Luogu P2522 [HAOI2011]Problem b 莫比乌斯反演

Posted 2021-12-30 jackpei

设$f(d)=\sum_i=1^N\sum_j=1^M[gcd(i,j)==d],\\F(n)=\sum_n|df(d)=\lfloor \fracNn \rfloor \lfloor \fracMn \rfloor$

则$f(n)$

$=\sum_n|d\mu(\fracnd)F(d)$

$=\sum_n|d\mu(\fracnd)\lfloor \fracNd \rfloor \lfloor \fracMd \rfloor$

设$d=kn$

$=\sum_k=1^min(\lfloor \fracNn \rfloor,\lfloor \fracMn \rfloor)\space\mu(k)\lfloor \fracNkn \rfloor \lfloor \fracMkn \rfloor$

所以对$\lfloor \fracNkn \rfloor \lfloor \fracMkn \rfloor$整除分块，对$\mu(k)$搞一个前缀和。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<vector>
#include<queue>
#include<map>
#include<set>
#define ll long long
#define R register ll
using namespace std;
namespace Fread 
    static char B[1<<15],*S=B,*D=B;
    #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
    inline int g() 
        R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch==‘-‘?-1:fix;
        do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
    
using Fread::g;
int n,a,b,c,d,x,cnt;
int mu[50010],pri[25000];
bool v[50010];
inline void MU(int n)  mu[1]=1;
    for(R i=2;i<=n;++i) 
        if(!v[i]) pri[++cnt]=i,mu[i]=-1;
        for(R j=1;j<=cnt&&i*pri[j]<=n;++j) 
            v[i*pri[j]]=true;
            if(i%pri[j]==0) break;
            mu[i*pri[j]]=-mu[i];
        
     for(R i=1;i<=n;++i) mu[i]+=mu[i-1];

inline ll calc(int a,int b)  R ret=0; a>b?swap(a,b):void(0);
    for(R l=1,r;l<=a;l=r+1) 
        r=min(a/(a/l),b/(b/l));
        ret+=(ll)(mu[r]-mu[l-1])*(a/l)*(b/l);
     return ret;

signed main() 
#ifdef JACK
    freopen("NOIPAK++.in","r",stdin);
#endif
    MU(50000); n=g(); while(n--)  R ans=0;
        a=g()-1,b=g(),c=g()-1,d=g(),x=g();
        printf("%lld\n",calc(b/x,d/x)-calc(a/x,d/x)-calc(b/x,c/x)+calc(a/x,c/x));
    

 

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2019.06.09