并不对劲的bzoj3994:loj2185:p3327[SDOI2015]约数个数和

题目大意

设d(x)为x的约数个数，(t)组询问，给定(n,m)((t,m,nleq5*10^4))，求$ sum^n_{i=1}summ_{j=1}d(i*j)$

题解

假设(nleq m)
设(i=p_1^{a_1}*p_2^{a_2}*...*p_k^{a_k},j=p_1^{b_1}*p_2^{b_2}*...*p_k^{b_k})
对于(i*j)的某个约数(x)，设(x=p_1^{c_1}*p_2^{c_2}*...*p_k^{c_k})，那么可以用两个数(e,f)表示(x)，当(c_qleq a_q)时(e)的(p_q)的指数为(c_q)，当(c_q> a_q)时(f)的(p_q)的指数为(c_q-a_q)
这样每个(x)都能对应到一对((e,f))上，每对满足(e|i,f|j,gcd(e,f)=1)的((e,f))也能对应到一个(x)上
所以就有(d(i,j)=sum_{e|i}sum_{f|j}[gcd(e,f)=1])
原式=$ sum^n_{i=1}summ_{j=1}sum_{e|i}sum_{f|j}[gcd(e,f)=1]( 把枚举)e,f(放到前面，得原式=)sum_{e=1}^{n}sum_{f=1}{m}lfloorfrac{n}{e} floorlfloorfrac{m}{f} floor[gcd(e,f)=1]( =)sum_{e=1}^{n}sum_{f=1}{m}lfloorfrac{n}{e} floorlfloorfrac{m}{f} floorsum_{i|e,i|f}mu(i)( =)sum_{i=1}^{{n}mu(i)sum_{i|e}}{n}{lfloorfrac{n}{e} floor}sum_{i|f}^{m}{lfloorfrac{m}{f} floor}( =)sum_{i=1}^{{n}mu(i)sum_{e=1}}{lfloorfrac{n}{i} floor}{lfloorfrac{n}{ei} floor}sum_{f=1}^{lfloorfrac{m}{i} floor}{lfloorfrac{m}{fi} floor}( 设)g(x)=sum_{i=1}^{x}{lfloorfrac{x}{i} floor}(，预处理)g(x)( 则原式=)sum_{i=1}^{n}{mu(i)g(n/i)g(m/i)}$
接下来整除分块就行了

代码

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define maxn 50010 
#define lim 50000
#define LL long long
using namespace std;
int read()
{
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-')ch=getchar();
	if(ch=='-')f=-1,ch=getchar();
	while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return x*f;
}
void write(LL x)
{
	if(x==0){putchar('0'),putchar('
');return;}
	int f=0;char ch[20];
	if(x<0)putchar('-'),x=-x;
	while(x)ch[++f]=x%10+'0',x/=10;
	while(f)putchar(ch[f--]);
	putchar('
');
	return;
}
int n,m,t,p[maxn],no[maxn],cnt; 
LL f[maxn],g[maxn],mu[maxn];
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	mu[1]=p[1]=no[1]=1;
	rep(i,2,lim)
	{
		if(!no[i])p[++cnt]=i,mu[i]=-1;
		for(int j=1;j<=cnt&&i*p[j]<=lim;j++)
		{
			no[i*p[j]]=1;
			if(i%p[j]==0){mu[i*p[j]]=0;break;}
			else mu[i*p[j]]=-mu[i];
		}
	}
	rep(i,1,lim)mu[i]+=mu[i-1];
	rep(i,1,lim)
	{
		for(int l=1,r=0;l<=i;l=r+1)
		{
			r=i/(i/l);
			f[i]+=(LL)(i/l)*(LL)(r-l+1);
		}
	}
	t=read();
	while(t--)
	{
		n=read(),m=read();LL ans=0;
		if(n>m)swap(n,m);
		for(int l=1,r=0;l<=n;l=r+1)
		{
			r=min(n/(n/l),m/(m/l));
			ans+=(mu[r]-mu[l-1])*f[n/l]*f[m/l];
		}
		write(ans);
	}
	return 0;
}