luogu2257 YY的GCD--莫比乌斯反演

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给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对

多组数据T = 10000

N, M <= 10000000

推式子

(sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=p])

(=sum_psum_{i=1}^{n/p}sum_{j=1}^{m/p}[gcd(i,j)=1])

(=sum_psum_{i=1}^{n/p}sum_{j=1}^{m/p}sum_{d|i,d|j}mu(d))

(=sum_{d=1}^nmu(d)sum_plfloorfrac n{dp} floorlfloorfrac m{dp} floor)

令(q=dp)

(=sum_{q=1}^n(sum_{p|q}mu(frac q p))lfloorfrac nq floorlfloorfrac mq floor)

(mu)线性筛

然后在对于质数枚举倍数求对于每个(i)的(sum_{p|i}mu(frac i p))

然后打数论分块就行了

#include <cstdio>
#include <functional>
using namespace std;

const int fuck = 10000000;
int prime[10000010], tot;
bool vis[10000010];
int mu[10000010], sum[10000010];

int main()
{
	mu[1] = 1;
	for (int i = 2; i <= fuck; i++)
	{
		if (vis[i] == false) prime[++tot] = i, mu[i] = -1;
		for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
		{
			vis[i * prime[j]] = true;
			if (i % prime[j] == 0) break;
			mu[i * prime[j]] = -mu[i];
		}
	}
	for (int i = 1; i <= tot; i++)
		for (int j = 1; j * prime[i] <= fuck; j++)
			sum[j * prime[i]] += mu[j];
	for (int i = 1; i <= fuck; i++)
		sum[i] += sum[i - 1];
	int t; scanf("%d", &t);
	while (t --> 0)
	{
		int n, m;
		long long ans = 0; //别忘了初始化。。。
		scanf("%d%d", &n, &m);
		if (n > m) {int t = m; m = n; n = t; }
		for (int i = 1, j; i <= n; i = j + 1)
		{
			j = min(n / (n / i), m / (m / i));
			ans += (sum[j] - sum[i - 1]) * (long long)(n / i) * (m / i);
		}
		printf("%lld
", ans);
	}
	return 0;
}