「ZJOI2014」力 FFT

FFTl裸题，小于的部分直接做，大于的部分倒序后再做就行了。

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1 << 18;
const double Pi = acos(-1.0);
struct cp {
    double x, y;
    cp() { x = y = 0; }
    cp(double x, double y) : x(x), y(y) {}
    inline cp operator+(const cp &o) const { return cp(x + o.x, y + o.y); }
    inline cp operator-(const cp &o) const { return cp(x - o.x, y - o.y); }
    inline cp operator*(const cp &o) const { return cp(x * o.x - y * o.y, x * o.y + o.x * y); }
} f[MAXN], g[MAXN], b[MAXN];
int rev[MAXN];
inline void DFT(cp *arr, int len, int flg) {
    for (int i = 0; i < len; ++i)
        if (i < rev[i])
            swap(arr[i], arr[rev[i]]);
    for (int i = 2; i <= len; i <<= 1) {
        cp wn = cp(cos(2 * Pi / i), flg * sin(2 * Pi / i));
        for (int j = 0; j < len; j += i) {
            cp w = cp(1, 0);
            for (int k = j; k < j + i / 2; ++k, w = w * wn) {
                cp x = arr[k], y = arr[k + i / 2] * w;
                arr[k] = x + y;
                arr[k + i / 2] = x - y;
            }
        }
    }
    if (flg == -1)
        for (int i = 0; i < len; ++i) arr[i].x /= len;
}
int n;
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%lf", &f[i].x), g[n - i + 1].x = f[i].x, b[i].x = 1.0 / i / i;
    int len = 1;
    while (len <= (n << 1)) len <<= 1;
    for (int i = 0; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (len >> 1));
    DFT(f, len, 1), DFT(g, len, 1), DFT(b, len, 1);
    for (int i = 0; i < len; ++i) f[i] = f[i] * b[i], g[i] = g[i] * b[i];
    DFT(f, len, -1), DFT(g, len, -1);
    for (int i = 1; i <= n; ++i) printf("%.3f\n", f[i].x - g[n - i + 1].x);
}