[Los] P2257 YY Valley of GCD

Topic links: https://www.luogu.org/problemnew/show/P2257

Solution: Mobius inversion, this is not a simple, entry Mobius inversion, need math, so hard ~ ~ ~

#include<bits/stdc++.h>
typedef long long int ll;
using namespace std;
const int maxn = 10000000 + 10;
int mu[maxn], vis[maxn], prim[maxn];	//莫比乌斯需要的数组，可以求出素数以及个数
int cnt = 0;							//有多少个素数
void get_mu(int n)
{
	mu[1] = 1;
	for (int i = 2; i <= n; i++)
	{
		if (!vis[i]) { prim[++cnt] = i; mu[i] = -1; }
		for (int j = 1; j <= cnt && prim[j] * i <= n; j++)
		{
			vis[prim[j] * i] = 1;
			if (i%prim[j] == 0)break;
			else mu[i*prim[j]] = -mu[i];
		}
	}
}
int sum[maxn],f[maxn];					//用来求前缀和与后面的预处理的结果
void solve()
{
	for (int i = 1; i <= cnt; i++)
		for (int j = 1; prim[i] * j <= 10000000; j++)
			f[j*prim[i]] += mu[j];
	for (int i = 1; i <= 10000000; i++)
		sum[i] = sum[i - 1] + f[i];
}

int main()
{
	get_mu(10000000);
	solve();
	int t;
	cin >> t;
	while (t--)
	{
		int n, m;
		cin >> n >> m;
		if (n > m)
		{
			n = n ^ m;
			m = n ^ m;
			n = n ^ m;
		}
		ll ans = 0;
		for (int l = 1, r; l <= n; l=r+1)
		{
			r = min(n / (n / l), m / (m / l));
			ans += (ll)(sum[r] - sum[l - 1])*(ll)(n / l)*(ll)(m / l);//这里不能加上（r-l+1），因为每次加上的值不一样，需要用前缀和处理
		}
		cout << ans << endl;
	}

	return 0;
}