(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
", 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;
}