zoukankan      html  css  js  c++  java
  • C

    Problem description

    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.

    解题思路:从起始点0出发到达终点x,输出这一过程所走的最少步数。做法:从大的步数往小的步数贪心即可,水过。

    AC代码:

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int main(){
     4     int x,tmp,step=0,s[5]={5,4,3,2,1};
     5     cin>>x;
     6     for(int i=0;i<5;++i)
     7         if(x>=s[i]){step+=(tmp=x/s[i]);x-=tmp*s[i];}
     8     cout<<step<<endl;
     9     return 0;
    10 }
  • 相关阅读:
    GridView合并表头多重表头
    C# 导出Excel或Word
    GridView的分页功能?
    如何在GridView中判断Radio被选中?
    GridView無數據時,顯示表頭
    Oracle replace函数使用
    获取数据后导出Excel
    Oracel用rownum实现真分页
    转载C#泛型集合—Dictionary<K,V>使用技巧
    临时向表插入有自增的字段的记录
  • 原文地址:https://www.cnblogs.com/acgoto/p/9150012.html
Copyright © 2011-2022 走看看