[luogu P1829] [国家集训队]Crash的数字表格 / JZPTAB

我丢 : https://www.luogu.com.cn/problem/P1829
$先假设n<m$
$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j)\\ =& \sum _{i=1}^n\sum _{j=1}^n\frac{i\times j}{\gcd (i,j)}\\ =& \sum _{d=1}^n\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor m/d\rfloor }[\gcd (i,j)=1]d\times i\times j(枚举d=gcd)\\ =& \sum _{d=1}^n\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor m/d\rfloor }\sum _{k|\gcd (i,j)}\mu (k)\times d\times i\times j\\ =& \sum _{d=1}^n\sum _{k=1}^{\lfloor n/d\rfloor }\sum _{i=1}^{\lfloor n/kd\rfloor }\sum _{j=1}^{\lfloor m/kd\rfloor }\mu (k)\times d\times ik\times jk\\ =& \sum _{T=1}^{n}T\times \sum _{i=1}^{\lfloor n/T\rfloor }i\times \sum _{j=1}^{\lfloor m/T\rfloor }j\times \sum _{k|T}\mu (k)\times k(设T=kd,k\ast k\ast d=T\ast k)\end{array}$

$显然\sum _{i=1}^{\lfloor n/T\rfloor }i和{\sum }_{k|T}\mu (k)\times k都是可以预处理的$
$时间复杂度貌似是O(nlnn)，理论上最慢要跑1.6s，然鹅常数不允许，还是Tle了$
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;
}

主要是倍数那里好像优化不了了，那就换一种方法推吧，枚举倍数
$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j)\\ =& \sum _{d=1}^n\sum _{k=1}^{\lfloor n/d\rfloor }\sum _{i=1}^{\lfloor n/kd\rfloor }\sum _{j=1}^{\lfloor m/kd\rfloor }\mu (k)\times d\times ik\times jk\\ =& \sum _{d=1}^{n}d\sum _{k=1}^{\lfloor n/d\rfloor }{k}^2\mu (k)\sum _{i=1}^{\lfloor n/kd\rfloor }i\sum _{j=1}^{\lfloor m/kd\rfloor }j\end{array}$
$$\begin{gathered} 这个东西\sum _{k=1}^{\lfloor n/d\rfloor }{k}^2\mu (k)\sum _{i=1}^{\lfloor n/kd\rfloor }i\sum _{j=1}^{\lfloor m/kd\rfloor }j \\ 看起来很整除分块 \end{gathered}$$
$设g(n,m)=\sum _{i=1}^{n}i\sum _{j=1}^{m}j,这个显然可以O(1)算$
$$\begin{gathered} 即\sum _{k=1}^{\lfloor n/d\rfloor }{k}^2\mu (k)\sum _{i=1}^{\lfloor n/kd\rfloor }i\sum _{j=1}^{\lfloor m/kd\rfloor } \\ =\sum _{k=1}^{\lfloor n/d\rfloor }{k}^2\mu (k)g(\lfloor n/dk\rfloor ,\lfloor m/dk\rfloor )显然可以整除分块 \end{gathered}$$

$再设f(n,m)=\sum _{k=1}^n{k}^2\mu (k)g(\lfloor n/k\rfloor ,\lfloor m/k\rfloor )$
$$\begin{gathered} 那么\sum _{d=1}^{n}d\sum _{k=1}^{\lfloor n/d\rfloor }{k}^2\mu (k)\sum _{i=1}^{\lfloor n/kd\rfloor }i\sum _{j=1}^{\lfloor m/kd\rfloor }j \\ =\sum _{d=1}^{n}d\times f(\lfloor n/d\rfloor ,\lfloor m/d\rfloor ) \end{gathered}$$

$就是两个乘除分块套在一起时间复杂度O({\sqrt{n}}^2)=O(n)$
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;
}