CodeForces 670 A. Holidays（模拟）

Description

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.

Sample Input
Input
14
Output
4 4
Input
2
Output
0 2
Hint
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 4 days 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.

题目链接：http://codeforces.com/problemset/problem/670/A

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题意：一周5天工作，2天休息 。现在有n天，但是你不知道第一天是星期几，问你最多放多少天，最少放多少天

分析：简单模拟。

7天为一个周期，如果求放假天数最小的时候就看成 1 2 3 4 5 6 7，

如果求放假天数最大就看成 6 7 1 2 3 4 5。

AC代码：

 

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#include<time.h>
#include<stack>
using namespace std;
#define N 1200000
#define INF 0x3f3f3f3f

int dp[N];
int a[N];

int main()
{
    int n,l,r;

    while(scanf("%d", &n) != EOF)
    {
        if(n<2)
            r=n,l=0;
        else if(n<=5)
            r=2,l=0;
        else if(n<=7)
            l=n-5,r=2;
        else
        {
            if(n%7<2)
            {
                l=n/7*2;
                r=n/7*2+n%7;
            }
            else if(n%7<=5)
            {
                 l=n/7*2;
                 r=n/7*2+2;
            }
            else if(n%7<7)
            {
                l=n/7*2+n%7-5;
                r=n/7*2+2;
            }
        }
        printf("%d %d\n",l,r);
    }
    return 0;
}