POJ 1284 Primitive Roots 解题报告（欧拉函数）

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#欧拉函数

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本文介绍了一个算法，用于计算一个给定奇素数的原根数量。通过使用欧拉函数phi来确定原根数目，该算法有效地解决了问题。

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2714		Accepted: 1532

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

解题报告：求一个奇素数的原根。原根的概率数论里说过，简单来说一个数原根的数量就是phi(phi(n))。

素数n的欧拉函数为n-1，所以本题直接求phi(n-1)即可。代码如下：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int epi(int n)
{
    int m=(int)(sqrt(n+0.0)+0.5);

    int phi=n;
    for(int i=2;i<=m;i++) if(n%i==0)
    {
        phi=phi*(i-1)/i;
        while(n%i==0) n/=i;
    }
    if(n>1) phi=phi*(n-1)/n;

    return phi;
}

int main()
{
    int n;
    while(~scanf("%d", &n))
        printf("%d\n", epi(n-1));
}