cout<<"Case "<<k<<":"<<endl;
        cout<<a1<<" + "<<b1<<" = ";
        if(TM)
            cout<<TM;
        for(i=lenc-1;i>=0;i--)
            cout<<c[i];
        cout<<endl;
        if(k!=t)
            cout<<endl;
    }
    return 0;
}