[CTSC2010]性能优化 [混合基FFT]

Posted 2022-07-08 qq62c30ac77b2a7

​​传送门​​

先来复习一下普通的FFT

对于混合基FFT, 有

然后就可以分治了，rev 数组的预处理有一些奇葩

#include<bits/stdc++.h>
#define N 2000050
using namespace std;
typedef long long ll;
int read()
  int x = 0, f = 1; char ch = 0;
  while(!isdigit(ch)) ch = getchar(); if(ch == -) f = -1;
  while(isdigit(ch)) x = x*10 + (ch-0), ch = getchar();
  return x * f;

int Mod, G;
ll add(ll a, ll b) return (a + b) % Mod;
ll mul(ll a, ll b) return (a * b) % Mod;
ll power(ll a, ll b) ll ans = 1;
  for(;b;b>>=1) if(b&1) ans = mul(ans, a); a = mul(a, a);
  return ans;

int n, c;
ll a[N], b[N], wn[N];
int rev[N], pcnt, prim[N];
ll tp[N];
void factor(int x)
  for(int i = 2; i*i <= x; i++)
    while(x % i == 0) prim[++pcnt] = i, x /= i;
   if(x > 1) prim[++pcnt] = x; 

int getG()
  factor(n);
  for(int i = 2;; i++)
    bool flag = true;
    for(int j = 1; j <= pcnt; j++) if(power(i, n/prim[j]) == 1)
      flag = false; break;
     if(flag) return i;
  

void Reverse(int *A)
  for(int i = pcnt, block = n; i >= 1; i--)
    for(int idx = 0, j = 0; j < n; j += block)
      for(int k = 0; k < prim[i]; k++)
        for(int l = 0; l < block; l += prim[i])
          tp[idx++] = A[j+k+l];
    for(int j = 0; j < n; j++) A[j] = tp[j];
    block /= prim[i];
  

void DFT(ll *a, int typ)
  for(int i = 0; i < n; i++) tp[i] = a[rev[i]];
  for(int i = 0; i < n; i++) a[i] = tp[i];
  for(int i = 1, block = 1; i <= pcnt; i++)
    int mid = block; block *= prim[i]; 
    ll wi = wn[n/block];
    for(int j = 0; j < n; j++) tp[j] = 0;
    for(int j = 0; j < n; j += block)
      ll wk = 1;
      for(int k = 0; k < block; k++)
        for(int l = k%mid, w=1; l < block; l+=mid, w = mul(w,wk))
          tp[j+k] = add(tp[j+k], mul(w, a[j+l]));
         wk = mul(wk, wi);
      
    
    for(int j = 0; j < n; j++) a[j] = tp[j];
  
  if(typ == -1)
    reverse(a+1, a+n);
    for(int i = 0, inv = power(n, Mod-2); i < n; i++) a[i] = mul(a[i], inv);
  

int main()
  n = read(), c = read(); Mod = n+1; G = getG();
  wn[0] = 1; wn[1] = G;
  for(int i = 2; i < n; i++) wn[i] = mul(wn[i-1], G);
  for(int i = 0; i < n; i++) a[i] = read();
  for(int i = 0; i < n; i++) b[i] = read();
  for(int i = 0; i < n; i++) rev[i] = i; Reverse(rev);
  DFT(a, 1); DFT(b, 1); 
  for(int i = 0; i < n; i++) a[i] = mul(a[i], power(b[i], c));
  DFT(a, -1);
  for(int i = 0; i < n; i++) cout << a[i] << \\n;
  return 0;