Codeforces Round #340 (Div. 2) A. Elephant 水题

探讨了在坐标轴上,大象从位置0出发到位置x的最短行走步数问题,每步可走1至5个单位。通过向上取整的简便方法,实现了高效的求解。

A. Elephant

题目连接:

http://www.codeforces.com/contest/617/problem/A

Descriptionww.co

An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point x(x > 0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.

Input

The first line of the input contains an integer x (1 ≤ x ≤ 1 000 000) — The coordinate of the friend's house.

Output

Print the minimum number of steps that elephant needs to make to get from point 0 to point x.

Sample Input

6

Sample Output

2

Hint

题意

给你一个数x,然后让你输出x/5的向上取整

题解:

输出(x+4)/5就好了

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    long long x;
    cin>>x;
    cout<<(x+4)/5<<endl;
}
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