「莫比乌斯反演」

1.YY的GCD

\[\sum_{x=1}^n\sum_{y=1}^m[gcd(x,y)==P] \]

\[\sum_{p}\sum_{x=1}^{n/p}\sum_{y=1}^{m/p}[gcd(x,y)==1] \]

\[\sum_p\sum_{d=1}^{n/p}\mu(d)\frac{n}{pd}\frac{m}{pd} \]

\[\sum_{T=1}^n\frac{n}{T}\frac{m}{T}\sum_{p|T}\mu(T/p) \]

设\(f[n]=\sum_{p|n}\mu(n/p)\),从\(f[i]\)转移到\(f[i*pr[j]]\).
\(i\% pr[j]\)
\(pr[j]\)作为p,贡献为\(\mu[i]\).
\(pr[j]\)作为n/p,贡献为\(-f[i]\).
\(i\% pr[j]==0\)
\(pr[j]\)作为p,贡献为\(\mu[i]\).
\(pr[j]\)作为n/p,贡献为0.

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e7+50;
int n,m;
ll pr[N],v[N],f[N],mu[N];
inline void Pre(){
	const int M=1e7;
	mu[1]=1;
	for(int i=2;i<=M;++i){
		if(!v[i]) v[i]=1,pr[++pr[0]]=i,mu[i]=-1;
		for(int j=1;j<=pr[0];++j){
			if(i*pr[j]>M) break;
			v[i*pr[j]]=1;
			if(i%pr[j]!=0) mu[i*pr[j]]=-mu[i];
			else{mu[i*pr[j]]=0;break;}
		}
	}
	for(int i=1;i<=pr[0];++i) for(int j=1;j*pr[i]<=M;++j) f[j*pr[i]]+=mu[j];
	for(int i=1;i<=M;++i) f[i]+=f[i-1];
}
void solve(){
	scanf("%d%d",&n,&m);
	if(n>m) swap(n,m);
	ll ans=0;
	for(int l=1,r;l<=n;l=r+1){
		r=min(n/(n/l),m/(m/l));
		ans+=(f[r]-f[l-1])*(n/l)*(m/l);
	}
	printf("%lld\n",ans);
}
int main(){
	Pre();
	int T;scanf("%d",&T);
	while(T--) solve();
	return 0;
}

2.数表

\[\sum_{i=1}^n\sum_{j=1}^m\sigma(gcd(i,j)) \]

\[\sum_{g=1}^n\sigma(g)\sum_{i=1}^{n/g}\sum_{g=1}^{m/g}[gcd(i,j)==1] \]

\[\sum_{d=1}^n\mu(d)\sum_{g=1}^{n/d}\sigma(g)\frac{n}{gd}\frac{m}{gd} \]

\[\sum_{T=1}^n\frac{n}{T}\frac{m}{T}\sum_{d|T}\mu(d)\sigma(T/d) \]

每次要求\(\sigma(x)<=a\)
设\(f[n]=\sum_{d|n}\mu(d)\sigma(n/d)\).
把所有二元组\((\sigma(n),n)\)升序,把询问升序.
增量构造,每次更新d的倍数.
所以总更新次数\(nln\)
因为需要查询区间和,单点修改,BIT维护.

#include<bits/stdc++.h>
using namespace std;
const int Query=2e4+50;
const int N=1e5+50;
const int M=1e5;
const int mod=(1ll<<31)-1;
int n,m;
int pri[N],v[N],mu[N],c[N],sum[N],ret[N];
inline void add(int p,int v){
	for(;p<N;p+=p&-p) c[p]+=v;
}
inline int get(int p,int ans=0){
	for(;p;p-=p&-p) ans+=c[p];
	return ans;
}
struct Node{int n,m,a,id;}q[N];
bool cmp(Node a,Node b){return a.a<b.a;}
#define pr pair<int,int>
#define fir first
#define sec second
pr ord[N];
inline void pre(){
	mu[1]=1;sum[1]=1;
	for(int i=2;i<=M;++i){
		if(!v[i]) v[i]=1,pri[++pri[0]]=i,mu[i]=-1,sum[i]=1+i;
		for(int j=1;j<=pri[0];++j){
			if(i*pri[j]>M) break;
			v[i*pri[j]]=1;
			if(i%pri[j]!=0) mu[i*pri[j]]=-mu[i],sum[i*pri[j]]=sum[i]*sum[pri[j]];
			else {mu[i*pri[j]]=0;sum[i*pri[j]]=sum[i]*sum[pri[j]]-sum[i/pri[j]]*pri[j];break;}
		}
	}
	for(int i=1;i<=M;++i) ord[i]=make_pair(sum[i],i);
	sort(ord+1,ord+M+1);
}
int main(){
	pre();
	int T;scanf("%d",&T);
	for(int i=1;i<=T;++i) q[i].id=i,scanf("%d%d%d",&q[i].n,&q[i].m,&q[i].a);
	sort(q+1,q+T+1,cmp);
	int it=1;
	for(int i=1;i<=T;++i){
		n=q[i].n,m=q[i].m;
		while(it<=M&&ord[it].fir<=q[i].a){
			for(int j=1;ord[it].sec*j<=M;++j) add(ord[it].sec*j,mu[j]*ord[it].fir);
			it++;
		}
		if(n>m) swap(n,m);
		int ans=0;
		for(int l=1,r;l<=n;l=r+1){
			r=min(n/(n/l),m/(m/l));
			ans+=(get(r)-get(l-1))*(n/l)*(m/l);
		}
		ret[q[i].id]=ans&mod;
	}
	for(int i=1;i<=T;++i) printf("%d\n",ret[i]);
	return 0;
}

3.DZY loves Math
由于一个神奇的性质,导致本来无法线性筛的本题异常简单.
套路不用再讲,不套路的地方又不会讲.
所以鸽掉了.
4.约数个数和
都做过一次还不会,太丢人了...

\[\sum_{i=1}^n\sum_{j=1}^md(ij) \]

\[d(ij)=\sum_{x|i}\sum_{y|j}[gcd(x,y)==1] \]

解释的话,因为i,j互质所以可以看作质因子的自由组合.

\[\sum_{i=1}^n\sum_{j=1}^m\sum_{x|i}\sum_{y|j}[gcd(x,y)==1] \]

\[\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)==1]\frac{n}{i}\frac{m}{j} \]

\[\sum_{d=1}^n\mu(d)\sum_{i=1}^{n/d}\frac{n}{id}\sum_{j=1}^{m/d}\frac{m}{jd} \]

到这里后发现按照原来的套路搞不动了.

\[\sum_{d=1}^n\mu(d)\sum_{i=1}^{n/d}\frac{\frac{n}{d}}{i}\sum_{j=1}^{m/d}\frac{\frac{m}{d}}{j} \]

设\(f[n]=\sum_{i=1}^n\frac{n}{i}\)

\[\sum_{d=1}^n\mu(d)f(n/d)f(m/d) \]

问题在求\(f[n]\).

\[f[n]=\sum_{i=1}^nd(i) \]

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=5e4+50;
ll n,m;
ll pr[N],v[N],mu[N],F[N];
inline void pre(){
	const int M=5e4;
	mu[1]=1;
	for(int i=2;i<=M;++i){
		if(!v[i]) pr[++pr[0]]=i,v[i]=1,mu[i]=-1;
		for(int j=1;j<=pr[0];++j){
			if(i*pr[j]>M) break;
			v[i*pr[j]]=1;
			if(i%pr[j]!=0) mu[i*pr[j]]=-mu[i];
			else {mu[i*pr[j]]=0;break;}
		}
	}
	for(int i=1;i<=M;++i) mu[i]+=mu[i-1];
	for(int D=1;D<=M;++D)
		for(int l=1,r;l<=D;l=r+1)
			r=D/(D/l),F[D]+=(r-l+1)*(D/l);
		
}
inline void solve(){
	scanf("%lld%lld",&n,&m);
	if(n>m) swap(n,m);
	ll ans=0;
	for(ll l=1,r;l<=n;l=r+1){
		r=min(n/(n/l),m/(m/l));
		ans+=1ll*(mu[r]-mu[l-1])*F[n/l]*F[m/l];
	}
	printf("%lld\n",ans);
	return ;
}
int main(){
	pre();
	int T;scanf("%d",&T);
	while(T--) solve();
	return 0;
}

5.数字表格

\[\prod_{i=1}^n\prod_{j=1}^mf[gcd(i,j)] \]

\[\prod_{d=1}^nf[d]^{\sum_{g=1}^{n/d}\mu(g)\frac{n}{gd}\frac{m}{gd}} \]

\[\prod_{g=1}^n\prod_{d=1}^{n/d}(f[d]^{\mu(g)})^{\frac{n}{gd}\frac{m}{gd}} \]

\[\prod_{T=1}^n\prod_{d|T}(f[d]^{\mu(T/d)})^{\frac{n}{T}\frac{m}{T}} \]

处理出前缀积就行了.

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e6+50;
const int mod=1e9+7;
ll n,m;
ll fib[N],mu[N],F[N],pr[N],v[N];
ll mgml(ll a,ll b,ll ans=1){
	b=(b%(mod-1)+mod-1)%(mod-1);
	for(;b;b>>=1,a=a*a%mod) if(b&1) ans=ans*a%mod;
	return ans;
}
inline void pre(){
	const int M=1e6;
	fib[0]=0,fib[1]=1,mu[1]=1;
	for(int i=0;i<=M;++i) F[i]=1;
	for(ll i=2;i<=M;++i){
		fib[i]=(fib[i-1]+fib[i-2])%mod;
		if(!v[i]) v[i]=1,pr[++pr[0]]=i,mu[i]=-1;
		for(int j=1;j<=pr[0];++j){
			if(i*pr[j]>M) break;
			v[i*pr[j]]=1;
			if(i%pr[j]!=0) mu[i*pr[j]]=-mu[i];
			else {mu[i*pr[j]]=0;break;}
		}
	}
	for(ll i=1;i<=M;++i){
		ll st[3]={mgml(fib[i],mod-2),1,fib[i]};
		for(ll j=1;i*j<=M;++j) F[i*j]=F[i*j]*st[mu[j]+1]%mod;
	}
	for(ll i=1;i<=M;++i) F[i]=F[i-1]*F[i]%mod;
}
inline void solve(){
	scanf("%lld%lld",&n,&m);
	if(n>m) swap(n,m);
	ll ans=1;
	for(ll l=1,r;l<=n;l=r+1){
		r=min(n/(n/l),m/(m/l));
		ans=ans*mgml(F[r]*mgml(F[l-1],mod-2)%mod,(n/l)*(m/l))%mod;
	}
	printf("%lld\n",ans);
	return;
}
int main(){
	pre();
	int T;scanf("%d",&T);
	while(T--) solve();
	return 0;
}

6.于神之怒加强版

\[\sum_{i=1}^n\sum_{j=1}^mgcd(i,j)^k \]

\[\sum_{g=1}^n\mu(g)\sum_{d=1}^{n/g}d^k\frac{n}{gd}\frac{m}{gd} \]

\[\sum_{T=1}^n\frac{n}{T}\frac{m}{T}\sum_{g|T}\mu(g)(\frac{T}{g})^k \]

设\(f[n]=\sum_{g|n}\mu(g)(\frac{n}{g})^k\).
由于是积性函数卷积的形式,所以依然是积性的.
\(i\% pr[j]\)

\[f[i*pr[j]]=f[i]*f[pr[j]] \]

\(i\% pr[j]==0\)
\(pr[j]\)只能呆在\(n/g\)里,放在g里就会使\(\mu(g)=0\).

\[f[i*pr[j]]=f[i]*E[pr[j]] \]

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=5e6+50;
const int mod=1e9+7;
ll n,m,k;
ll v[N],pr[N],mu[N],F[N],E[N];
ll mgml(ll a,ll b,ll ans=1){
	for(;b;b>>=1,a=a*a%mod) if(b&1) ans=ans*a%mod;
	return ans;
}
inline void pre(){
	const int M=5e6;
	mu[1]=1;E[1]=1; F[1]=1;
	for(int i=2;i<=M;++i){
		if(!v[i]) v[i]=1,pr[++pr[0]]=i,mu[i]=-1,E[i]=mgml(i,k),F[i]=E[i]-1;
		for(int j=1;j<=pr[0];++j){
			if(i*pr[j]>M) break;
			v[i*pr[j]]=1;
			E[i*pr[j]]=E[i]*E[pr[j]]%mod;
			if(i%pr[j]!=0) mu[i*pr[j]]=-mu[i],F[i*pr[j]]=F[i]*F[pr[j]]%mod;
			else {mu[i*pr[j]]=0;F[i*pr[j]]=F[i]*E[pr[j]]%mod;break;}
		}
	}
//	for(int i=1;i<=M;++i) printf("F[%d]=%lld\n",i,F[i]);
	for(int i=1;i<=M;++i) F[i] =(F[i]+F[i-1])%mod;
}
inline void solve(){
	scanf("%lld%lld",&n,&m);
	if(n>m) swap(n,m);
	ll ans=0;
	for(ll l=1,r;l<=n;l=r+1){
		r=min(n/(n/l),m/(m/l));
		ans=(ans+1ll*(F[r]-F[l-1]+mod)*(n/l)%mod*(m/l)%mod)%mod;
	}
	printf("%lld\n",ans);
	return;
}
int main(){
	int T;scanf("%d%lld",&T,&k);
	pre();
	while(T--) solve();
	return 0;
}

7.jzptab

\[\sum_{i=1}^n\sum_{j=1}^mlcm(i,j) \]

\[\sum_{i=1}^n\sum_{j=1}^m\frac{ij}{gcd(i,j)} \]

\[\sum_{d=1}^nd\sum_{i=1}^{n/d}i\sum_{j=1}^{m/d}j[gcd(i,j)==1] \]

\[\sum_{T=1}^n\sum_{i=1}^{n/T}i\sum_{j=1}^{m/T}j\ T\sum_{g|T}\mu(g)g \]

设\(f[n]=\sum_{g|n}\mu(g)g\)
\(f=\mu * Id\)
\(i\% pr[j]\)
\(f[i*pr[j]]=f[i]*f[pr[j]]\)
\(i\% pr[j]==0\)
\(pr[j]\)只能在\(n/g\)里.
\(f[i*pr[j]]=f[i]\)

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e7+50;
const int mod=1e8+9;
int n,m;
int pr[N],v[N],mu[N],f[N];
inline void pre(){
	mu[1]=1;
	f[1]=1;
	const int M=1e7;
	for(int i=2;i<=M;++i){
		if(!v[i]) v[i]=1,pr[++pr[0]]=i,mu[i]=-1,f[i]=1-i;
		for(int j=1;j<=pr[0];++j){
			if(i*pr[j]>M) break;
			v[i*pr[j]]=1;
			if(i%pr[j]!=0) mu[i*pr[j]]=-mu[i],f[i*pr[j]]=f[pr[j]]*f[i];
			else {mu[i*pr[j]]=0;f[i*pr[j]]=f[i];break;}
		}
	}
	for(int i=1;i<=M;++i) f[i]=1ll*i*f[i]%mod;
	for(int i=1;i<=M;++i) f[i]=(f[i]+f[i-1])%mod;
}
inline int calc(int D){
	return 1ll*((1ll+(n/D))*(n/D)/2%mod)*((1ll+(m/D))*(m/D)/2%mod)%mod;
}
inline void solve(){
	scanf("%lld%lld",&n,&m);
	if(n>m) swap(n,m);
	int ans=0;
	for(int l=1,r;l<=n;l=r+1){
		r=min(n/(n/l),m/(m/l));
		ans=(ans+1ll*calc(l)*(f[r]-f[l-1])%mod+mod)%mod;
	}
	printf("%lld\n",(ans+mod)%mod);
}
signed main(){
	pre();
	int T;scanf("%lld",&T);
	while(T--) solve();
	return 0;
}

8.一个人的数论
再看这道题就感觉简单多了.
这不是你颓题解的理由

\[\sum_{i=1}^n[gcd(i,n)==1]i^k \]

\[\sum_{i=1}^ni^k\sum_{d|i\&\&d|n}\mu(d) \]

\[\sum_{d|n}\mu(d)d^k\sum_{i=1}^{n/d}i^k \]

\[\frac{1}{k}\sum_{i=0}^kC(k+1,i)B_i\sum_{d|n}\mu(d)d^k(\frac{n}{d})^{k+1-i} \]

\[\frac{n^k}{k}\sum_{i=0}^kC(k+1,i)B_i\sum_{d|n}\mu(d)(\frac{n}{d})^{1-i} \]

\(\mu*id^{1-i}\)是积性的.

\[ans=\frac{n^k}{k}\sum_{i=0}^kC(k+1,i)B_i\prod_{j=1}^{pr[0]}((pr[j]^{c[j]})^{1-i}-(pr[j]^{c[j]-1})^{1-i}) \]

#include<bits/stdc++.h>
#define ll long long
#define int long long
using namespace std;
const int mod=1e9+7;
int k,w,p[1500],c[1500];
ll a[105][105];
ll mgml(ll a,ll b,ll ans=1){
	b=(b%(mod-1)+mod-1)%(mod-1);
	for(;b;b>>=1,a=a*a%mod) if(b&1) ans=ans*a%mod;
	return ans;
}
int L;
inline void gauss(){
	for(int i=1;i<=L;++i){
		int maxi=i;
		for(int j=i+1;j<=L;++j) if(a[j][i]>a[maxi][i]) maxi=j;
		swap(a[maxi],a[i]);
		if(!a[i]) continue;
		ll rate=mgml(a[i][i],mod-2);
		for(int j=1;j<=L+1;++j) a[i][j]=a[i][j]*rate%mod;
		for(int j=1;j<=L;++j){
			if(j==i) continue;
			ll rate=a[j][i];
			for(int d=1;d<=L+1;++d) a[j][d]=(a[j][d]-rate*a[i][d]%mod+mod)%mod;
		}
	}
}
inline int rd(register int x=0,register char ch=getchar(),register int f=1){
	while(!isdigit(ch)) f=ch=='-'?-1:1,ch=getchar();
	while(isdigit(ch)) x=(x<<1)+(x<<3)+ch-48,ch=getchar();
	return x*f;
}
signed main(){
	k=rd(); w=rd(); L=k+1;
	for(int i=1;i<=w;++i) p[i]=rd(),c[i]=rd();
	ll pre=0;
	for(ll i=1;i<=k+1;++i){
		pre=(pre+mgml(i,k))%mod;
		a[i][L+1]= pre;
		for(ll j=1,powi=i;j<=L;++j,powi=1ll*powi*i%mod) a[i][j]=powi;
	}
	gauss();
	ll ans=0;
	for(int i=1;i<=L;++i){
		ll tmp=1;
		for(int j=1;j<=w;++j){
			tmp=tmp*mgml(p[j],i*c[j])%mod*(1-mgml(p[j],k-i)+mod)%mod;
		}
		ans=(ans+a[i][L+1]*tmp%mod+mod)%mod;
	}
	printf("%lld\n",ans);
	return 0;
}