【BZOJ 3642】Phi的反函数

http://www.lydsy.com/JudgeOnline/problem.php?id=3643

因为$$\varphi(n)=\prod_i p_i^{{k_i-1}(p_i-1),n=\prod_ip_i}{k_i}$$

直接根据这个式子暴搜即可。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1 << 16;
bool notp[N];
int prime[N], num = 0, ans;
void shai() {
	for (int i = 2; i < N; ++i) {
		if (!notp[i])
			prime[++num] = i;
		for(int j = 1; j <= num && prime[j] * i < N; ++j) {
			notp[i * prime[j]] = true;
			if (i % prime[j] == 0)
				break;
		}
	}
}
int n;
bool zhi(int x) {
	for(int i = (int) sqrt(x); i >= 2; --i)
		if (x % i == 0) return false;
	return true;
}
ll dfs(int tmp, int k) {
	if (k == 1) return 1;
	ll ans = -1, t, bas;
	int m;
	for(int i = tmp; i <= num && prime[i] - 1 <= k; ++i)
		if (k % (prime[i] - 1) == 0) {
			m = k / (prime[i] - 1); bas = prime[i];
			t = dfs(i + 1, m);
			if (t != -1 && (ans == -1 || ans > bas * t)) ans = bas * t;
			while (m % prime[i] == 0) {
				m /= prime[i], bas *= prime[i];
				t = dfs(i + 1, m);
				if (t != -1 && (ans == -1 || ans > bas * t)) ans = bas * t;
			}
		}
	if (k >= N && zhi(k + 1) && (ans == -1 || ans > 1ll + k))
		ans = 1ll + k;
	return ans > 2147483647 ? -1 : ans;
}
int main() {
	shai();
	scanf("%d", &n);
	printf("%lld\n", dfs(1, n));
	return 0;
}