Elephant(CodeForces - 617A)_codeforces617a elephant-优快云博客

本文链接：https://blog.youkuaiyun.com/ITzz_qqzi/article/details/114132569

一只大象要从坐标点0出发去朋友家，坐标为x(x>0)。大象每步可以前进1到5个位置。问题是大象至少需要多少步才能到达朋友家。输入包含一个整数x，输出大象到达x所需的最小步数。例如，当x=5时，大象只需一步就能走完。思路是优先每次走5步，剩余不足5步时额外增加一步。

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Elephant

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.

Examples

Input
5

Output
1

Input
12

Output
3

Note

In the first sample the elephant needs to make one step of length 5 to reach the point x.

In the second sample the elephant can get to point x if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach x in less than three moves.

思路很简单，要求最少的步骤数，即尽可能每次走5步，当剩余距离不足5时，总步骤数加1即可。

/*by BFU zq
  2021/2/26 14:47*/
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

int main()
{
	int m, n, x, ans = 0;
	cin >> x;
	
	m = x / 5;
	n = x % 5;
	
	if (n == 0)
	{
		ans = m;
	}
	else
	{
		ans = m + 1;
	}
	
	cout << ans << endl;
	
	return  0;
}