[洛谷P2257] YY的GCD

前言

板题2号

题目

洛谷

讲解

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

则 (f(k)=sum_{i=1}^{frac{min(n,m)}{p}}mu(i)lfloordfrac{n}{ip} floorlfloordfrac{m}{ip} floor)

现 (p) 为任意质数，即求 (sum_{pin prime}sum_{i=1}^{frac{min(n,m)}{p}}mu(i)lfloordfrac{n}{ip} floorlfloordfrac{m}{ip} floor)

换元后得到 (sum_{pin prime}sum_{p|t}mu(dfrac{t}{p})lfloordfrac{n}{t} floorlfloordfrac{m}{t} floor)

我们来枚举(t)，得到 (sum_{t=1}^{min(n,m)}sum_{p|t}mu(dfrac{t}{p})lfloordfrac{n}{t} floorlfloordfrac{m}{t} floor)

把与 (p) 无关的两项提出来：

(sum_{t=1}^{min(n,m)}lfloordfrac{n}{t} floorlfloordfrac{m}{t} floorsum_{p|t}mu(dfrac{t}{p}))

令 (s(t)=sum_{p|t}mu(dfrac{t}{p}))

预处理 (s) 即可

代码

int mu[MAXN],prime[MAXN],pn,s[MAXN];
bool vis[MAXN];
void sieve(int x)
{
	mu[1] = 1;
	for(int i = 2;i <= x;++ i)
	{
		if(!vis[i]) prime[++pn] = i,mu[i] = -1;
		for(int j = 1;j <= pn && i * prime[j] <= x;++ j)
		{
			vis[i * prime[j]] = 1;
			if(i % prime[j] == 0) break;
			mu[i * prime[j]] = -mu[i];
		}
	}
	for(int i = 1;i <= pn;++ i)
		for(int j = prime[i];j <= x;j += prime[i])
			s[j] += mu[j / prime[i]];
	for(int i = 1;i <= x;++ i) s[i] += s[i-1];
}

LL solve(int x,int y)
{
	if(x > y) swap(x,y);
	LL ret = 0;
	for(int l = 1,r;l <= x;l = r+1)
	{
		r = Min(x/(x/l),y/(y/l));
		ret += 1ll * (s[r] - s[l-1]) * (x/l) * (y/l);
	}
	return ret;
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	sieve(10000019); 
	for(int T = Read(); T ;-- T)
	{
		n = Read(); m = Read();
		Put(solve(n,m),'
');
	}
	return 0;
}