分治NTT（洛谷P4721）

2024-01-26 08:28

文章标签 分治洛谷 ntt p4721

本文主要是介绍分治NTT（洛谷P4721），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目路径：P4721 【模板】分治 FFT - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

思路：

如果用NTT一个一个求时间复杂度是O（n^2）。

可以发现，i<j,f[i]会对f[j]有贡献，而f[j]对f[i]没有贡献，考虑分治。

对于区间（l,r）,当我们算出f[l]-f[mid](mid=(l+r)>>1)，我们可以算出(l,mid)对后边区间贡献，再算（mid+1，r）内的f值。

代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<unordered_map>
using namespace std;
#define LL long long
const int N = 4e5 + 10;
const LL mod = 998244353;
const double PI = acos(-1);
LL a[N], b[N], f[N], g[N];
int n,tot,bit;
int rel[N];
LL gi,gg=3;
LL quick(LL a, LL b, LL mod)
{
   LL ans = 1;
   while (b)
   {
       if (b & 1) ans = ans * a % mod;
       b = b >> 1;
       a = a * a % mod;
   }
   return ans;
}
void ntt(LL a[], int op)
{
   for (int i = 0; i < tot; i++)
       if (i < rel[i]) swap(a[i], a[rel[i]]);
   for (int m = 2; m<= tot; m=m<<1)
   {
       LL g1 = quick((op == 1) ? gg : gi, (mod - 1) / m,mod);
       for (int i = 0; i < tot; i += m)
       {
           LL gk = 1;
           for (int j = 0; j < m / 2; j++)
           {
               LL x = a[i + j], y = a[i + j + m / 2]*gk%mod;
               a[i + j] = (x + y) % mod;
               a[i + j + m / 2] = (x - y + mod) % mod;
               gk = gk * g1 % mod;
           }
       }
   }
}
void cbd(int l,int r)
{
   if (l == r)    return;
   int mid = (l + r) >> 1;
   cbd(l, mid);
   tot = 1;
   bit = 0;
   while (tot <= (mid - l) + (r - l)) {
       tot <<= 1;
       ++bit;
   }
   for (int i = 0; i < tot; i++) a[i] = b[i] = 0;
   for (int i = l; i <= mid; i++) a[i-l] = f[i];
   for (int i = 0; i <= r - l; i++) b[i] = g[i];
   for (int i = 0; i < (1 << bit); ++i)
   {
       rel[i] = (rel[i >> 1] >> 1) | ((i & 1) << (bit - 1));
   }
   ntt(a, 1);
   ntt(b, 1);
   for (int i = 0; i < tot; i++) a[i] = a[i] * b[i];
   ntt(a, -1);
   LL gk = quick(tot, mod - 2, mod);
   for (int i = 0; i < tot; i++)
       a[i] = (a[i] * gk % mod + mod) % mod;
   for (int i = mid + 1; i <= r; i++) f[i] = (f[i] + a[i - l]) % mod;
   cbd(mid + 1, r);
}
int main() {
   cin >> n;
   for (int i = 1; i < n; i++) cin >> g[i];
   gi = quick(gg, mod - 2, mod);
   f[0] = 1;
   cbd(0,n-1);
   for (int i = 0; i < n; i++) cout << f[i] << " ";
   cout << endl;
   return 0;
}