bzoj 3527: [Zjoi2014]力 快速傅里叶变换

题意：

给出n个数q_i，给出F_j的定义如下： 
 
令E_i=F_i/q_i，求E_i. 

　　fft的那一堆东西还是背不到啊。。。这次写虽说完全自己写的，但是还是在参见了以前fft程序的情况下调了很久，主要在如下几点写错：1、非递归中内层数组调用中下表忘掉加k 2、每次转换乘的那个数是cos(...)+isin(...)，不要记混了，且里面是(a/b*2*PI) 3、pp[]没有每次清零这一些逗B错误。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define MAXN 1100000
#define PI 3.1415926535897932384
int m,n;
struct Complex
{
        double x,y;
        Complex(){};
        Complex(double x,double y=0):x(x),y(y){};
        Complex operator +(Complex a)
        {
                return Complex(x+a.x,y+a.y);
        }
        Complex operator -(Complex a)
        {
                return Complex(x-a.x,y-a.y);
        }
        Complex operator *(Complex a)
        {
                return Complex(x*a.x-y*a.y,x*a.y+y*a.x);
        }
};
Complex ww[MAXN][2];
void dft(Complex g[],int len,bool d)
{
        Complex t;
        for (int i=1;i<len;i<<=1)
        {
                for (int j=0;j<len;j+=(i<<1))
                {
                        for (int k=0;k<i;k++)
                        {
                                t=g[k+j];
                                g[k+j]=g[k+j]+g[k+j+i]*ww[k * (n/(i<<1))][d];
                                g[k+j+i]=t-g[k+j+i]*ww[k * (n/(i<<1))][d];
                        }
                }
        }
}
int pp[MAXN];
Complex g1[MAXN],g2[MAXN];
void fft(double s1[],double s2[],int m,double res[])
{
        int i,j,k,x;
        n=m;
        while (n != (n&(-n)))n-=n&(-n);
        n<<=2;
        memset(pp,0,sizeof(pp));
        for (i=0;i<n;i++)
        {
                for (x=1,j=n>>1;j;j>>=1,x<<=1)
                {
                        pp[i]+=((i&j)!=0)*x;
                }
        }
        for (i=0;i<n;i++)g1[pp[i]]=s1[i];
        for (i=0;i<n;i++)g2[pp[i]]=s2[i];
        for (i=0;i<=n;i++)
        {
                ww[i][0]=Complex(cos(2*PI*i/n),-sin(2*PI*i/n));
                ww[i][1]=Complex(ww[i][0].x,-ww[i][0].y);
        }
        dft(g1,n,0);
        dft(g2,n,0);
        for (i=0;i<n;i++)g2[i]=g1[i]*g2[i];
        for (i=0;i<n;i++)g1[pp[i]]=g2[i];
        dft(g1,n,1);
        for (i=n;i>=0;i--)
                res[i]=g1[i].x/n;
}
double q1[MAXN],q2[MAXN],a[MAXN];
double r1[MAXN],r2[MAXN];
double f[MAXN];
double s1[MAXN],s2[MAXN];
int main()
{
        //freopen("input.txt","r",stdin);
        //freopen("output.txt","w",stdout);
        int n;
        int i,j,k;
        scanf("%d",&n);
        int x,y,z;
        for (i=0;i<n;i++)
                scanf("%lf",q1+i),q2[n-i-1]=q1[i];
        for (i=1;i<n;i++)
                a[i]=1.0/i/i;
        fft(q1,a,n,r1);
        fft(q2,a,n,r2);
        for (i=0;i<n;i++)
        {
                f[i]+=r1[i];
                f[i]-=r2[n-i-1];
        }
        for (i=0;i<n;i++)
        {
                printf("%.5lf ",f[i]);
        }

}