n行，第i行输出Ei。与标准答案误差不超过1e-2即可。

#include<bits/stdc++.h>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define ls i<<1#define rs ls | 1#define mid ((ll+rr)>>1)#define pii pair<int,int>#define MP make_pairtypedef long long LL;const long long INF = 1e18+1LL;const double pi = acos(-1.0);const int N = 5e5+10, M = 1e3+20,inf = 2e9,mod = 1e9+7;struct Complex {    double r , i ;    Complex () {}    Complex ( double r , double i ) : r ( r ) , i ( i ) {}    Complex operator + ( const Complex& t ) const {        return Complex ( r + t.r , i + t.i ) ;    }    Complex operator - ( const Complex& t ) const {        return Complex ( r - t.r , i - t.i ) ;    }    Complex operator * ( const Complex& t ) const {        return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ;    }} ;void FFT ( Complex y[] , int n , int rev ) {    for ( int i = 1 , j , t , k ; i < n ; ++ i ) {        for ( j = 0 , t = i , k = n >> 1 ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ;        if ( i < j ) swap ( y[i] , y[j] ) ;    }    for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) {        Complex wn = Complex ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ;        for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) {            for ( int i = k ; i < n ; i += s ) {                y[i + ds] = y[i] - ( t = w * y[i + ds] ) ;                y[i] = y[i] + t ;            }        }    }    if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;}double q[N],num[N];Complex s[N],t[N];int n;int main() {    scanf("%d",&n);    for(int i = 1; i <= n; ++i)        scanf("%lf",&q[i]);    for(int i = 0; i < n-1; ++i)        num[i] = (double)-1.0/(1.0*(n-i-1)*(n-i-1));    num[n-1] = 0;    for(int i = n; i < 2*n-1; ++i)        num[i] = (double)1.0/(1.0*(i-n+1)*(i-n+1));    int n1 = 1;    for(n1=1;n1<2*n-1;n1<<=1);    for(int i = 0; i < 2*n-1; ++i) s[i] = Complex(num[i],0);    for(int i = 2*n-1; i < n1; ++i) s[i] = Complex(0,0);    for(int i = 0; i < n; ++i)t[i] = Complex(q[i+1],0);    for(int i = n; i < n1; ++i) t[i] = Complex(0,0);    FFT(s,n1,1);FFT(t,n1,1);    for(int i = 0; i < n1; ++i) t[i] = t[i]*s[i];    FFT(t,n1,-1);    int cnt = 1;    for(int i = n-1; i < 2*n-1; ++i) {        printf("%.3f\n",t[i].r);    }    return 0;}