PAT A1058 A+B in Hogwarts

PAT A1058 A+B in Hogwarts

题目描述：

　　If you are a fan of Harry Potter, you would know the world of magic has its own currency system -- as Hagrid explained it to Harry, "Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Sickle, it's easy enough." Your job is to write a program to compute A+B where A and B are given in the standard form of Galleon.Sickle.Knut (Galleon is an integer in [0,10​7​​], Sickle is an integer in [0, 17), and Knut is an integer in [0, 29)).

　　Input Specification:
　　Each input file contains one test case which occupies a line with A and B in the standard form, separated by one space.

　　Output Specification:
　　For each test case you should output the sum of A and B in one line, with the same format as the input.

　　Sample Input:
　　3.2.1 10.16.27

　　Sample Output:
　　14.1.28

参考代码：

/****************************************************
PAT A1058 A+B in Hogwarts
****************************************************/
#include <iostream>

using namespace std;

const int RATIO[2] = { 17, 29 };  //存储进制信息

int main() {
    char point;
    int num[3][3]{ 0 };           //num[0], num[1], num[2]分别代表a, b, a + b
    
    //读入加数a, b
    for (int i = 0; i < 2; ++i) {
        cin >> num[i][0] >> point >> num[i][1] >> point >> num[i][2];
    }

    //计算a + b
    int carry = 0;
    for (int i = 2; i > 0; --i) {
        num[2][i] = (num[0][i] + num[1][i] + carry) % RATIO[i - 1];
        carry = (num[0][i] + num[1][i] + carry) / RATIO[i - 1];
    }
    num[2][0] = num[0][0] + num[1][0] + carry;

    //输出a + b
    for (int i = 0; i < 3; ++i) {
        cout << num[2][i];
        if (i != 2) cout << point;
    }

    return 0;
}

注意事项：

　　1：本题和乙级中的1037 在霍格沃茨找零钱：https://www.cnblogs.com/mrdragon/p/11402980.html相似，如果按照B1037中的做法的话需要注意应该将Galleon、allByKunt的类型改为long long，但是这种算法会大大增加程序的运行时间。