Codeforces Round #340 (Div. 2)(A) 模拟

A. Elephant

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 test(s)

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.

题意：大象在0位置，他要到x点，每次可以选择走1，2，3，4，5步，问最少走几步

题解：直接模拟一下就可以了

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <sstream>
#include <algorithm>
#define LL long long
#define N 100000
using namespace std;

int main()
{
#ifdef CDZSC
freopen("i.txt","r",stdin);
#endif
int n;
while(~scanf("%d",&n))
{
int sum=n/5;
int num=n%5;
for(int i=4;i>=1;i--)
{
if(num>=i)
{
num-=i;
sum++;
}
}
printf("%d\n",sum);
}
return 0;
}