A + B Problem II HDU - 1002 高精度

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

#include<iostream>
using namespace std;
int main()
{
    int T, n = 0;
    cin >> T;
    while (T--)
    {
        char a[1005], b[1005];int  c[1005];
        cin >> a >> b;
        int A, B, C, D;
        A = strlen(a);
        B = strlen(b);
        int temp1 = 0;
        int i=A, j=B, k=0;
        while(1)
        {
            
            C = (a[i-1] - '0'); D =( b[j-1] - '0');//转换成int型
            if (i <= 0)C =0; if (j <= 0)D =0;
            c[k] = ( temp1 + C  + D )% 10;//记录
            temp1 =( temp1 + C  + D )/ 10;//进位
            i--;j--;
            if (i <= 0 && j <= 0)break;
            k++;
        }
        c[k + 1] = '\0';
        cout << "Case " << ++n << ":" << endl;//注意输出格式
        cout << a << " + " << b << " = ";
        for (int i = k; i>=0; i--)//将c倒序输出
            cout <<c[i];
        cout<< endl;
        if (T != 0)cout <<endl;//两个输出之间有一个空行
    }
    return 0;
}