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;
}

Codeforces Round #340 (Div. 2) A. Elephant 水题的更多相关文章

  1. Codeforces Round #340 (Div. 2) B. Chocolate 水题

    B. Chocolate 题目连接: http://www.codeforces.com/contest/617/problem/D Descriptionww.co Bob loves everyt ...

  2. Codeforces Round #340 (Div. 2) D. Polyline 水题

    D. Polyline 题目连接: http://www.codeforces.com/contest/617/problem/D Descriptionww.co There are three p ...

  3. Codeforces Round #185 (Div. 2) B. Archer 水题

    B. Archer Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/312/problem/B D ...

  4. Codeforces Round #360 (Div. 2) A. Opponents 水题

    A. Opponents 题目连接: http://www.codeforces.com/contest/688/problem/A Description Arya has n opponents ...

  5. Codeforces Round #190 (Div. 2) 水果俩水题

    后天考试,今天做题,我真佩服自己... 这次又只A俩水题... orz各路神犇... 话说这次模拟题挺多... 半个多小时把前面俩水题做完,然后卡C,和往常一样,题目看懂做不出来... A: 算是模拟 ...

  6. Codeforces Round #256 (Div. 2/A)/Codeforces448A_Rewards(水题)解题报告

    对于这道水题本人觉得应该应用贪心算法来解这道题: 下面就贴出本人的代码吧: #include<cstdio> #include<iostream> using namespac ...

  7. Codeforces Round #338 (Div. 2) A. Bulbs 水题

    A. Bulbs 题目连接: http://www.codeforces.com/contest/615/problem/A Description Vasya wants to turn on Ch ...

  8. Codeforces Round #282 (Div. 1) A. Treasure 水题

    A. Treasure Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/494/problem/A ...

  9. Codeforces Round #327 (Div. 2) B. Rebranding 水题

    B. Rebranding Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/problem ...

随机推荐

  1. EhCache 分布式缓存/缓存集群

    开发环境: System:Windows JavaEE Server:tomcat5.0.2.8.tomcat6 JavaSDK: jdk6+ IDE:eclipse.MyEclipse 6.6 开发 ...

  2. cp: omitting directory”错误的解释和解决办法

    在linux下拷贝的时候有时候会出现cp:omitting directory的错误 ,例如 cp:omitting directory "bbs" 说明bbs目录下面还有目录,不 ...

  3. Unity 5 引擎收费版和免费版的区别(转)

    最新Unity 5的Professional Edition(收费版)具备全新而强大的功能,除了全局动态光照或是最新的基于物理的着色器之外,也把原本分开销售的Team License放入,并含有12个 ...

  4. bzoj3064 CPU监控

    今天终于写了一道正常的题 思路是这样的: 1.普通线段树add,set不变,并改为下放标记版本 2.past_addv 记录一个区间内可能的addv值的最大值 3.past_setv 记录一个区间被s ...

  5. 如何解决grails2.3.2中不能运行fork模式

    升级到grails 2.3.2之后,运行时报如下的异常: Exception in thread "main" Error | Forked Grails VM exited wi ...

  6. 求职基础复习之冒泡排序c++版

    代码中在第一层循环中增加一个bool值,是为了防止在排序完成后还继续无谓的比较,最多会有(n-1)*(n-2)/2次循环. #include<iostream> using namespa ...

  7. NET知识大纲

    第一部分 C#编程基础 1.(30)变量.运算符(+.-.*./.++.--.括号.==.!=.>.<.>=.<=.&&.||).流程控制(if.while.f ...

  8. effective c++:private继承

    如果class间使用private继承关系,编译器就不会自动的将派生类转换为基类,而且private继承而来的成员都变为private属性. private继承意味着根据某物实现出,当我们想要避免重复 ...

  9. 数据结构(11) -- 邻接表存储图的DFS和BFS

    /////////////////////////////////////////////////////////////// //图的邻接表表示法以及DFS和BFS //////////////// ...

  10. Java学习日志-01-Hello World

    1.安装JDK1.7 2.安装eclipse 3.eclipse上写第一个java程序-hello world 先建工程,再建包,养成良好的习惯,然后新建类 若不先建立包,可能会提示"The ...