欧拉函数 原根 POJ 1284 Primitive Roots

Primitive Roots

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 <= 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

题目大意就是给出一个奇素数，求出他的原根的个数,定义n的原根x满足条件0<x<n,并且有集合{ (xⁱ mod n) | 1 <= i <=n-1 } 和集合{ 1, ..., n-1 }相等

关于这道题。如果知道欧拉函数的话，看出的答案是phi(n-1)其实也不难

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

AC代码：

View Code

 1 #include<stdio.h>

 2 int main()

 3 {

 4     int p,rea,i;

 5     while(scanf("%d",&p)!=EOF)

 6     {

 7         p--;

 8         rea=p;

 9         for(i=2;i*i<=p;i++)

10         {

11             if(p%i==0)

12             {

13                 rea=rea-rea/i;

14                 do{

15                     p/=i;

16                 }while(p%i==0);

17             }

18         }

19         if(p>1)

20             rea=rea-rea/p;

21         printf("%d\n",rea);

22     }

23     return 0;

24 }