Codeforces 617A Elephant

解决一个关于大象如何用最少步骤从0位置到达x位置的问题,其中x为朋友家的位置。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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 1234 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 35 and 4. There are other ways to get the optimal answer but the elephant cannot reach x in less than three moves.


#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
int main()
{
    int g;
    while(~scanf("%d",&g))
    {
        int sum=g/5;
        if(g%5!=0)
            sum++;
        printf("%d\n",sum);
    }
    return 0;
}


 



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值