Problem I : Frog's Jumping


From: DHUOJ, 2017060309
<a type="button" class="btn btn-default" href="/solution/submit.html?problemId=5292">Submit</a> (Out of Contest)
<center> Time Limit: 1 s </center>

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.
 

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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

 

Author: handoku



题意就是青蛙每次跳k，然后下一个位置是（当前位置+k）% n ，问你怎么选k才能使青蛙把所有的数都跳了一遍之后再返回0位置

若n和k有约数，那么一定要他们的约数的整数倍的地方是没有办法走到的（即n和k没有公约数 即互质）。

显然 只需要计算这个数n的欧拉函数即可。


#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
    int n,i,temp;  
    while(scanf("%d",&n)!=EOF)  
    {  
        temp=n;  
        for(i=2;i*i<=n;i++)  
        {  
          if(n%i==0)  
          {  
              while(n%i==0) n=n/i;  
              temp=temp/i*(i-1);  
          }  
          if(n<i+1)  
              break;  
        }  
        if(n>1)  
            temp=temp/n*(n-1);  
        printf("%d\n",temp);  
    }  
    return 0;
}