GCD与莫比乌斯反演的勾当

时间：2024-06-19 12:33:20

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;

}

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