BZOJ2322:[BeiJing2011]梦想封印(&&BZOJ3793) 线性基
BZOJ3628:[JLOI2014]天天酷跑 dp

BZOJ3771:Triple FFT+容斥原理

shinbokuow posted @ Feb 04, 2015 09:09:19 AM in BZOJ with tags FFT 容斥原理 , 613 阅读

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;
}

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