1058. A+B in Hogwarts (20)

1058. A+B in Hogwarts (20)

时间限制 50 ms 内存限制 65536 kB 代码长度限制 16000 B

判题程序 Standard 作者 CHEN, Yue

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, 107], 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

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <fstream>
#include <string>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;

typedef struct
{
unsigned long long G;
unsigned int S;
unsigned int K;
}M;

int main()
{
#ifdef _DEBUG
freopen("data.txt", "r+", stdin);
//fstream cin("data.txt");
#endif // _DEBUG

M A, B, res;
res.G = res.S = res.K = 0;
scanf("%llu.%u.%u %llu.%u.%u", &A.G, &A.S, &A.K, &B.G, &B.S, &B.K);
res.K = (A.K + B.K >= 29) ? (res.S += 1, A.K + B.K - 29) : (A.K + B.K);
res.S = (res.S + A.S + B.S >= 17) ? (res.G += 1, res.S + A.S + B.S - 17) : (res.S + A.S + B.S);  //特别注意这里的进位，第一遍被我写丢了res.S，没考虑之前进上来的位。
res.G += A.G + B.G;
printf("%llu.%u.%u", res.G, res.S, res.K);
#ifdef _DEBUG
//cin.close();
#ifndef _CODEBLOCKS
system("pause");
#endif // !_CODEBLOCKS
#endif // _DEBUG

return 0;
}