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\(n,m\)(\(t,m,n\leq5*10^4\))，求$ \sum^n_{i=1}\summ_{j=1}d(i*j)$

题解

假设\(n\leq m\)
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\(c_q\leq a_q\)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}}\lfloor\frac{n}{e}\rfloor\lfloor\frac{m}{f}\rfloor[gcd(e,f)=1]\( =\)\sum_{e=1}^{{n}\sum_{f=1}}\lfloor\frac{n}{e}\rfloor\lfloor\frac{m}{f}\rfloor\sum_{i|e,i|f}\mu(i)\( =\)\sum_{i=1}^{{n}\mu(i)\sum_{i|e}}\rfloor}\sum_{i|f}^{m}{\lfloor\frac{m}{f}\rfloor}\( =\)\sum_{i=1}^{{n}\mu(i)\sum_{e=1}}\rfloor}{\lfloor\frac{n}{ei}\rfloor}\sum_{f=1}^{\lfloor\frac{m}{i}\rfloor}{\lfloor\frac{m}{fi}\rfloor}\( 设\)g(x)=\sum_{i=1}^{x}{\lfloor\frac{x}{i}\rfloor}\(，预处理\)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('\n');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('\n');
	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;
}