大整数加法

时间限制：3000 ms | 内存限制：65535 KB | 难度：3

描述

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

A,B must be positive.

输入

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.

输出

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.

样例输入

2
1 2zz
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

南阳理工ACM题目-103：A+B Problem II

#include <iostream>
#include <string>
using namespace std;
    
int main()
{
    char str[1002];
    string str1,str2,ts;
    int n,a,b,k,id,m=1;
    cin >> n;
    while (n--)
    {
        cin >> str1 >> str2;
        k = id = 0;
        cout << "Case " << m ++ << ":" << endl << str1 << " + " << str2 << " = ";
        if (str1.length() < str2.length())
        {
        ts = str1;
        str1 = str2;
        str2 = ts;
        }
        a = str1.length();
        b = str2.length();
        for (int i = a-1,j=b-1; j>=0; --i,--j)
        {
        k = (str1[i] - '0') + (str2[j] - '0') + k;
        str[id++] = k % 10;
        k /= 10;
        }
        for (int i = a - b - 1; i >= 0; --i)
        {
        k = (str1[i] - '0') + k;
        str[id++] = k % 10;
        k /= 10;
        }
        if (k > 0)cout << k;
        for (int i = id - 1; i >= 0; --i)
        {
        cout << char(str[i] + '0');
        }
        cout << endl;
    }
    //system("pause");
    return 0;
}