数论分块

模板

即求 $\sum _{i=1}^n\lfloor \frac{n}{i}\rfloor$的值

$y=n/i\mathrm{的}\mathrm{图}\mathrm{像}$

奇妙题解

参考

$$\sum _{i=1}^n\lfloor \frac{n}{i}\rfloor =2\sum _{i=1}^{\lfloor \sqrt{n}\rfloor }\lfloor \frac{n}{i}\rfloor -\lfloor \sqrt{n}{\rfloor }^2$$

int h(cs int x)
{
	int s=sqrt(x),re=0;
	for(ri int i(1);i<=s;++i) re+=x/i;
	return re*2-s*s;
}

(还没验证正确性
upd10.29:AC

一般化

参考

某段数满足：

$$\because \lfloor \frac{n}{l}\rfloor =\lfloor \frac{n}{l+1}\rfloor =\lfloor \frac{n}{l+2}\rfloor =\dots =\lfloor \frac{n}{r-1}\rfloor =\lfloor \frac{n}{r}\rfloor$$

$$\therefore \lfloor \frac{n}{l}\rfloor \le \frac{n}{r}<\lfloor \frac{n}{l}\rfloor +1$$

$$\therefore r\le \lfloor \frac{n}{\lfloor \frac{n}{l}\rfloor }\rfloor ,r_{max}=\lfloor \frac{n}{\lfloor \frac{n}{l}\rfloor }\rfloor$$

$$\therefore \sum _{i=l}^{r}f(\lfloor \frac{n}{i}\rfloor )=(\lfloor \frac{n}{\lfloor \frac{n}{l}\rfloor }\rfloor -l+1)\times \lfloor \frac{n}{l}\rfloor$$

int h(cs int x)
{
	if(!x) return 0;
	int re=0,j;
	for(ri int i(1);i<=x;i=j+1)
		j=x/(x/i),re+=(j-i+1)*(x/i);
	return re;
}

P2261 [CQOI2007]余数求和

参考

$$ans=\sum _{i=1}^{n}k\mathrm{mod}i$$

$$\because a\mathrm{mod}b=a-b\times \lfloor \frac{a}{b}\rfloor$$

$$\therefore ans=\sum _{i=1}^{n}k-i\times \lfloor \frac{k}{i}\rfloor =n\times k-\sum _{i=1}^{n}i\times \lfloor \frac{k}{i}\rfloor$$

然后整除分块

signed main()
{
	read(n),read(k);
	for(ri int l(1),r;l<=n;l=r+1)
	{
		if(k/l) r=min(k/(k/l),n);
		else r=n;
		ans-=(k/l)*(r-l+1)*(l+r)>>1;
	}
	wt(ans+n*k);
	return 0;
}