【BZOJ4803】逆欧拉函数

【BZOJ4803】逆欧拉函数

题面

bzoj

题解

题目是给定你\(\varphi(n)\)要求前\(k\)小的\(n\)。

设\(n=\prod_{i=1}^k{p_i}^{c_i}\)

则\(\varphi(n)=\prod_{i=1}^k{p_i}^{c_i-1}(p_i-1)\)

然后我们猜一下这个\(n\)不是很多，事实上\(n\)不超过\(50w\)个。

考虑暴力\(dfs\)出所有的\(n\)：

首先筛出\(\sqrt{\varphi(n)}\)内的素数

对于当前\(dfs\)的值\(phi\)

看\(phi\)中的约数有没有\(筛出的素数-1\)

若有，假设该素数为\(p\)

去除\(phi\)中的所有\(p\)，之后再将\(dfs\)的\(n\)累乘上\(p\)

在每一次递归开头用\(miller\)_\(Rabin\)判断\(phi+1\)是否为素数，如果是，则直接加进答案就行了

想一想,为什么?

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <ctime>
using namespace std;
typedef long long ll;
const int MAX_N = 1e7 + 5;
const int T = 10;
bool is_prime[MAX_N];
int prime[MAX_N], num, K;
ll N = 1e7, ans[MAX_N], cnt_ans;
void sieve() {
	for (int i = 1; i <= N; i++) is_prime[i] = 1;
	is_prime[1] = 0;
	for (int i = 2; i <= N; i++) {
		if (is_prime[i]) prime[++num] = i;
		for (int j = 1; prime[j] * i <= N && j <= num; j++) {
			is_prime[i * prime[j]] = 0;
			if (!(i % prime[j])) break;
		}
	}
}
ll fmul(ll x, ll y, ll Mod) {
	ll res = 0;
	while (y) {
		if (y & 1ll) res = (res + x) % Mod;
	    y >>= 1ll;
		x = (x + x) % Mod;
	}
	return res;
}
ll fpow(ll x, ll y, ll Mod) {
	ll res = 1;
	while (y) {
		if (y & 1ll) res = fmul(res, x, Mod);
		y >>= 1ll;
		x = fmul(x, x, Mod);
	}
	return res;
}
bool Test(ll a, ll n) {
	ll r = 0, t = n - 1, m;
	while ((t & 1ll) == 0) ++r, t >>= 1ll;
	m = (n - 1) / (1ll << r);
	for (int i = 0; i < r; i++) if (fpow(a, (1ll << i) * m, n) == n - 1) return 1;
	if (fpow(a, m, n) == 1) return 1;
	return 0;
}
bool Miller_Rabin(ll n) {
	if (n == 2ll) return 1;
	if (n < 2ll || ((n & 1ll) == 0)) return 0;
	for (int i = 1; i <= T; i++) {
		ll a = rand() % (n - 2) + 2;
		if (fpow(a, n - 1, n) != 1) return 0;
		if (!Test(a, n)) return 0;
	}
	return 1;
}
void solve(ll phi, ll n, int lst) {
	if (phi + 1 > prime[num] && Miller_Rabin(phi + 1))
		ans[++cnt_ans] = n * (phi + 1);
	for (int i = lst; i; i--) {
		if (!(phi % (prime[i] - 1))) {
			ll t1 = phi / (prime[i] - 1), t2 = n, t3 = 1ll;
			while (!(t1 % t3)) {
				t2 *= prime[i];
				solve(t1 / t3, t2, i - 1);
				t3 *= prime[i];
			}
		}
	}
	if (phi == 1ll) ans[++cnt_ans] = n;
}
int main () {
	srand(time(NULL));
	sieve();
	cin >> N >> K;
	solve(N, 1ll, num);
	sort(&ans[1], &ans[cnt_ans + 1]);
	for (int i = 1; i < K; i++) printf("%lld ", ans[i]);
	printf("%lld\n", ans[K]);
	return 0;
}