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  • 【BZOJ4836】[Lydsy1704月赛]二元运算(分治FFT)

    传送门

    显然权值转下标后分治FFTFFT即可

    利用MTTMTT的技巧可以做到DFTDFT次数减半

    卡到bzoj rk2bzoj rk2
    不知到第一是怎么做到2k代码跑那么快的

    #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 N=200005;
    ll ans[N];
    int A[N],B[N];
    namespace Poly{
    cs int C=18,L=(1<<C)+1;
    cs double pi=acos(-1);
    struct plx{
        double x,y;
        plx(double _x=0,double _y=0):x(_x),y(_y){}
        friend inline plx operator +(cs plx &a,cs plx &b){
            return plx(a.x+b.x,a.y+b.y);
        }
        friend inline plx operator -(cs plx &a,cs plx &b){
            return plx(a.x-b.x,a.y-b.y);
        }
        friend inline plx operator *(cs plx &a,cs plx &b){
            return plx(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
        }
        friend inline plx operator /(cs plx &a,cs int &b){
            return plx(a.x/b,a.y/b);
        }
        inline plx conj(){return plx(x,-y);}
    };
    plx *w[C+1];
    int rev[L];
    inline void init_w(){
        for(int i=1;i<=C;i++)w[i]=new plx[(1<<(i-1))+1];
        plx wn=plx(cos(pi/(1<<(C-1))),sin(pi/(1<<(C-1))));
        w[C][0]=plx(1,0);
        for(int i=1;i<(1<<(C-1));i++){
            if(i&31)w[C][i]=w[C][i-1]*wn;
            else w[C][i]=plx(cos(i*pi/(1<<(C-1))),sin(i*pi/(1<<(C-1))));
        }
        for(int i=C-1;i;i--)
        for(int j=0;j<(1<<(i-1));j++)w[i][j]=w[i+1][j<<1];
    }
    inline void init_rev(int lim){
        for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
    }
    inline void fft(plx *f,int lim,int kd){
        for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
        plx a0,a1;
        for(int mid=1,l=1;mid<lim;mid<<=1,l++)
        for(int i=0;i<lim;i+=mid<<1)
        for(int j=0;j<mid;j++)
        a0=f[i+j],a1=f[i+j+mid]*w[l][j],f[i+j]=a0+a1,f[i+j+mid]=a0-a1;
        if(kd==-1){
            reverse(f+1,f+lim);
            for(int i=0;i<lim;i++)f[i]=f[i]/lim;
        }
    }
    #define poly vector<plx>
    inline poly operator *(poly a,poly b){
        int deg=a.size()+b.size()-1,lim=1;
        while(lim<deg)lim<<=1;
        init_rev(lim);
        a.resize(lim),b.resize(lim);
        fft(&a[0],lim,1),fft(&b[0],lim,1);
        for(int i=0;i<lim;i++)a[i]=a[i]*b[i];
        fft(&a[0],lim,-1),a.resize(deg);
        return a;
    }
    inline void mul(int *A,int *B,int l,int r){
        static plx a[L],b[L],c[L],da,db,dc,dd;
        int mid=(l+r)>>1,deg=r-l+1,lim=1;
        while(lim<deg)lim<<=1;
        init_rev(lim);
        for(int i=0;i<lim;i++)a[i]=plx(0,0),b[i]=plx(0,0),c[i]=plx(0,0);
        for(int i=l;i<=mid;i++)a[i-l].x=A[i],b[mid-i].y=B[i];
        for(int i=mid+1;i<=r;i++)b[i-mid-1].x=A[i],a[i-mid-1].y=B[i];
        fft(a,lim,1),fft(b,lim,1);
        for(int i=0;i<lim;i++){
            int j=(lim-i)&(lim-1);
            da=(a[i]+a[j].conj())*plx(0.5,0);
            dd=(a[j].conj()-a[i])*plx(0,0.5);
            db=(b[i]+b[j].conj())*plx(0.5,0);
            dc=(b[j].conj()-b[i])*plx(0,0.5);
            c[i]=(da*dd)+((db*dc)*plx(0,1));
        }
        fft(c,lim,-1);
        for(int i=0;i<deg;i++)ans[l+mid+1+i]+=(ll)(c[i].x+0.5),ans[i+1]+=(ll)(c[i].y+0.5);
    }
    }
    using Poly::init_w;
    using Poly::mul;
    inline void calc(int l,int r){
        if(r-l+1<=32){
            for(int i=l;i<=r;i++)
            for(int j=l;j<=r;j++){
                if(i<j)ans[i+j]+=1ll*A[i]*B[j];
                else ans[i-j]+=1ll*A[i]*B[j];
            }return;
        }
        int mid=(l+r)>>1;     
        mul(A,B,l,r);
        calc(l,mid),calc(mid+1,r);
    }
    int n,m,q;
    inline void solve(){
        memset(ans,0,sizeof(ans));
        memset(A,0,sizeof(A));
        memset(B,0,sizeof(B));
        int mx=0;
        n=read(),m=read(),q=read();
        for(int i=1,x;i<=n;i++)x=read(),chemx(mx,x),A[x]++;
        for(int i=1,x;i<=m;i++)x=read(),chemx(mx,x),B[x]++;
        calc(0,mx);
        while(q--)cout<<ans[read()]<<'
    ';
    }
    int main(){
        #ifdef Stargazer
        freopen("lx.in","r",stdin);
        #endif
        init_w();
        int T=read();
        while(T--)solve();
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328318.html
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