BZOJ3771:Triple FFT+容斥原理
Feb 04, 2015 09:09:19 AM
思路:
首先第一次看的时候认为是神题.
再一次看的时候卧槽这不是FFT裸题么.
然后写的时候发现有点坑...
(FFT都告诉你了难道还不会么)
#include<cmath> #include<cstdio> #include<cctype> #include<cstring> #include<iostream> #include<algorithm> using namespace std; typedef double f2; static const f2 pi=acos(-1.0); struct Complex{ f2 u,v; Complex(){} Complex(f2 _u,f2 _v):u(_u),v(_v){} Complex operator+(const Complex&B)const{return Complex(u+B.u,v+B.v);} Complex operator-(const Complex&B)const{return Complex(u-B.u,v-B.v);} Complex operator*(const Complex&B)const{return Complex(u*B.u-v*B.v,u*B.v+v*B.u);} Complex operator*(const f2&p)const{return Complex(p*u,p*v);} Complex operator/(const f2&p)const{return Complex(u/p,v/p);} inline void operator=(const f2&p){u=p,v=0;} inline void operator+=(const Complex&B){u+=B.u,v+=B.v;} inline void operator-=(const Complex&B){u-=B.u,v-=B.v;} inline void operator*=(const Complex&B){f2 uu=u*B.u-v*B.v,vv=u*B.v+v*B.u;u=uu,v=vv;} inline void operator*=(const f2&p){u*=p,v*=p;} inline void operator/=(const f2&p){u/=p,v/=p;} }; inline int revbit(int x,int b){ int res=0;for(int i=0;i<b;++i)if((x>>i)&1)res+=1<<(b-1-i);return res; } inline void fft(Complex A[],int n,int rev){ register int i,j,k; int bit=0,tmp=n;for(;tmp^1;tmp>>=1)++bit; int ano;for(i=0;i<n;++i){ano=revbit(i,bit);if(i<ano)swap(A[i],A[ano]);} Complex wn,w,t; for(k=2;k<=n;k<<=1) for(wn=Complex(cos(2*pi/k),rev*sin(2*pi/k)),i=0;i<n;i+=k) for(w=1,j=0;j<k/2;++j,w*=wn) t=w*A[i+j+k/2],A[i+j+k/2]=A[i+j]-t,A[i+j]+=t; if(rev==-1)for(i=0;i<n;++i)A[i]/=n; } #define N 131072 inline void mul(Complex A[],Complex B[],Complex C[]){ static Complex sa[N],sb[N]; memcpy(sa,A,N*sizeof(Complex)),memcpy(sb,B,N*sizeof(Complex)); fft(sa,N,1),fft(sb,N,1); for(int i=0;i<N;++i)C[i]=sa[i]*sb[i]; fft(C,N,-1); } Complex A[N],A2[N],A3[N]; Complex mul2[N],mul3[N],mul3_2[N]; long long res[N]; long long up(const f2&p){ return(long long)(p+.5); } int main(){ int n;scanf("%d",&n);register int i,j; int x;while(n--)scanf("%d",&x),A[x]=1,A2[2*x]=1,A3[3*x]=1; mul(A,A,mul2); mul(mul2,A,mul3); mul(A,A2,mul3_2); for(int i=0;i<N;++i)mul2[i]-=A2[i],mul2[i]*=.5; for(int i=0;i<N;++i)mul3_2[i]-=A3[i]; for(int i=0;i<N;++i)mul3[i]-=A3[i]+mul3_2[i]*3,mul3[i]/=6; for(int i=0;i<N;++i)res[i]+=up(A[i].u)+up(mul2[i].u)+up(mul3[i].u); for(int i=0;i<N;++i)if(res[i]!=0)printf("%d %lld\n",i,res[i]); return 0; }