n行，第i行输出Ei。与标准答案误差不超过1e-2即可。

 1 #include <cmath> 2 #include <cstdio> 3 #include <iostream> 4  5 const int siz = 600005; 6 const double pi = acos(-1); 7  8 inline double sqr(double x) { 9     return x * x;10 }11 12 int n, rev[siz];13 double num[siz];14 15 struct complex16 {17     double r, i;18 19     inline complex(double a = 0, double b = 0)20         : r(a), i(b) {};21     inline complex operator + (const complex &a) {22         return complex(r + a.r, i + a.i);23     }24     inline complex operator - (const complex &a) {25         return complex(r - a.r, i - a.i);26     }27     inline complex operator * (const complex &a) {28         return complex(r*a.r - i*a.i, r*a.i + i*a.r);29     }30 }a[siz], b[siz];31 32 inline void FFT(complex *s, int k)33 {34     for (int i = 0; i < n; ++i)35         if (i < rev[i])36             std::swap(s[i], s[rev[i]]);37     38     for (int h = 2; h <= n; h <<= 1)39     {40         complex wn(cos(2*pi*k/h), sin(2*pi*k/h));41         for (int i = 0; i < n; i += h)42         {43             complex w(1, 0);44             for (int j = 0; j < (h >> 1); ++j, w = w*wn)45             {46                 complex t = s[i + j + (h >> 1)]*w;47                 s[i + j + (h >> 1)] = s[i + j] - t;48                 s[i + j] = s[i + j] + t;49             }50         }51     }52     53     if (k == -1)54         for (int i = 0; i < n; ++i)55             s[i].r = s[i].r / n;56 }57 58 signed main(void)59 {60     scanf("%d", &n); --n;61     62     for (int i = 0; i <= n; ++i)63         scanf("%lf", &a[i].r);64         65     for (int i = 0; i < n; ++i)66         b[i].r = -1.0 / sqr(n - i);67         68     for (int i = n + 1; i <= 2*n; ++i)69         b[i].r = -1.0 * b[2*n - i].r;70         71     int m = n, k = n << 2, len = 0;72     73     for (n = 1; n <= k; n <<= 1)++len;74     75     for (int i = 0; i < n; ++i)76         rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));77         78     FFT(a, 1);79     FFT(b, 1);80     81     for (int i = 0; i <= n; ++i)82         a[i] = a[i] * b[i];83     84     FFT(a, -1);85     86     for (int i = m; i <= 2*m; ++i)87         printf("%lf\n", a[i].r);88 }