codeforces 670A Holidays_codeforces 966e may holidays-优快云博客

本文介绍了一个简单的算法问题，即计算火星一年中可能的最少和最多的休息日数量。火星的一年有n天，但同样遵循5个工作日加2天休息的原则。输入为火星一年的总天数，输出是最少和最多休息日的数量。

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On the planet Mars a year lasts exactly n days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 1 000 000) — the number of days in a year on Mars.

Output

Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.

        Examples
      
          input
        
14

          output
        
4 4

          input
        
2

          output
        
0 2

Note

In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4days off .

In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off.

水题，看细心程度。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int w = n % 7, t = n / 7, p, q;
    if(w == 6)
        p = 2 * t + 1;
    else
        p = 2 * t;

    if(w == 1)
        q = 2 * t + 1;
    else if(w >= 2)
        q = 2 * t + 2;
    else
        q = 2 * t;
    cout << p << " " << q << endl;
    return 0;
}