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 { (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

Source

贾怡@pku

 

 1 /*

 2 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）

 3 p为素数，当然φ(p)=p-1,因此就有φ(p-1)个原根

 4 

 5 

 6 数学差的人，很伤不起...

 7 */

 8 

 9 #include<stdio.h>

10 

11 

12 int Euler(int n)

13 {

14     int i,temp=n;

15     for(i=2;i*i<=n;i++)

16     {

17         if(n%i==0)

18         {

19             while(n%i==0)

20             n=n/i;

21             temp=temp/i*(i-1);

22         }

23     }

24     if(n!=1)

25     temp=temp/n*(n-1);

26     return temp;

27 }

28 

29 int main()

30 {

31     int n;

32     while(scanf("%d",&n)>0)

33     {

34         printf("%d\n",Euler(n-1));

35     }

36     return 0;

37 }