2017年上海金马五校程序设计竞赛：Problem I : Frog's Jumping （找规律）-优快云博客

本文介绍了一种关于青蛙在荷叶上跳跃的游戏算法。通过分析青蛙如何选择合适的步长k，确保在n次跳跃后能经过所有荷叶并返回起点。文章提供了具体的实现代码，并解释了如何筛选符合条件的k值。

Description

There are n lotus leaves floating like a ring on the lake, which are numbered 0, 1, ..., n-1 respectively. The leaf 0 and n-1 are adjacent.

The frog king wants to play a jumping game. He stands at the leaf 0 initially. For each move, he jumps k (0 < k < n) steps forward. More specifically, if he is standing at the leaf x, the next position will be the leaf (x + k) % n.

After n jumps, he wants to go through all leaves on the lake and go back to the leaf 0 finally. He can not figure out how many different k can be chosen to finish the game, so he asks you for help.

Input

There are several test cases (no more than 25).

For each test case, there is a single line containing an integer n (3 ≤ n ≤ 1,000,000), denoting the number of lotus leaves.

Output

For each test case, output exactly one line containing an integer denoting the answer of the question above.

Sample Input

4
5

Sample Output

2
4

分析：

一些荷叶漂浮在湖面上其编号是0~n-1,有一只青蛙初始再0位置，它每次可以跳过K个位置（0<K<n），最终跳n次回到0.求1~n-1中有多少个K值满足在n此次跃中可以把每片荷叶就跳一次。

i 从 2~n-1 . 如果 n%i 取余等于0.则 i 和 i 的倍数全都不满足要求。

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
using namespace std;
int vis[1000000];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        long long ans = n-1;
        memset(vis,0,sizeof(vis));
        for(int i = 2; i <= n/2; i++)///后一半中的如果不能够走的话，肯定有个倍数在前一半出现过
        {
            if(n%i == 0)
            {
                for(int j = i; j < n; j+=i)///往后找整数倍
                {
                    if(vis[j]==0)
                    {
                        vis[j] = 1;
                        ans--;
                    }
                }
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/cmmdc/p/6941249.html