BZOJ2194: 快速傅立叶之二

【传送门:BZOJ2194


简要题意:

  给出位置为0到n-1的a数组和b数组,定义$C[k]=\sum_{k<=i<n}a[i]*b[i-k]$,求出C数组


题解:

  题目即正解系列

  将这个式子变为$C[k]=\sum_{i=k}^{n-1}a[i]*b[i-k]$

  要使得为卷积形式,就将b倒过来,设$b'[i]=b[n-1-i]$

  得到$C[k]=\sum_{i=k}^{n-1}a[i]*b'[n-1-i+k]$

  这样子就可以用FFT求卷积了,所有输出n-1到2n-2的数就行了


参考代码:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
const double PI=acos(-1.0);
struct Complex
{
    double r,i;
    Complex(){}
    Complex(double _r,double _i){r=_r;i=_i;}
    friend Complex operator + (const Complex &x,const Complex &y){return Complex(x.r+y.r,x.i+y.i);}
    friend Complex operator - (const Complex &x,const Complex &y){return Complex(x.r-y.r,x.i-y.i);}
    friend Complex operator * (const Complex &x,const Complex &y){return Complex(x.r*y.r-x.i*y.i,x.r*y.i+x.i*y.r);}
}a[410000],b[410000];
int R[410000];
int n;
void fft(Complex *y,int len,int on)
{
    for(int i=0;i<len;i++) if(i<R[i]) swap(y[i],y[R[i]]);
    for(int i=1;i<len;i<<=1)
    {
        Complex wn(cos(PI/i),sin(on*PI/i));
        for(int j=0;j<len;j+=(i<<1))
        {
            Complex w(1,0);
            for(int k=0;k<i;k++,w=w*wn)
            {
                Complex u=y[j+k];
                Complex v=w*y[j+k+i];
                y[j+k]=u+v;
                y[j+k+i]=u-v;
            }
        }
    }
    if(on==-1) for(int i=0;i<=len;i++) y[i].r/=len;
}
void calc(int n)
{
    int L=0,m=2*n;
    for(n=1;n<=m;n<<=1) L++;
    memset(R,0,sizeof(R));
    for(int i=0;i<n;i++) R[i]=(R[i>>1]>>1)|(i&1)<<(L-1);
    fft(a,n,1);
    fft(b,n,1);
    for(int i=0;i<=n;i++) a[i]=a[i]*b[i];
    fft(a,n,-1);
}
int main()
{
    scanf("%d",&n);n--;
    for(int i=0;i<=n;i++) scanf("%lf%lf",&a[i].r,&b[n-i].r);
    calc(n);
    for(int i=n;i<=2*n;i++) printf("%d\n",int(a[i].r+0.5));
    return 0;
}

 

渺渺时空,茫茫人海,与君相遇,幸甚幸甚
posted @ 2018-05-02 19:21  Star_Feel  阅读(134)  评论(0编辑  收藏  举报