poj1284——Primitive Roots（欧拉函数）

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

看了会欧拉函数的内容，准备自己独立做一个题目，结果就碰上这种题，想了半天想不出来，百度了才知道还有个定理。。。我果然不适合搞数学

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

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <map>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 100
using namespace std;
int euler(int n)
{
    int ans=n,m=sqrt(n+0.5);
    for(int i=2;i<=m;++i)
    {
        if(n%i==0)
        {
            ans=ans/i*(i-1);
            while(n%i==0)
                n/=i;
        }
    }
    if(n!=1)
        ans=ans/n*(n-1);
    return ans;
}
int main()
{
    int p;
    while(~scanf("%d",&p))
    {
        printf("%d\n",euler(p-1));
    }
    return 0;
}