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 { (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
使用到了原根定理：如果p有原根，则它恰有φ(φ(p))个不同的原根，p为素数，当然φ(p)=p-1,因此就有φ(p-1)个原根 代码如下：
#include <iostream> using namespace std; int phi(int ); int main() { int p,ans; while(cin>>p) { ans=phi(p-1); cout<<ans<<endl; } return 0; } int phi(int n) { int res=n; for(int i=2;i*i<=n;i++) { if(n%i==0) res=res-res/i; while(n%i==0) n=n/i; } if(n>1) res=res-res/n; return res; }