poj

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

233179

Sample Output

10824

题目让求一个奇素数的原根的个数

有定理：如果p为素数，那么素数p一定存在原根，又因为当模p有原根时，它有φ(φ(p))个原根，那么对于素数p的原根的个数为φ(p−1)。

具体定义见百科：原根，http://blog.csdn.net/muxidreamtohit/article/details/8026611

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define M 65540int phi[M];void init() //预先求出所有的欧拉函数{    int i,j;    for(i=2;i<M;i++)        phi[i]=0;    phi[1]=1;    for(i=2;i<M;i++)        if(!phi[i])    {        for(j=i;j<M;j+=i)        {            if(!phi[j])                phi[j]=j;            phi[j]=phi[j]/i*(i-1);        }    }}int main(){    init();    int n;    while(scanf("%d",&n)!=EOF)    {        printf("%d\n",phi[n-1]);    }    return 0;}