(Relax 数论1.8)POJ 1284 Primitive Roots(欧拉函数的应用: 以n为模的本原根的个数phi(n-1))

/*

 * POJ_2407.cpp

 *

 *  Created on: 2013年11月19日

 *      Author: Administrator

 */



#include <iostream>

#include <cstdio>

#include <cstring>



using namespace std;



typedef long long ll;



const int maxn = 1000015;



bool u[maxn];

ll su[maxn];

ll num;



ll gcd(ll a, ll b) {

	if (b == 0) {

		return a;

	}



	return gcd(b, a % b);

}



void prepare() {//欧拉筛法产生素数表

	ll i, j;

	memset(u, true, sizeof(u));



	for (i = 2; i <= 1000010; ++i) {

		if (u[i]) {

			su[++num] = i;

		}



		for (j = 1; j <= num; ++j) {

			if (i * su[j] > 1000010) {

				break;

			}



			u[i * su[j]] = false;



			if (i % su[j] == 0) {

				break;

			}

		}

	}

}



ll phi(ll x) {//欧拉函数,用于求[1,x)中与x互质的整数的个数

	ll ans = 1;

	int i, j, k;

	for (i = 1; i <= num; ++i) {

		if (x % su[i] == 0) {

			j = 0;

			while (x % su[i] == 0) {

				++j;

				x /= su[i];

			}



			for (k = 1; k < j; ++k) {

				ans = ans * su[i] % 1000000007ll;

			}

			ans = ans * (su[i] - 1) % 1000000007ll;

			if (x == 1) {

				break;

			}

		}

	}



	if (x > 1) {

		ans = ans * (x - 1) % 1000000007ll;

	}



	return ans;

}



int main() {

	prepare();

	int n;

	while (scanf("%d", &n) != EOF) {

		printf("%lld\n", phi(n-1));//以n为模的本原根的个数为phi(n-1)。而,[1,n]与n互质的整数的个数为phi(n)

	}



	return 0;

}