Luogu P2522 [HAOI2011]Problem b

题意

(n) 组询问，每组给定 (a,b,c,d,k)，求 (sumlimits_{i=a}^{b}sumlimits_{j=c}^{d}[gcd(i,j)=k])。

( exttt{Data Range:}1leq n,kleq 5 imes 10^4,1leq aleq bleq 5 imes 10^4,1leq cleq dleq 5 imes 10^4)

题解

首先考虑二维差分一下，发现我们只需要求出类似于 (sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)=k]) 的东西就好了（这里的 (n) 不再是数据组数）

然后考虑推一下式子

[sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)=k]=sumlimits_{i=1}^{leftlfloorfrac{n}{k} ight floor}sumlimits_{j=1}^{leftlfloorfrac{m}{k} ight floor}[gcd(i,j)=1] ]

然后发现可以莫反一下

[sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)=k]=sumlimits_{i=1}^{leftlfloorfrac{n}{k} ight floor}sumlimits_{j=1}^{leftlfloorfrac{m}{k} ight floor}sumlimits_{dmidgcd(i,j)}mu(d) ]

然后交换一下求和变量得到

[sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)=k]=sumlimits_{dmidgcd(i,j)}mu(d)leftlfloorfrac{n}{dk} ight floorleftlfloorfrac{m}{dk} ight floor ]

然后线性筛一下 (mu) 的前缀和就可以整除分块做了。

代码

#include<bits/stdc++.h>
using namespace std;
typedef int ll;
typedef long long int li;
const ll MAXN=2e5+51;
ll ptot,n,u,v,w,x,kk;
ll np[MAXN],prime[MAXN],mu[MAXN],prefix[MAXN];
inline ll read()
{
    register ll num=0,neg=1;
    register char ch=getchar();
    while(!isdigit(ch)&&ch!='-')
    {
        ch=getchar();
    }
    if(ch=='-')
    {
        neg=-1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        num=(num<<3)+(num<<1)+(ch-'0');
        ch=getchar();
    }
    return num*neg;
}
inline void sieve(ll limit)
{
    np[1]=mu[1]=1;
    for(register int i=2;i<=limit;i++)
    {
        if(!np[i])
        {
            prime[++ptot]=i,mu[i]=-1;
        }
        for(register int j=1;i*prime[j]<=limit;j++)
        {
            np[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    for(register int i=1;i<=limit;i++)
    {
        prefix[i]=prefix[i-1]+mu[i];
    }
}
inline ll calc(ll n,ll m)
{
    ll res=0;
    for(register int l=1,r;l<=min(n,m);l=r+1)
    {
        r=min(n/(n/l),m/(m/l));
        res+=(prefix[r]-prefix[l-1])*(n/l/kk)*(m/l/kk);
    }
    return res;
}
int main()
{
    n=read(),sieve(2e5+10);
    for(register int i=1;i<=n;i++)
    {
        u=read(),v=read(),w=read(),x=read(),kk=read();
        printf("%d
",calc(v,x)-calc(u-1,x)-calc(v,w-1)+calc(u-1,w-1));
    }
}