Codeforces Round #350 (Div. 2) A

A. Holidays
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 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.

题意: 假设一年有n天 问一年内的休息天的数量(一周7天 5+2) 的最小值与最大值

题解:最小值---从周一开始算

         最大值---从周六开始算

   比赛的时候这题都gg了 没处理好 菜

 1 #include<bits/stdc++.h>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<queue>
 6 #include<stack>
 7 #include<map> 
 8 #define ll __int64
 9 #define pi acos(-1.0)
10 using namespace std;
11 int n;
12 int main()
13 {
14     while(scanf("%d",&n)!=EOF)
15     {
16     int s1;
17     if(n>=7)
18     {
19       s1=(n/7)*2;
20       if(n%7==6)
21       s1+=1; 
22     }
23     else
24     {
25      if(n==6)
26      s1=1;
27      else
28      s1=0;
29     }
30     int s2=2;
31     if(n-2>=7)
32     { 
33       s2+=((n-2)/7*2);
34       if((n-2)%7==6)
35          s2+=1;
36     }
37     else
38     {
39      if(n-2==6)
40      s2+=1;
41      else
42      s2+=0;
43     }
44     if(n>2)
45     printf("%d %d
",s1,s2);
46     if(n==1)
47     printf("0 1
");
48     if(n==2)
49     printf("0 2
");
50 }
51     return 0; 
52 } 
原文地址:https://www.cnblogs.com/hsd-/p/5464441.html