HDU-1002 A + B Problem 2(高精度加法)

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

---

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

vector<char> bigNumPlus(vector<char> & num1, vector<char> & num2)
{
    int extra = 0;
    for (size_t i = 0; i < num1.size(); ++i) {
        int sum = (num1[i] - '0') + (num2[i] - '0') + extra;
        extra = sum / 10;
        num1[i] = '0' + sum % 10;
    }
    if (extra & 1) num1[num1.size() - 1] = '1';
    reverse(num1.begin(), num1.end());

    return num1;
}

template <typename T>
ostream & operator<<(ostream & os, const vector<T> & v)
{
    size_t i = 0;
    if (v[0] == '0') i = 1; 
    for ( ; i < v.size(); ++i) {
        os << v[i];
    }

    return os;
}

int main()
{
    std::ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int caseNum, num = 0;
    cin >> caseNum;
    while (caseNum--) {
        ++num;
        string str1, str2;
        cin >> str1 >> str2;
        int length = max(str1.size(), str2.size()) + 1; //可能存在进位，所以多留出一位

        vector<char> num1(length, '0'), num2(length, '0');
        for (size_t i = 0; i < str1.size(); ++i) {
            num1[str1.size() - 1 - i] = str1[i];
        }

        for (size_t i = 0; i < str2.size(); ++i)
            num2[str2.size() - 1 - i] = str2[i];

        cout << "Case " << num << ":" << endl;
        cout << str1 << " + " << str2 << " = " << bigNumPlus(num1, num2) << endl;
        if(caseNum != 0) cout << endl;
    }

    return 0;
}