P3327 [SDOI2015]约数个数和 莫比乌斯反演

P3327 [SDOI2015]约数个数和 莫比乌斯反演

链接

luogu

思路

第一个式子我也不会，luogu有个证明，自己感悟吧。

\[d(ij)=\sum\limits_{x|i}\sum\limits_{y|j}[gcd(x,y)==1]
\]

\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{x|i}\sum\limits_{y|j}[gcd(x,y)==1]
\]

\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}{\left \lfloor \frac{n}{i} \right \rfloor \left \lfloor \frac{m}{j} \right \rfloor \left [ gcd(i,j)==1 \right ]}
\]

\[f(x)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}{\left \lfloor \frac{n}{i} \right \rfloor \left \lfloor \frac{m}{j} \right \rfloor \left [ gcd(i,j)==x \right ]}
\]

\[g(x)=\sum\limits_{x|d} f(d)
\]

\[g(x)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}{\left \lfloor \frac{n}{i} \right \rfloor \left \lfloor \frac{m}{j} \right \rfloor \left [ x|gcd(i,j)\right ]}
\]

\[g(x)=\sum\limits_{i=1}^{\frac{n}{x}}\sum\limits_{j=1}^{\frac{m}{x}}{\left \lfloor \frac{n}{x*i} \right \rfloor \left \lfloor \frac{m}{x*j} \right \rfloor }
\]

\[g(x)=\sum\limits_{i=1}^{\frac{n}{x}}{\left \lfloor \frac{n}{x*i} \right \rfloor }\sum\limits_{j=1}^{\frac{m}{x}}{\left \lfloor \frac{m}{x*j} \right \rfloor }
\]

\[g(x)=\sum\limits_{i=1}^{N}{\left \lfloor \frac{N}{i} \right \rfloor }\sum\limits_{j=1}^{M}{\left \lfloor \frac{M}{j} \right \rfloor }(N=n/x,M=m/x)
\]

整除分块预处理，O（1）查询g(x)

\[f(x)=\sum\limits_{x|d}\mu(\frac{d}{n})g(d)
\]

所求$$f(1)=\sum\limits_{d=1}^{min(m,n)}\mu(d)g(d)$$

g是可以整除分块的

其他

改马蜂，加空格

代码

#include <bits/stdc++.h>
const int N = 5e5+7;
using namespace std;
int read() {
	int x = 0, f = 1; char s = getchar();
	for (;s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
	for (;s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
	return x * f;
}
int n, m, T;
int pri[N], vis[N], tot, mu[N], g[N];
void Euler(int limit) {
	mu[1] = 1;
	for (int i = 2; i <= limit; ++i) {
		if (!vis[i]) {
			pri[++tot] = i;
			mu[i] = -1;
		}
		for (int j = 1; j <= tot && i * pri[j] <= limit; ++j) {
			vis[i * pri[j]] = 1;
			if (i % pri[j] == 0) {
				mu[i * pri[j]] = 0;
				break;
			}
			mu[i * pri[j]] = -mu[i];
		}
	}
	for (int i = 1; i <= limit; ++i) {
		for (int l = 1, r; l <= i; l = r + 1) {
			r = i / (i / l);
			g[i] += (r - l + 1) * (i / r);
		}
		mu[i] += mu[i - 1];
	}
}
void solve() {
	n = read(), m = read();
	if (n > m) swap(n, m);
	long long ans = 0;
	for (int l = 1, r; l <= n; l = r + 1) {
		r = min(n / (n / l), m / (m / l));
		ans += 1LL * (mu[r] - mu[l-1]) * (1LL * g[n/l] * g[m/l]);
	}
	printf("%lld\n", ans);
}
int main() {
	Euler(50000);
	int T = read();
	while (T--) solve();
	return 0;
}