luogu2257 YY的GCD——莫比乌斯反演

注意一下处理前缀和的时候的trick

#include <bits/stdc++.h>
using namespace std;
#define N 10000005
bool vis[N];
int miu[N];
typedef long long LL;
LL sum[N],ans,g[N];
int prime[N];
int cnt,T,n,m;
void Miu()
{
	miu[1] = 1;
	for (int i=2;i<=N-5;i++)
	{
		if (!vis[i])
		{
			prime[++cnt] = i;
			miu[i] = -1;
		}
		for (int j=1;j<=cnt && prime[j] * i <= N-5;j++)
		{
			vis[prime[j] * i] = 1;
			if (i % prime[j] == 0) break;
			else miu[i * prime[j]] = -miu[i];
		}
	}
	//前缀和也可以预处理出来
	for (int i=1;i<=cnt;i++)
		for (int j=1;j * prime[i] <= N-5;j++)
			g[j * prime[i]] += (LL)miu[j];
	for (int i=1;i<=N-5;i++)
		sum[i] = sum[i-1] + g[i];
}
int main()
{
	scanf("%d", &T);
	Miu();
	while (T--)
	{
		scanf("%d%d", &n, &m);
		if (n > m) swap(n,m);
		ans = 0;
		for (int l = 1,r;l<=n;l=r+1)
		{
			r = min(n / (n / l), m / (m / l));	
			ans += 1LL * (n / l) * (m / l) * (sum[r] - sum[l-1]); 
		}
		printf("%lld\n", ans);
	}
}