poj 1284 Primitive Roots

Primitive Roots Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4267   Accepted: 2501 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

提示

题目：

给出一个奇素数p(就是不包含2的素数)，求出0到p-1有几个原根。

原根的定义：x的1到p-1次方对p取余所得的值的集合，恰好等于1到p-1的集合。

思路：

已知定理我们可知道p的原根个数为φ(p-1)。这里的φ(x)表示的是欧拉函数，求的是小于x的数中与x互质的数的数目。

示例程序

#include <cstdio>
using namespace std;
int f(int n)
{
    int i,ans;
    ans=n;
    for(i=2;n>=i*i;i++)
    {
        if(n%i==0)
        {
            ans=ans-ans/i;
            do
            {
                n=n/i;
            }while(n%i==0);
        }
    }
    if(n!=1)
    {
        ans=ans-ans/n;
    }
    return ans;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d\n",f(n-1));
    }
    return 0;
}