【Luogu P2257】YY 的 GCD

题目

求:

\[\sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) \in \mathbb P]
\]

有 \(T\) 组数据, \(T\le 10^4, n, m\le 10^7\)

分析

莫比乌斯反演:

\[\begin{align*}
& \sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) \in \mathbb P]\\
= & \sum_{p \in \mathbb P, p\le \min(n, m)}\sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) = p]\\
\end{align*}
\]

设 \(f(x) = \sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) = x]\), $F(x) = \sum_{i = 1}^n \sum_{j = 1}^m [x\ |\gcd(i, j)]=\left\lfloor \frac nx\right\rfloor\left\lfloor \frac mx\right\rfloor $

则有:

\[F(x) = \sum_{x|d, d\le \min(n, m)} f(d) \Rightarrow f(x) = \sum_{x|d, d\le\min(n, m)} \mu(\frac dx)F(d)
\]

故有:

\[\begin{align*}
& \sum_{p \in \mathbb P, p\le \min(n, m)}\sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) = p]\\
= & \sum_{p \in \mathbb P, p\le \min(n, m)} f(p) \\
= & \sum_{p \in \mathbb P, p\le \min(n, m)} \sum_{p|d,d\le \min(n, m)} \mu(\frac dp) \left\lfloor \frac nd\right\rfloor\left\lfloor \frac md\right\rfloor \\
= & \sum_{d = 1}^{\min(n, m)} \sum_{p \in \mathbb P,p|d, p\le \min(n, m)} \mu(\frac dp) \left\lfloor \frac nd\right\rfloor\left\lfloor \frac md\right\rfloor \\
= & \sum_{d = 1}^{\min(n, m)} \left\lfloor \frac nd\right\rfloor\left\lfloor \frac md\right\rfloor \sum_{p \in \mathbb P,p|d, p\le \min(n, m)} \mu(\frac dp)
\end{align*}
\]

设 $g(x) = \sum_{p \in \mathbb P,p|x, p\le \min(n, m)} \mu(\frac xp) $

求法(暴力出奇迹, 经测试下面算法耗时不到 \(0.5 s\)):

void get_g(int n) {
	for(int i = 1; i <= n; i++) {
		int tmp = i;
		while(tmp != 1) {
			g[i] += mobius[i / s_factor[tmp]];
			int p = s_factor[tmp];
			if(tmp % (p * p) == 0) break;
			for(; tmp % p == 0; tmp /= p);
		}
		g_prefix[i] = g_prefix[i - 1] + g[i];
	}
}

有上式等于:

\[\sum_{d = 1}^{\min(n, m)} \left\lfloor \frac nd\right\rfloor\left\lfloor \frac md\right\rfloor g(d)
\]

对于 \(\left\lfloor \frac nd\right\rfloor\left\lfloor \frac md\right\rfloor\) 相同的值整除分块即可.

代码

#include <bits/stdc++.h>

typedef long long Int64;

const int kMaxSize = 1e7 + 5;

int s_factor[kMaxSize], prime[kMaxSize], mobius[kMaxSize], g[kMaxSize],
	g_prefix[kMaxSize], prime_tot = 0;
bool isn_prime[kMaxSize];

void euler_sieve(int n) {
	mobius[1] = 1;
	isn_prime[0] = isn_prime[1] = true;
	for(int i = 2; i <= n; i++) {
		if(!isn_prime[i]) {
			prime[prime_tot++] = i;
			s_factor[i] = i;
			mobius[i] = -1;
		}
		for(int j = 0; j < prime_tot && i * prime[j] <= n; j++) {
			isn_prime[i * prime[j]] = true;
			s_factor[i * prime[j]] = prime[j];
			mobius[i * prime[j]] = -mobius[i];
			if(i % prime[j] == 0) {
				mobius[i * prime[j]] = 0;
				break;
			}
		}
	}
}

void get_g(int n) {
	for(int i = 1; i <= n; i++) {
		int tmp = i;
		while(tmp != 1) {
			g[i] += mobius[i / s_factor[tmp]];
			int p = s_factor[tmp];
			if(tmp % (p * p) == 0) break;
			for(; tmp % p == 0; tmp /= p);
		}
		g_prefix[i] = g_prefix[i - 1] + g[i];
	}
}

int main() {
	euler_sieve(1e7);
	get_g(1e7);
	int t;
	scanf("%d", &t);
	while(t--) {
		int n, m;
		Int64 ans = 0;
		scanf("%d%d", &n, &m);
		if(!n) break;
		for(int l = 1, r; l <= n && l <= m; l = r + 1) {
			r = std::min(n / (n / l), m / (m / l));
			ans += (Int64)(n / l) * (m / l) * (g_prefix[r] - g_prefix[l - 1]);
		}
		printf("%lld\n", ans);
	}
	return 0;
}