约数个数和

前言

越来越熟练

题目

洛谷

讲解

原式 (=sum_{i=1}^nsum_{j=1}^md(ij))

(=sum_{i=1}^nsum_{j=1}^msum_{x|i}sum_{y|j}[gcd(x,y)=1]) 这一步可谓是至关重要！

改写一下：(sum_{i=1}^nsum_{j=1}^mlfloorfrac{n}{i} floorlfloorfrac{m}{j} floor[gcd(i,j)=1])

令 (f(x)=sum_{i=1}^nsum_{j=1}^mlfloorfrac{n}{i} floorlfloorfrac{m}{j} floor[gcd(i,j)=x])

且 (F(x)=sum_{i=1}^nsum_{j=1}^mlfloorfrac{n}{i} floorlfloorfrac{m}{j} floor[x|gcd(i,j)])

(Rightarrow F(x)=sum_{i=1}^{frac{n}{x}}sum_{j=1}^frac{m}{x}lfloorfrac{n}{xi} floorlfloorfrac{m}{xj} floor)

(Rightarrow f(x)=sum_{x|d}mu(frac{d}{x})F(d))

我们所求的即为 (f(1))

(f(1)=sum_{d=1}^{min(n,m)}mu(d)sum_{i=1}^{frac{n}{d}}sum_{j=1}^frac{m}{d}lfloorfrac{n}{di} floorlfloorfrac{m}{dj} floor)

(Rightarrow f(1)=sum_{d=1}^{min(n,m)}mu(d)sum_{i=1}^{frac{n}{d}}lfloorfrac{n}{di} floorsum_{j=1}^frac{m}{d}lfloorfrac{m}{dj} floor)

令 (g(x)=sum_{i=1}^xlfloorfrac{x}{i} floor)

(Rightarrow f(1)=sum_{d=1}^{min(n,m)}mu(d)g(lfloorfrac{n}{d} floor)g(lfloorfrac{m}{d} floor))

预处理 (g(x)) 即可

代码

int prime[MAXN],pn,s[MAXN],mu[MAXN],g[MAXN];
bool vis[MAXN]; 
void sieve(int x)
{
	s[1] = 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];
		}
		s[i] = s[i-1] + mu[i];
	}
	for(int i = 1;i <= x;++ i)
	{
		LL ret = 0;
		for(int l = 1,r;l <= i;l = r+1)
		{
			r = i / (i/l);
			ret += 1ll * i / l * (r-l+1);
		}
		g[i] = ret;
	}
} 

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	sieve(50000);
	for(int T = Read(); T ;-- T)
	{
		n = Read(); m = Read();
		if(n > m) swap(n,m);
		LL ans = 0;
		for(int l = 1,r;l <= n;l = r+1)
		{
			r = Min(n/(n/l),m/(m/l));
			ans += 1ll * (s[r] - s[l-1]) * g[n / l] * g[m / l];
		}
		Put(ans,'
');
	}
	return 0;
}