[luogu P1829] [國家集訓隊]Crash的數字表格 / JZPTAB

我丟 : https://www.luogu.com.cn/problem/P1829
$先假設n < m$
$\begin{aligned} & \sum_{i=1}^n\sum_{j=1}^m lcm(i,j)\\ =& \sum_{i=1}^n\sum_{j=1}^n \frac{i \times j}{\gcd(i,j)} \\ =& \sum_{d=1}^n\sum_{i=1}^{\lfloor n/d \rfloor}\sum_{j=1}^{\lfloor m/d \rfloor}[\gcd(i,j)=1]d\times i \times j (枚舉d=gcd)\\ =& \sum_{d=1}^n\sum_{i=1}^{\lfloor n/d \rfloor}\sum_{j=1}^{\lfloor m/d \rfloor}\sum_{k|\gcd(i,j)}\mu(k) \times d\times i \times j \\ =& \sum_{d=1}^n\sum_{k=1}^{\lfloor n/d\rfloor}\sum_{i=1}^{\lfloor n/kd\rfloor}\sum_{j=1}^{\lfloor m/kd\rfloor}\mu(k)\times d \times ik \times jk \\ =& \sum_{T=1}^{n}T\times\sum_{i=1}^{\lfloor n/T \rfloor}i\times\sum_{j=1}^{\lfloor m/T \rfloor}j\times\sum_{k|T}\mu(k)\times k(設T=kd,k*k*d=T*k) \end{aligned}$

$\large 顯然\sum\limits_{i=1}^{\lfloor n/T \rfloor}i 和\sum_{k|T}\mu(k)\times k都是可以預處理的$
$時間複雜度貌似是 O(n \ ln\ n)，理論上最慢要跑1.6s，然鵝常數不允許，還是Tle了$
code:

#include<bits/stdc++.h>
#define N 10000005
#define mod 20101009
#define int long long
using namespace std;
int prime[N], vis[N], mu[N], s[N], sz;
void init() {
	mu[1] = 1;
	for(int i = 2; i < N; i ++) {
		if(!vis[i]) {
			prime[++ sz] = i;
			mu[i] = mod - 1;
		}
		for(int j = 1; j <= sz && prime[j] * i < N; j ++) {
			vis[prime[j] * i] = 1;
			if(i % prime[j] == 0) break;
			else mu[i * prime[j]] = mod - mu[i];
		}
	}	
	for(int i = 1; i < N; i ++)
		for(int j = i; j < N; j += i)
			s[j] = (s[j] + mu[i] * i % mod) % mod;
}
int calc(int x) {
	return (x * (x + 1) / 2) % mod;
}
int n, m;
signed main() {
	init();
	scanf("%lld%lld", &n, &m);
	int ans = 0;
	for(int T = 1; T < N; T ++) {
		int ha = 0;
		ans += T * calc(n / T) % mod * calc(m / T) % mod * s[T] % mod, ans %= mod;
	}
	printf("%lld", ans);	
	return 0;
}

主要是倍數那裏好像優化不了了，那就換一種方法推吧，枚舉倍數
$\begin{aligned} & \sum_{i=1}^n\sum_{j=1}^m lcm(i,j)\\ =& \sum_{d=1}^n\sum_{k=1}^{\lfloor n/d\rfloor}\sum_{i=1}^{\lfloor n/kd\rfloor}\sum_{j=1}^{\lfloor m/kd\rfloor}\mu(k)\times d \times ik \times jk \\ =& \sum_{d=1}^nd\sum_{k=1}^{\lfloor n/d\rfloor}k^2\mu(k)\sum_{i=1}^{\lfloor n/kd\rfloor}i\sum_{j=1}^{\lfloor m/kd\rfloor}j\\ \end{aligned}$
$\large 這個東西\sum\limits_{k=1}^{\lfloor n/d\rfloor}k^2\mu(k)\sum\limits_{i=1}^{\lfloor n/kd\rfloor}i\sum\limits_{j=1}^{\lfloor m/kd\rfloor}j\\ 看起來很整除分塊$
$\large設g(n,m)=\sum\limits_{i=1}^{ n}i\sum\limits_{j=1}^{m}j,這個顯然可以O(1)算$
$即\large \sum\limits_{k=1}^{\lfloor n/d\rfloor}k^2\mu(k)\sum\limits_{i=1}^{\lfloor n/kd\rfloor}i\sum\limits_{j=1}^{\lfloor m/kd\rfloor}\\=\sum\limits_{k=1}^{\lfloor n/d\rfloor}k^2\mu(k)g(\lfloor n/dk\rfloor,\lfloor m/dk\rfloor)顯然可以整除分塊$

$\large 再設f(n, m)=\sum\limits_{k=1}^{n}k^2\mu(k)g(\lfloor n/k\rfloor,\lfloor m/k\rfloor)$
$\large 那麼\sum\limits_{d=1}^nd\sum\limits_{k=1}^{\lfloor n/d\rfloor}k^2\mu(k)\sum\limits_{i=1}^{\lfloor n/kd\rfloor}i\sum\limits_{j=1}^{\lfloor m/kd\rfloor}j\\ \ \ \ \ =\sum\limits_{d=1}^nd\times f(\lfloor n/d \rfloor, \lfloor m/d \rfloor)$

$就是兩個乘除分塊套在一起時間複雜度 O( \sqrt n^2)=O(n)$
禁止套娃

code:

#include<bits/stdc++.h>
#define N 10000005
#define mod 20101009
#define int long long
using namespace std;
int prime[N], vis[N], mu[N], s[N], sz;
inline void init() {
	mu[1] = 1;
	for(int i = 2; i < N; i ++) {
		if(!vis[i]) {
			prime[++ sz] = i;
			mu[i] = mod - 1;
		}
		for(register int j = 1; j <= sz && prime[j] * i < N; j ++) {
			vis[prime[j] * i] = 1;
			if(i % prime[j] == 0) break;
			else mu[i * prime[j]] = mod - mu[i];
		}
	}	
	for(int i = 1; i < N; i ++) s[i] = (s[i - 1] + i * i % mod * mu[i]) % mod; 
}
int g(int x, int y) {
	return (x * (x + 1) / 2 % mod) * (y * (y + 1) / 2 % mod) % mod;
}
int f(int x, int y) {
	int ret = 0;
	for(int l = 1, r = 0; l <= x; l = r + 1) {
		r = min(x / (x / l), y / (y / l));
		ret += (s[r] - s[l - 1] + mod) * g(x / l, y / l) % mod, ret %= mod;
	}
	return ret;
}
int n, m;
signed main() {
	init();
	scanf("%lld%lld", &n, &m);
	if(n > m) swap(n, m);
	int ans = 0;
	for(int l = 1, r = 0; l <= n; l = r + 1) {
		r = min(n / (n / l), m / (m / l));
		ans += (r - l + 1) * (l + r) / 2 % mod * f(n / l, m / l) % mod, ans %= mod;
	}
	printf("%lld", ans);	
	return 0;
}