POJ 1284：Primitive Roots（素数原根的个数）

Primitive Roots

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 5709 Accepted: 3261

Description

We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set ${ (x_i mod p) | 1 \leq i \leq p-1 }$ 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

题意

给出一个正素数p，求p的原根的个数

思路

原根的定义： 对于两个正整数$(a,m)=1$，由欧拉定理可知：存在$d\leq m-1$。比如说欧拉函数$d=φ(m)$，即小于等于$m$的正整数与$m$互质的正整数的个数，使得$a^d\equiv1 (mod m)$。由此，在$(a,m)=1$时，定义$a$对模$m$的指数$\delta m(a)$为使$a^d\equiv1(mod m)$成立的最小正整数$d$。由前知$\delta m(a)$一定小于等于$φ(m)$，若$\delta m(a)=φ(m)$，则称$a$为$m$的原根
原根个数定理： 如果$p$有原根，则它恰有$φ(φ(p))$个不同的原根，$p$为素数时，$φ(p)=p-1$,因此就有$φ(p-1)$个原根

AC代码

#include <iostream>
using namespace std;
int Eular(int n)
{
    int eu=n;
    for (int i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            eu-=eu/i;
            while(n%i==0)
                n/=i;
        }
    }
    if(n>1)      //n本身也是个质因子
        eu-=eu/n;
    return eu; 
}
int main(int argc, char const *argv[])
{
    int p;
    while(cin>>p)
    {
        cout<<Eular(p-1)<<endl;
    }
    return 0;
}