洛谷 - YY的GCD

莫比乌斯反演

P2257 YY的GCD - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

题意

有 $T\;(1<=T<=10^4)$ 组数据，每组给定两个正整数 $n,m\;(1<=n,m<=10^7)$

求 $1<=i<=n,\;1<=j<=m$ 的 $(i,j)$ 中 $\gcd(i,j)$ 为质数的有多少对

思路

$$\begin{aligned} & ans=\sum\limits_{p\in primes}\sum\limits_{i=1}^n\sum\limits_{j=1}^m[\gcd(i,j)==p]\\ &\;\;\;\;\;\;\;=\sum\limits_{p\in primes}\sum\limits_{i=1}^{\lfloor\frac np\rfloor }\sum\limits_{j=1}^{\lfloor\frac mp\rfloor }[\gcd(i,j)==1]\\ &\;\;\;\;\;\;\;=\sum\limits_{p\in primes}\sum\limits_{d=1}^n\mu(d)*\lfloor\frac n{d*p}\rfloor*\lfloor\frac m{d*p}\rfloor\\ &令\; T=d*p\\ &\;\;\;\;\;\;\;=\sum\limits_{T=1}^n\sum\limits_{p\mid T}\mu(\frac Tp)*\lfloor \frac nT\rfloor*\lfloor \frac mT\rfloor\\ &令\;h(T)=\sum\limits_{p\mid T}\mu(\frac Tp)\\ &枚举\;p,\;根据调和级数可在 O(nlnn) 预处理出h(T) \end{aligned}$$

代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>

using namespace std;
#define endl "\n"

typedef long long ll;
typedef pair<int, int> PII;
const int N = 1e7 + 10;
int n;
int pr[N / 5], p[N], pe[N], cnt;
int mu[N], h[N];
ll s[N];
void get_primes(int n)
{
	p[1] = 1;
	for (int i = 2; i <= n; i++)
	{
		if (!p[i])
		{
			p[i] = i;
			pr[++cnt] = i;
			pe[i] = i;
		}
		for (int j = 1; j <= cnt && pr[j] <= n / i; j++)
		{
			p[i * pr[j]] = pr[j];
			if (p[i] == pr[j])
			{
				pe[i * pr[j]] = pe[i] * pr[j];
				break;
			}
			pe[i * pr[j]] = pr[j];
		}
	}
}

void presolve(int n)
{
	mu[1] = s[1] = 1;
	for (int i = 2; i <= n; i++)
	{
		if (i == pe[i])
			mu[i] = (i == p[i] ? -1 : 0);
		else
			mu[i] = mu[i / pe[i]] * mu[pe[i]];
	}
	for (int i = 1; i <= cnt; i++)
	{
		for (int j = pr[i]; j <= n; j += pr[i])
			h[j] += mu[j / pr[i]];
	}
	for (int i = 1; i <= n; i++)
		s[i] = s[i-1] + h[i];
}

int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	get_primes(N - 10);
	presolve(N - 10);
	int T;
	cin >> T;
	while(T--)
	{
		int n, m;
		cin >> n >> m;
		if (n > m) swap(n, m);
		ll ans = 0;
		for (int l = 1; l <= n; l++)
		{
			int r = min(n / (n / l), m / (m / l));
			ans += (s[r] - s[l-1]) * (n / l) * (m / l);
			l = r;
		}
		cout << ans << endl;
	}
    return 0;
}