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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：
1 #include<stdio.h> 2 #include<string.h> 3 #include<math.h> 4 #include<queue> 5 #include<algorithm> 6 #include<time.h> 7 #include<stack> 8 using namespace std; 9 #define N 1200000 10 #define INF 0x3f3f3f3f 11 12 int dp[N]; 13 int a[N]; 14 15 int main() 16 { 17 int n,l,r; 18 19 while(scanf("%d", &n) != EOF) 20 { 21 if(n<2) 22 r=n,l=0; 23 else if(n<=5) 24 r=2,l=0; 25 else if(n<=7) 26 l=n-5,r=2; 27 else 28 { 29 if(n%7<2) 30 { 31 l=n/7*2; 32 r=n/7*2+n%7; 33 } 34 else if(n%7<=5) 35 { 36 l=n/7*2; 37 r=n/7*2+2; 38 } 39 else if(n%7<7) 40 { 41 l=n/7*2+n%7-5; 42 r=n/7*2+2; 43 } 44 } 45 printf("%d %d\n",l,r); 46 } 47 return 0; 48 }
CodeForces 670 A. Holidays（模拟）
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原文地址：http://www.cnblogs.com/weiyuan/p/5789363.html
