GCD与莫比乌斯反演的勾当

机房最后一个学懵逼钨丝的人

机房最后一个学懵逼钨丝的人

题目一

链接

题目没找到
求$\sum_{1}^{n}\sum_{1}^{m}gcd(i,j)$

式子

$\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} gcd(i,j)$
$\sum\limits_{k=1}^{min(N,M)} k*\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} [gcd(i,j)==k]=\sum\limits_{k=1}^{min(N,M)} k*\sum\limits_{i=1}^{\frac{N}{k}}\sum\limits_{j=1}^{\frac{M}{k}} [gcd(i,j)==1]$
$看这部分\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} [gcd(i,j)==k]$
设$g(d)=\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{M} [gcd(i,j)==d]$
$f(n)=\sum\limits_{n|d}g(d)=[\frac{N}{d}][\frac{M}{d}]$（显然，不说了）
$g(n)=\sum\limits_{n|d}\mu(\frac{d}{n})f(d)$
$g(1)=\sum\limits_{i=1}^{min(N,M)}\mu(i)[\frac{N}{i}][\frac{M}{i}]$
$ans=\sum\limits_{k=1}^{min(N,M)} k*\sum\limits_{i=1}^{min(N/k,M/k)}\mu(i)[\frac{N/k}{i}][\frac{M/k}{i}]$
除法分块嵌套复杂度$O（n）$

注意

除数不为0

代码

/*
segima gcd n,m
*/
#include <algorithm>
#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int N=15000007,mod=1e9+7;
int read() {
    int x=0,f=1;char s=getchar();
    for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
    for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
    return x*f;
}
int pri[N/10],mu[N],cnt;
bool vis[N];
void Euler(int n) {
    vis[1]=mu[1]=1;
    for(int i=2;i<=n;++i) {
        if(!vis[i]) pri[++cnt]=i,mu[i]=-1;
        for(int j=1;j<=cnt&&i*pri[j]<=n;++j) {
            vis[ i*pri[j] ] = 1;
            if(i % pri[j] == 0) {
                mu[ i*pri[j] ] = 0;
                break;
            } else
                mu[ i*pri[j] ] = -mu[i];
        }
    }
    for(int i=1;i<=n;++i) mu[i]+=mu[i-1];
}
int g(int N,int M,int n) {
    int ans=0;
    for(int l=1,r=1;l<=n;l=r+1) {
        r=min(N/(N/l),M/(M/l));
        ans=(ans+1LL*(mu[r]-mu[l-1]+mod)%mod*(N/l)%mod*(M/l)%mod)%mod;
    }
    return ans;
}
ll calc(int a) {return 1LL*a*(a+1)%mod*500000004%mod;}
int main() {
    int N=read(),M=read(),n=min(N,M),ans=0;
    Euler(n);
    for(int l=1,r=1;l<=n;l=r+1) {
        r=min(N/(N/l),M/(M/l));
        ans=(1LL*ans+(1LL*(calc(r)-calc(l-1))%mod+mod)%mod
                *1LL*g(N/l,M/l,min(N/l,M/l)))%mod;
    }
    cout<<ans<<"\n";
    //  ans=0;
    // for(int i=1;i<=N;++i) {
    //  for(int j=1;j<=M;++j) {
    //      ans=(ans+__gcd(i,j))%mod;
    //  }
    // }
    // cout<<ans;
    return 0;
}

题目

类似的差不多=多组询问的T1
https://www.luogu.org/problemnew/show/P2257

链接&&问题

$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[gcd(i,j)==pri]$

公式

$\sum\limits_{k=1}^{N}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[gcd(i,j)==k]$
扣出$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[gcd(i,j)==k]$
啊哈，熟悉的套路
$ 此处省略，在T1或T2都有$
$ans=\sum\limits_{p=pri}^{min(N,M)} \sum\limits_{i=1}^{N/p}\mu(i)[\frac{N/p}{i}][\frac{M/p}{i}]$
A了？1000组询问T飞了
$\sum\limits_{p=pri} \sum\limits_{i=1}^{N/p}\mu(i)[\frac{N/p}{i}][\frac{M/p}{i}]$
$\sum\limits_{p=pri} \sum\limits_{i=1}^{N/p}\mu(\frac{k}{p})[\frac{N}{k}][\frac{M}{k}]$
$\sum\limits_{k=1}^{N}\sum\limits_{p=pri,p|k} \mu(\frac{k}{p})[\frac{N}{k}][\frac{M}{k}]$
$z(k)=\sum\limits_{p=pri,p|k} \mu(\frac{k}{p})$
线性筛预处理出z的前缀和就可以了
它的证明不错
https://orzsiyuan.com/articles/problem-Luogu-2257-YY-GCD/

代码

#include <algorithm>
#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int N=10000007,mod=1e9+7;
int read() {
    int x=0,f=1;char s=getchar();
    for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
    for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
    return x*f;
}
int pri[N/10],mu[N],cnt,z[N];
bool vis[N];
void Euler(int n) {
    vis[1]=mu[1]=1;
    for(int i=2;i<=n;++i) {
        if(!vis[i]) pri[++cnt]=i,mu[i]=-1,z[i]=mu[1];
        for(int j=1;j<=cnt&&i*pri[j]<=n;++j) {
            vis[ i*pri[j] ] = 1;
            if(i % pri[j] == 0) {
                mu[ i*pri[j] ] = 0;
                z[ i*pri[j] ] = mu[i];
                break;
            } else {
                mu[ i*pri[j] ] = -mu[i];
                z[ i*pri[j] ] = -z[i] + mu[i];
            }
        }
    }
    for(int i=1;i<=n;++i) z[i]+=z[i-1];
}
int g(int N,int M,int n) {
    ll ans=0;
    for(int l=1,r=1;l<=n;l=r+1) {
        r=min(N/(N/l),M/(M/l));
        ans=ans+1LL*(z[r]-z[l-1])*(N/l)*(M/l);
    }
    return ans;
}
int main() {
    Euler(10000000);
    int T=read();
    while(T--) {
        int a=read(),b=read();
        printf("%lld\n",g(a,b,min(a,b)));
    }
    return 0;
}

bzoj1101

链接

https://www.lydsy.com/JudgeOnline/problem.php?id=1101

式子

同楼上，或许，比他还要简单的多

代码

#include <algorithm>
#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
const int N=100007,mod=1e9+7;
int read() {
    int x=0,f=1;char s=getchar();
    for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
    for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
    return x*f;
}
int pri[N/10],mu[N],cnt;
bool vis[N];
void Euler(int n) {
    vis[1]=mu[1]=1;
    for(int i=2;i<=n;++i) {
        if(!vis[i]) pri[++cnt]=i,mu[i]=-1;
        for(int j=1;j<=cnt&&i*pri[j]<=n;++j) {
            vis[ i*pri[j] ] = 1;
            if(i % pri[j] == 0) {
                mu[ i*pri[j] ] = 0;
                break;
            } else
                mu[ i*pri[j] ] = -mu[i];
        }
    }
    for(int i=1;i<=n;++i) mu[i]+=mu[i-1];
}
int g(int N,int M,int n) {
    ll ans=0;
    for(int l=1,r=1;l<=n;l=r+1) {
        r=min(N/(N/l),M/(M/l));
        ans=ans+1LL*(mu[r]-mu[l-1])*(N/l)*(M/l);
    }
    return ans;
}
int main() {
    // freopen("11.in","r",stdin);
    // freopen("a.out","w",stdout);
    Euler(100000);
    int T=read();
    while(T--) {
        int a=read(),b=read(),c=read();
        printf("%lld\n",g(a/c,b/c,min(a/c,b/c)));
    }
    return 0;
}