[PAT A1058]A+B in Hogwarts_综合

[PAT A1058]A+B in Hogwarts

题目描述

1058 A+B in Hogwarts （20 分）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)).

输入格式

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

输出格式

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

输入样例

3.2.1 10.16.27

输出样例

14.1.28

解析

这道题目很简单，属于送分题，具体题目翻译过来就是给定两个货币数（霍格沃兹货币格式，具体格式进制格式可以看题，也可以参考我的博客，下面）
[PAT B1037]在霍格沃茨找零钱

这里要强调一点的就是，我使用的是在B1037中的另一种我没有写出来的方法，就是把货币全部换成零钱，相加后再换回去，这样虽然很简单暴力，但是有可能导致溢出，所以数据使用的是long long型，不信？哈哈那可以去试一试，我是有一组数据过不了的

#include<iostream>
using namespace std;
const int Galleon = 17 * 29;
const int Sickle = 29;
int main()
{
    long long a1, b1, c1, a2, b2, c2;scanf("%lld.%lld.%lld %lld.%lld.%lld", &a1, &b1, &c1, &a2, &b2, &c2);c1 += a1*Galleon + b1*Sickle;c2 += a2*Galleon + b2*Sickle;c1 += c2;a1 = c1 / Galleon;b1 = c1%Galleon / Sickle;c1 = c1%Sickle;cout << a1 << "." << b1 << "." << c1;return 0;
}