「Luogu2257」YY的GCD

Posted 2021-02-12 lizbaka

「Luogu2257」YY的GCD

蒟蒻的第一道莫反

跟着题解推的式子，但还是记录一下过程吧

本文可能在一定程度上存在谬误，请谨慎分析

若发现文中有错误，如您愿意，恳请您向我指出，不胜感激

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problem

Solution

题目要求：

[ans=sum_{i=1}^Nsum_{j=1}^M[gcd(i,j)in prime]]

令(f(p)=sum_{i=1}^Nsum_{j=1}^M[gcd(i,j)=p](pin prime))

再令(g(p)=sum_{i=1}^Nsum_{j=1}^M[p|gcd(i,j)](pin prime))

于是有

[g(n)=sum_{n|d}f(d)]

反演后可得

[f(n)=sum_{n|d}mu(frac{d}{n})g(d)]

又知(g(d)=lfloorfrac{N}{d} floorlfloorfrac{M}{d} floor)

于是有

[ans=sum_{nin prime}f(n)=sum_{nin prime}sum_{n|d}mu (frac{d}{n})lfloorfrac{N}{d} floorlfloorfrac{M}{d} floor\=sum_{n|d}lfloorfrac{N}{d} floorlfloorfrac{M}{d} floorsum_{nin prime}mu(frac{d}{n})\=sum_{d=1}^{min(N,M)}lfloorfrac{N}{d} floorlfloorfrac{M}{d} floorsum_{n|d,nin prime}mu(frac{d}{n})]

令(sum(d)=sum_{n|d,nin prime}mu(frac{d}{n}))，预处理(sum(d))

那么答案即

[ans=sum_{d=1}^{min(M,N)}lfloorfrac{N}{d} floorlfloorfrac{M}{d} floor sum(d)]

(sum sum(d))仍可以利用前缀和优化，(sumlfloorfrac{N}{d} floorlfloorfrac{M}{d} floor)利用整除分块优化，最终时间复杂度为(O(Tsqrt{min(N,M)}+k))，(k)为预处理复杂度

Code

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define maxn 10000005
#define N 10000000
using namespace std;
typedef long long ll;

template <typename T> void read(T &t)
{
    t=0;int f=0;char c=getchar();
    while(!isdigit(c)){f|=c=='-';c=getchar();}
    while(isdigit(c)){t=t*10+c-'0';c=getchar();}
    if(f)t=-t;
}

int T;
int n,m;
int pri[maxn],pcnt,nop[maxn];
int mu[maxn];
ll sum[maxn],up;

void GetPrime()
{
    nop[1]=1,mu[1]=1;
    for(register int i=2;i<=N;++i)
    {
        if(!nop[i])pri[++pcnt]=i,mu[i]=-1; 
        for(register int j=1;j<=pcnt && i*pri[j]<=N;++j)
        {
            nop[i*pri[j]]=1;
            if(i%pri[j]==0)break;
            else mu[i*pri[j]]=-mu[i];
        }
    }
    for(register int i=1;i<=pcnt;++i)
        for(register int j=1;pri[i]*j<=N;++j)
            sum[pri[i]*j]+=mu[j];
    for(register int i=1;i<=N;++i)
        sum[i]+=sum[i-1];
}

ll Calc()
{
    ll re=0;
    for(register int l=1,r;l<=up;l=r+1)
    {
        r=min(n/(n/l),m/(m/l));
        re+=1ll*(n/l)*(m/l)*(sum[r]-sum[l-1]);
    }
    return re;
}

int main()
{
    read(T);
    GetPrime();
    while(T--)
    {
        read(n),read(m);
        up=min(n,m);
        printf("%lld
",Calc());
    }
    return 0;
}