【BZOJ】【2194】快速傅里叶之二

FFT

　　c[k]=sigma a[i]*b[i-k] 这个形式不好搞……

　　而我们熟悉的卷积的形式是这样的 c[k]=sigma a[i]*b[k-i]也就是【下标之和是定值】

　　所以我们将a数组反转一下就可以卷积了=。=

/**************************************************************
    Problem: 2194
    User: Tunix
    Language: C++
    Result: Accepted
    Time:2008 ms
    Memory:48148 kb
****************************************************************/
 
//BZOJ 2194
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
using namespace std;
int getint(){
    int v=0,sign=1; char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') sign=-1; ch=getchar();}
    while(isdigit(ch))  {v=v*10+ch-'0'; ch=getchar();}
    return v*sign;
}
/*******************template********************/
const int N=1000010;
const double pi=acos(-1);
struct comp{
    double r,i;
//  comp(){}
    comp(double _=0.0,double __=0.0) : r(_),i(__){}
    comp operator + (const comp &b)const{return comp(r+b.r,i+b.i);}
    comp operator - (const comp &b)const{return comp(r-b.r,i-b.i);}
    comp operator * (const comp &b)const{return comp(r*b.r-i*b.i,r*b.i+i*b.r);}
}a[N],b[N],c[N];
void FFT(comp *a,int n,int type){
    for(int i=1,j=0;i<n-1;++i){
        for(int s=n;j^=s>>=1,~j&s;);
        if(i<j) swap(a[i],a[j]);
    }
    for(int m=1;m<n;m<<=1){
        double u=pi/m*type; comp wm(cos(u),sin(u));
        for(int i=0;i<n;i+=(m<<1)){
            comp w(1,0);
            rep(j,m){
                comp &A=a[i+j+m], &B=a[i+j], t=w*A;
                A=B-t; B=B+t; w=w*wm;
            }
        }
    }
    if (type==-1) rep(i,n) a[i].r/=n;
}
  
int main(){
    int n=getint();
    rep(i,n){
        a[n-i-1].r=getint();
        b[i].r=getint();
    }
    int len=1;
    for(len=1;len<=n<<1;len<<=1);
    FFT(a,len,1); FFT(b,len,1);
    rep(i,len) c[i]=a[i]*b[i];
    FFT(c,len,-1);
    D(i,n-1,0) printf("%d
",int(c[i].r+0.5) );
    return 0;
}