Crash的数字表格

前言

板题三号

题目

洛谷

讲解

求 (sum_{i=1}^nsum_{j=1}^mfrac{i*j}{gcd(i,j)})

(sum_{i=1}^nsum_{j=1}^msum_{d|i,d|j,gcd(i/d,j/d)=1}frac{i*j}{d})

(sum_{d=1}^{min(n,m)}dsum_{i=1}^{lfloorfrac{n}{d} floor}sum_{j=1}^{lfloorfrac{m}{d} floor}[gcd(i,j)=1]*i*j)

令 (f(n,m)=sum_{i=1}^{n}sum_{j=1}^{m}[gcd(i,j)=1]*i*j)

(sum_{d=1}^{min(n,m)}sum_{d|i}^nsum_{d|j}^mmu(d)i*j)

(sum_{d=1}^{min(n,m)}mu(d)*d^2sum_{i=1}^{lfloorfrac{n}{d} floor}sum_{j=1}^{lfloorfrac{m}{d} floor}i*j)

令 (g(n,m)=sum_{i=1}^nsum_{j=1}^mi*j=frac{n*(n+1)}{2}*frac{m*(m+1)}{2})

(f(n,m)=sum_{d=1}^{min(n,m)}mu(d)*d^2*g(n,m))

前半段可以用前缀和，然后可以数论分块，后面的 (g(n,m)) 可以 (O(1)) 算

然后我们代回最原始的式子

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

同样也是数论分块！

所以这道题就是数论分块套数论分块

代码

int s[MAXN],mu[MAXN],prime[MAXN],pn;
bool vis[MAXN];
void sieve(int x)
{
	s[1] = mu[1] = 1;
	for(register 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] + 1ll * mu[i] * i * i % MOD + MOD) % MOD;
	}
}
int g(int x,int y){return (1ll * x * (x+1) / 2 % MOD) * (1ll * y * (y+1) / 2 % MOD) % MOD;}
int f(int x,int y)
{
	int ret = 0;
	if(x > y) swap(x,y);
	for(int l = 1,r;l <= x;l = r+1)
	{
		r = Min(x/(x/l),y/(y/l));
		ret = (ret + 1ll * (s[r] - s[l-1]) * g(x/l,y/l)) % MOD;
	}
	return ret;
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	n = Read(); m = Read();
	if(n > m) swap(n,m);
	sieve(n);
	for(int l = 1,r;l <= n;l = r+1)
	{
		int L = n/l,R = m/l;
		r = Min(n/L,m/R);
		ans = (ans + 1ll * (r-l+1) * (l+r) / 2 % MOD * f(L,R)) % MOD;
	}
	Put((ans + MOD) % MOD);
	return 0;
}