[ZJOI2014]力[卷积,FFT]

题目

给出个数，给出的定义如下：

令，求.

数据范围:

分析

先考虑的情况,即有四个数

考虑序列,.考虑卷积m即为上述序列相乘的形式,以序列分别为生成函数的前k项系数(其余项系数为0),答案即为的系数,以FFT优化即可.注意精度.

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

const int MAXN = 1e6 + 5;
const double PI = acos(-1.0);
int N;
double p[MAXN];
complex F[MAXN], G[MAXN], C[MAXN];
void FFT(int len, complex *a, int type);

int main(){
  ios::sync_with_stdio(false);
  cin >> N;
  int i;
  for(i = 1; i <= N; i++) cin >> p[i];
  for(i = 0; i <= N - 1; i++) F[i] = complex(p[i + 1], 0);
  for(i = 0; i <= N - 2; i++) G[i] = complex(-1.0 / ((N - 1.0 - i) * (N - 1.0 - i)), 0);
  for(i = N; i <= 2 * N - 2; i++) G[i] = complex(1.0 / ((N - 1.0 - i) * (N - 1.0 - i)), 0);
  G[N - 1] = complex(0, 0);
  int limit = 1;
  while(limit <= N + 2 * N - 1) limit <<= 1;
  FFT(limit, F, 1), FFT(limit, G, 1);
  for(i = 0; i <= limit; i++)
    C[i] = F[i] * G[i];
  FFT(limit, C, -1);
  for(i = N - 1; i <= 2 * N - 2; i++)
    printf("%.6lf\n", C[i].real() / limit);
  return 0;
}

void bit_reverse(int limit, complex *a);
void FFT(int len, complex *a, int type){
  int i, j, l;
  bit_reverse(len, a);
  for(l = 2; l <= len; l <<= 1){
    int mid = l >> 1;
    complex wn = complex(cos(2.0 * PI / l), type * sin(2.0 * PI / l));
    for(i = 0; i < len; i += l){
      complex w = complex(1, 0);
      for(j = 0; j < mid; j++, w = w * wn){
        complex x = a[i + j], y = w * a[i + j + mid];
        a[i + j] = x + y;
        a[i + j + mid] = x - y;
      }
    }
  }
}

void bit_reverse(int limit, complex *a){
  int i, j, l;
  for(i = 0, j = 0; i < limit; i++){
    if(i > j) swap(a[i], a[j]);
    for(l = limit >> 1; (j & l); j ^= l, l >>= 1);
    j ^= l;
  }
}