BZOJ2194: 快速傅立叶之二

给A，B，求C，$C_k=\sum_{i=0}^{n-1}A_iB_{i-k}$。

B反过来就是模板了。能不能算一道题呢？

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<complex>
#include<math.h>
//#include<iostream>
using namespace std;

int n,m,wei;
#define maxn 262222
const double pi=acos(-1);
typedef complex<double> cp;
cp a[maxn],b[maxn],c[maxn]; int rev[maxn];

void dft(cp *a,int n,int type)
{
    for (int i=0;i<n;i++) if (i<rev[i]) {cp t=a[i]; a[i]=a[rev[i]]; a[rev[i]]=t;}
    for (int i=1;i<n;i<<=1)
    {
        cp base=cp(cos(pi/i),type*sin(pi/i));
        for (int j=0,p=i<<1;j<n;j+=p)
        {
            cp t=cp(1,0);
            for (int k=0;k<i;k++,t=t*base)
            {
                cp tmp=t*a[j+k+i];
                a[j+k+i]=a[j+k]-tmp;
                a[j+k]+=tmp;
            }
        }
    }
}

void mul(cp *a,cp *b,cp *c)
{
    for (int i=0;i<n;i++)
    {
        rev[i]=0;
        for (int j=0;j<wei;j++) rev[i]|=((i>>j)&1)<<(wei-j-1);
    }
    dft(a,n,1); dft(b,n,1); for (int i=0;i<n;i++) c[i]=a[i]*b[i];
    dft(c,n,-1); for (int i=0;i<n;i++) c[i]/=n;
}

int main()
{
    scanf("%d",&n); n--;
    for (int i=0,x,y;i<=n;i++) scanf("%d",&x),a[i]=x,scanf("%d",&y),b[n-i]=y;
    m=n+n; for (n=1,wei=0;n<=m;n<<=1,wei++); mul(a,b,c);
    for (int i=(m>>1);i<=m;i++) printf("%d\n",(int)(c[i].real()+0.5));
    return 0;
}