POJ 1284：Primitive Roots 求原根的数量

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3381		Accepted: 1980

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

一个数m的1次方，2次方，3次方到n-1次方 mod n 得到的数值各不相同，就说m是n的原根。

一个数是数n的原根就必然与n-1互质，所以求n的原根的数量即是求欧拉函数n-1。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

long long euler(long long n)
{
	long long res = n, a = n;
	for (long long i = 2; i*i <= a; i++)
	{
		if (a%i == 0)
		{
			res = res / i*(i - 1);
			while (a%i == 0)a /= i;
		}
	}
	if (a > 1)res = res / a*(a - 1);
	return res;
}

int main()
{
	long long n;
	while (cin >> n)
	{
		cout << euler(n-1) << endl;
	}
	return 0;
}