POJ 1284 Primitive Roots 数论原根。

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2479		Accepted: 1385

Description

We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xⁱ 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

Source

贾怡@pku

 

/*
定理1：如果p有原根，则它恰有φ(φ(p))个不同的原根（无论p是否为素数都适用）
p为素数，当然φ(p)=p-1,因此就有φ(p-1)个原根


数学差的人，很伤不起...
*/

#include<stdio.h>


int Euler(int n)
{
    int i,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!=1)
    temp=temp/n*(n-1);
    return temp;
}

int main()
{
    int n;
    while(scanf("%d",&n)>0)
    {
        printf("%d\n",Euler(n-1));
    }
    return 0;
}