(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;
}