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=;i*i<=n;i++)
     {
         if(n%i==)
         {
             while(n%i==)
             n=n/i;
             temp=temp/i*(i-);
         }
     }
     if(n!=)
     temp=temp/n*(n-);
     return temp;
 }
 
 int main()
 {
     int n;
     while(scanf("%d",&n)>)
     {
         printf("%d\n",Euler(n-));
     }
     return ;
 }