loj6221. 幂数!

题意

求出(N(N leq {10}^ {11}))以内的幂数的个数及和。
幂数满足其所有质因数都至少有两个。

题解

不说了,又斯波了。
两个性质:
1.答案较小
2.幂数一定是(a ^ 2 b ^ 3)的形式

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 1;
ll n, ans1, ans2, a[N], b[N];
vector <ll> c;
bool sqrfree (ll x) {
	ll s = sqrtl(x + 0.5);
	return s * s != x;
}
int main () {
	cin >> n;
	a[0] = a[1] = 1;
	for (int i = 2; 1ll * i * i <= n; ++i) {
		a[++a[0]] = 1ll * i * i;
	}
	b[0] = b[1] = 1;
	for (int i = 2; 1ll * i * i * i <= n; ++i) {
		if (sqrfree(1ll * i * i * i)) {
			b[++b[0]] = 1ll * i * i * i;
		}
	}
	for (int i = 1; i <= a[0]; ++i) {
		for (int j = 1; j <= b[0] && a[i] * b[j] <= n; ++j) {
			c.emplace_back(a[i] * b[j]);
		}
	}
	sort(c.begin(), c.end());
	n = unique(c.begin(), c.end()) - c.begin();
	for (int i = 0; i < n; ++i) {
		++ans1;
		ans2 += c[i];
	}
	cout << ans1 << endl;
	cout << ans2 << endl;
	return 0;
}
原文地址:https://www.cnblogs.com/psimonw/p/11246293.html