POJ1284 Primitive Roots [欧拉函数，原根]

　　题目传送门

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5434		Accepted: 3072

Description

We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (x mod p) | 1 <= i <= p-1 } is equal to { 1, ..., p-1 }. For example, the consecutive powers of 3 modulo 7 are 3, 2, 6, 4, 5, 1, and thus 3 is a primitive root modulo 7. 
Write a program which given any odd prime 3 <= p < 65536 outputs the number of primitive roots modulo p. 

Input

Each line of the input contains an odd prime numbers p. Input is terminated by the end-of-file seperator.

Output

For each p, print a single number that gives the number of primitive roots in a single line.

Sample Input

23

31

79

Sample Output

10

8

24

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　　分析：

　　一句话题意：求原根的个数。

　　首先，如果知道原根的相关知识，那就可以直接上欧拉函数的板子了。关于原根的知识，请参考这里。

　　Code：

//It is made by HolseLee on 11th July 2018

//POJ 1284

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<algorithm>

#include<iostream>

#include<iomanip>

using namespace std;

const int N=1e5+;

int n,phi[N],top,q[];

bool vis[N];

void ready()

{

  phi[]=;

  for(int i=;i<N;i++){

    if(!vis[i])phi[q[++top]=i]=i-;

    for(int j=,k;j<=top&&(k=i*q[j])<N;j++){

      vis[k]=true;

      if(i%q[j])phi[k]=phi[i]*(q[j]-);

      else {phi[k]=phi[i]*q[j];break;}

    }

  }

}

int main()

{

  ios::sync_with_stdio(false);

  ready();

  while(cin>>n){

    printf("%d\n",phi[n-]);}

  return ;

}