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.
问题链接：https://vjudge.net/problem/CodeForces-617A
题意：一头大象决定去拜访他的朋友。
原来大象的家位于0点，他的朋友的家位于坐标线的X(X>0)点。
一步之内，大象就能向前移动1、2、3、4或5个位置。
确定，他到达朋友家所需要的最小步骤是多少。
输入的第一行包含一个整数X（1≤X≤1000000）——朋友家的坐标。
打印出大象从0点到X点所需的最小步数。
题解：朋友家坐标为X,大象只能移动1或2或3或4或5步
如果x<=5,大象移动一步即可；
如果x>5并且x是5的倍数，移动步数为 x/5;
如果x>5且x不是5的倍数，移动步数为x/5+1;
AC代码：
#include<iostream>
using namespace std;
int main()
{
int x,n;
int steps=0;
cin >> x;
if (x <= 5)cout << "1";
else
if (x % 5 == 0)
{
n = x / 5;
cout << n;
}
else
{
n = x / 5 + 1;
cout << n;
}
}
