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  • 【LOJ #6076】「2017 山东一轮集训 Day6」三元组(莫比乌斯反演 / 三元环计数)

    传送门

    莫反得到i=1aj=1bk=1calcm(i,j)blcm(j,k)clcm(i,k)sum_{i=1}^asum_{j=1}^bsum_{k=1}^cfrac{a}{lcm(i,j)}frac b{lcm(j,k)}frac c{lcm(i,k)}
    然后按照旧试题那道题的做法三元环计数即可

    #include<bits/stdc++.h>
    using namespace std;
    #define cs const
    #define re register
    #define pb push_back
    #define pii pair<int,int>
    #define ll long long
    #define fi first
    #define se second
    #define bg begin
    cs int RLEN=1<<20|1;
    inline char gc(){
        static char ibuf[RLEN],*ib,*ob;
        (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
        return (ib==ob)?EOF:*ib++;
    }
    inline int read(){
        char ch=gc();
        int res=0;bool f=1;
        while(!isdigit(ch))f^=ch=='-',ch=gc();
        while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
        return f?res:-res;
    }
    template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
    template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
    cs int mod=1e9+7;
    inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
    inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
    inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
    inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
    inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
    inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
    inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
    inline int Inv(int x){return ksm(x,mod-2);}
    inline int fix(int x){return (x<0)?x+mod:x;}
    inline int gcd(int a,int b){return b?gcd(b,a%b):a;}
    cs int N=50005;
    int a,b,c,lim,m,deg[N];
    struct edge{
    	int u,v,w;
    	edge(int a=0,int b=0,int c=0):u(a),v(b),w(c){}
    }E[N*20+1];
    vector<pii> e[N];
    int mu[N],pr[N],tot;
    bitset<N>vis;
    inline void init(cs int len=N-5){
    	mu[1]=1;
    	for(int i=2;i<=len;i++){
    		if(!vis[i])pr[++tot]=i,mu[i]=-1;
    		for(int j=1;j<=tot&&i*pr[j]<=len;j++){
    			vis[i*pr[j]]=1;
    			if(i%pr[j]==0)break;
    			mu[i*pr[j]]=-mu[i];
    		}
    	}
    }
    inline ll calc1(int u,int lcm){
    	return (1ll*(a/lcm)*(b/lcm)*(c/u)+1ll*(a/lcm)*(b/u)*(c/lcm)+1ll*(a/u)*(b/lcm)*(c/lcm));
    }
    inline ll calc2(int x,int y,int z){
    	return (1ll*(a/x)*(b/y)*(c/z)+1ll*(a/x)*(b/z)*(c/y)+1ll*(a/y)*(b/x)*(c/z)+1ll*(a/y)*(b/z)*(c/x)+1ll*(a/z)*(b/x)*(c/y)+1ll*(a/z)*(b/y)*(c/x));
    }
    int vt[N];
    int main(){
    	#ifdef Stargazer
    	freopen("lx.in","r",stdin);
    	#endif
    	init();
    	a=read(),b=read(),c=read();
    	lim=max(max(a,b),c);
    	ll res=0;
    	for(int g=1;g<=lim;g++)if(mu[g])
    	for(int i=1,l=lim/g;i<=l;i++)if(mu[i*g])
    	for(int j=i+1,l2=lim/g/i;j<=l2;j++)if(mu[j*g]&&gcd(i,j)==1){
    		int u=i*g,v=j*g,lcm=i*j*g;
    		deg[u]++,deg[v]++,E[++m]=edge(u,v,lcm);
    		res+=calc1(u,lcm)*mu[u]*mu[v]*mu[u]+calc1(v,lcm)*mu[v]*mu[v]*mu[u];
    	}
    	for(int i=1;i<=lim;i++)res+=1ll*mu[i]*(a/i)*(b/i)*(c/i);
    	for(int u,v,i=1;i<=m;i++){
    		u=E[i].u,v=E[i].v;
    		if(deg[u]>deg[v])e[u].pb(pii(v,E[i].w));
    		else e[v].pb(pii(u,E[i].w));
    	}
    	for(int u=1;u<=lim;u++)if(mu[u]){
    		for(pii &x:e[u])vt[x.fi]=x.se;
    		for(pii &x:e[u])if(mu[x.fi]){
    			int v=x.fi;
    			for(pii &z:e[v])if(mu[z.fi]){
    				int w=z.fi;if(!vt[w])continue;
    				res+=calc2(vt[w],x.se,z.se)*mu[u]*mu[v]*mu[w];
    			}
    		}
    		for(pii &x:e[u])vt[x.fi]=0;
    	}
    	cout<<(res%mod+mod)%mod<<'
    ';
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328282.html
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