## [BZOJ3527][ZJOI2014]力：FFT

#### 分析

$E_i=\sum_{j<i}{\frac{q_i}{(i-j)^2}}-\sum_{j>i}{\frac{q_i}{(i-j)^2}}$

$a:q_1 \quad q_2 \quad q_3 \quad q_4 \quad q_5$

$b:-\frac{1}{16} \quad -\frac{1}{9} \quad -\frac{1}{4} \quad -\frac{1}{1} \quad 0 \quad \frac{1}{1} \quad \frac{1}{4} \quad \frac{1}{9} \quad \frac{1}{16}$

FFT即可。

#### 代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <complex>
#define rin(i,a,b) for(int i=(a);i<=(b);i++)
#define rec(i,a,b) for(int i=(a);i>=(b);i--)
using std::cin;
using std::cout;
using std::endl;
typedef long long LL;
typedef std::complex<double> Complex;

const int MAXN=100005;
const double pi=3.14159265358979;
int n,m,len;
int rev[MAXN<<3];
Complex a[MAXN<<3],b[MAXN<<3];

inline void fft(Complex *c,int dft){
rin(i,0,n-1) if(i<rev[i])
std::swap(c[i],c[rev[i]]);
for(int mid=1;mid<n;mid<<=1){
int r=(mid<<1);
Complex u=(Complex){cos(pi/mid),dft*sin(pi/mid)};
for(int l=0;l<n;l+=r){
Complex v=1;
for(int i=0;i<mid;i++,v*=u){
Complex x=c[l+i],y=v*c[l+mid+i];
c[l+i]=x+y;
c[l+mid+i]=x-y;
}
}
}
if(dft==-1) rin(i,0,n-1)
c[i]/=n;
}

int main(){
scanf("%d",&n);
n--;
rin(i,0,n){
double x;
scanf("%lf",&x);
a[i]=x;
}
m=(n<<1);
rin(i,0,m){
if(i<n) b[i]=-1.0/(n-i)/(n-i);
else if(i==n) b[i]=0;
else b[i]=1.0/(i-n)/(i-n);
}
int nn=n;
for(m+=n,n=1;n<=m;n<<=1) len++;
rin(i,1,n-1) rev[i]=((rev[i>>1]>>1)|((i&1)<<(len-1)));
fft(a,1);
fft(b,1);
rin(i,0,n-1) a[i]*=b[i];
fft(a,-1);
n=nn;
rin(i,n,n+n) printf("%.10lf\n",a[i].real());
return 0;
}


posted on 2018-11-23 09:24  ErkkiErkko  阅读(66)  评论(0编辑  收藏

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