洛谷 P3911

题目大意

简洁易懂,我就不说了

solution

我们来观察这个式子, (lcm)怎么操作啊? 那我们就转化成(gcd)

(sumlimits_{i = 1}^nsumlimits_{j = 1}^n lcm(A_i, A_j) iff sumlimits_{i = 1}^nsumlimits_{j = 1}^n dfrac{A_iA_j}{gcd(A_i, A_j)})

那还是很难处理啊, 那我们怎么办呢, 我们是不是可以转化成值域来做啊,记录一下每个数的个数.

我们发现这样是可行的, 毕竟 (A_i in [1, 50000])

我们设 (N) 为值域的最大值, (C_i) 是每个元素的个数, 那我们要求的就变成了:

(sumlimits_{i = 1}^Nsumlimits_{j = 1}^N dfrac{i imes j imes C_i imes C_j}{gcd(A_i, A_j)})

这不就是莫比乌斯反演的板子题么, 那我们来推一下反演过程:

枚举公约数:

(sumlimits_{d = 1}^Nsumlimits_{i = 1}^Nsumlimits_{j = 1}^N dfrac{ij}{d}[gcd(A_i, A_j) == d] imes C_i imes C_j)

将后两重循环除以 (d) , (i) 和 (j) 变成 (id) 和 (jd) :

(sumlimits_{d = 1}^Nsumlimits_{i = 1}^{leftlfloorfrac{N}{d} ight floor}sumlimits_{j = 1}^{leftlfloorfrac{N}{d} ight floor} ijd[gcd(A_i, A_j) == 1] imes C_{id} imes C_{jd})

我们伟大的莫比乌斯函数来了:

(sumlimits_{d = 1}^Nsumlimits_{i = 1}^{leftlfloorfrac{N}{d} ight floor}sumlimits_{j = 1}^{leftlfloorfrac{N}{d} ight floor}sumlimits_{k|gcd(i, j)} mu(k)ijd imes C_{id} imes C_{jd})

我们发现 (gcd(i,j) == 1), 共有 (leftlfloorfrac{N}{d} ight floor) 个, 然后改变一下 (k) 的枚举方式:

(sumlimits_{d = 1}^Nsumlimits_{i = 1}^{leftlfloorfrac{N}{d} ight floor}sumlimits_{j = 1}^{leftlfloorfrac{N}{d} ight floor}sumlimits_{k}^{leftlfloorfrac{N}{d} ight floor} mu(k)ijd imes C_{id} imes C_{jd})

交换一下枚举顺序:

(sumlimits_{d = 1}^Nsumlimits_{k}^{leftlfloorfrac{N}{d} ight floor}sumlimits_{i = 1}^{leftlfloorfrac{N}{d} ight floor}sumlimits_{j = 1}^{leftlfloorfrac{N}{d} ight floor} mu(k)ijd imes C_{id} imes C_{jd})

然后将最后两重循环除以 (k), (id, jd) 就变成了 (idk, jdk) :

(sumlimits_{d = 1}^Nsumlimits_{k}^{leftlfloorfrac{N}{d} ight floor}sumlimits_{i = 1}^{leftlfloorfrac{N}{dk} ight floor}sumlimits_{j = }^{leftlfloorfrac{N}{dk} ight floor} mu(k)ijdk^2 imes C_{idk} imes C_{jdk})

我们令 (T = dk) :

(sumlimits_{T = 1}^Nsumlimits_{k | T}sumlimits_{i = 1}^{leftlfloorfrac{N}{T} ight floor}sumlimits_{j = 1}^{leftlfloorfrac{N}{T} ight floor} mu(k)k T ij imes C_{iT} imes C_{jT})

然后交换一下枚举顺序:

(sumlimits_{T = 1}^N T sumlimits_{i = 1}^{leftlfloorfrac{N}{T} ight floor}sumlimits_{j = 1}^{leftlfloorfrac{N}{T} ight floor} ij imes C_{iT} imes C_{jT}sumlimits_{k | T}mu(k)k)

把 (ij) 的两维合并:

(sumlimits_{T = 1}^N T sumlimits_{i = 1}^{leftlfloorfrac{N}{T} ight floor} (i imes C_{iT})^2sumlimits_{k | T}mu(k)k)

最后我们预处理出 (mu(k)k) 就做完了.

Code:

/**
*    Author: Alieme
*    Data: 2020.9.8
*    Problem: P3911
*    Time: O()
*/
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

#define int long long
#define rr register

#define inf 1e9
#define MAXN 100010

using namespace std;

inline int read() {
	int s = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void print(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) print(x / 10);
	putchar(x % 10 + 48);
}

int tot, n, ans;

int c[MAXN], mu[MAXN], prime[MAXN], sum[MAXN];

bool vis[MAXN];

inline void init() {  // 预处理mu和公式
	mu[1] = 1;
	for (rr int i = 2; i <= 50000; i++) {   // 线性筛预处理mu
		if (!vis[i]) prime[++tot] = i, mu[i] = -1;
		for (rr int j = 1; j <= tot; j++) {
			if (i * prime[j] > 50000) break;
			vis[i * prime[j]] = 1;
			mu[i * prime[j]] = -mu[i];
			if (i % prime[j] == 0) {
				mu[i * prime[j]] = 0;
				break;
			}
		}
	}
	for (rr int i = 1; i <= 50000; i++)  // 预处理 mu(k)*k
		for (rr int j = i; j <= 50000; j += i) 
			sum[j] += mu[i] * i;
}

signed main() {
	init();
	n = read();
	for (rr int i = 1; i <= n; i++) c[read()]++;
	for (rr int i = 1; i <= 50000; i++) {          // 暴力求解
		int s = 0;	
		for (rr int j = 1; j <= 50000 / i; j++) s += c[i * j] * j;
		ans += i * s * s * sum[i];
	}
	print(ans);
}