多项式开方

描述

给定多项式 g\left(x\right) ，求 f\left(x\right) ，满足：

f^{2}\left(x\right)\equiv g\left(x\right) \pmod{x^{n}}

解法

倍增法

假设现在已经求出了 g\left(x\right) 在模 x^{\left\lceil\frac{n}{2}\right\rceil} 意义下的平方根 f_{0}\left(x\right) ，则有：

\begin{aligned} f_{0}^{2}\left(x\right)&\equiv g\left(x\right) &\pmod{x^{\left\lceil\frac{n}{2}\right\rceil}}\\ f_{0}^{2}\left(x\right)-g\left(x\right)&\equiv 0 &\pmod{x^{\left\lceil\frac{n}{2}\right\rceil}}\\ \left(f_{0}^{2}\left(x\right)-g\left(x\right)\right)^{2}&\equiv 0 &\pmod{x^{n}}\\ \left(f_{0}^{2}\left(x\right)+g\left(x\right)\right)^{2}&\equiv 4f_{0}^{2}\left(x\right)g\left(x\right) &\pmod{x^{n}}\\ \left(\frac{f_{0}^{2}\left(x\right)+g\left(x\right)}{2f_{0}\left(x\right)}\right)^{2}&\equiv g\left(x\right) &\pmod{x^{n}}\\ \frac{f_{0}^{2}\left(x\right)+g\left(x\right)}{2f_{0}\left(x\right)}&\equiv f\left(x\right) &\pmod{x^{n}}\\ 2^{-1}f_{0}\left(x\right)+2^{-1}f_{0}^{-1}\left(x\right)g\left(x\right)&\equiv f\left(x\right) &\pmod{x^{n}} \end{aligned}

倍增计算即可。

时间复杂度

T\left(n\right)=T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log{n}\right)

还有一种常数较小的写法就是在倍增维护 f\left(x\right) 的时候同时维护 f^{-1}\left(x\right) 而不是每次都求逆。

当 \left[x^{0}\right]g\left(x\right)\neq 1 时，可能需要使用二次剩余来计算 \left[x^{0}\right]f\left(x\right) 。

洛谷模板题 P5205 【模板】多项式开根 参考代码 
  1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
#include <bits/stdc++.h>
using namespace std;

const int maxn = 1 << 20, mod = 998244353;

int a[maxn], b[maxn], g[maxn], gg[maxn];

int qpow(int x, int y) {  //快速幂
  int ans = 1;

  while (y) {
    if (y & 1) {
      ans = (long long)1 * ans * x % mod;
    }
    x = (long long)1 * x * x % mod;
    y >>= 1;
  }
  return ans;
}

int inv2 = qpow(2, mod - 2);  //逆元

inline void change(int *f, int len) {
  for (int i = 1, j = len / 2; i < len - 1; i++) {
    if (i < j) {
      swap(f[i], f[j]);
    }

    int k = len / 2;
    while (j >= k) {
      j -= k;
      k /= 2;
    }
    if (j < k) {
      j += k;
    }
  }
}

inline void NTT(int *f, int len, int type) {  // NTT
  change(f, len);

  for (int q = 2; q <= len; q <<= 1) {
    int nxt = qpow(3, (mod - 1) / q);
    for (int i = 0; i < len; i += q) {
      int w = 1;

      for (int k = i; k < i + (q >> 1); k++) {
        int x = f[k];
        int y = (long long)1 * w * f[k + (q / 2)] % mod;

        f[k] = (x + y) % mod;
        f[k + (q / 2)] = (x - y + mod) % mod;
        w = (long long)1 * w * nxt % mod;
      }
    }
  }

  if (type == -1) {
    reverse(f + 1, f + len);
    int iv = qpow(len, mod - 2);

    for (int i = 0; i < len; i++) {
      f[i] = (long long)1 * f[i] * iv % mod;
    }
  }
}

inline void inv(int deg, int *f, int *h) {  //求逆元
  if (deg == 1) {
    h[0] = qpow(f[0], mod - 2);
    return;
  }

  inv(deg + 1 >> 1, f, h);

  int len = 1;
  while (len < deg * 2) {  //倍增
    len *= 2;
  }

  copy(f, f + deg, gg);
  fill(gg + deg, gg + len, 0);

  NTT(gg, len, 1);
  NTT(h, len, 1);
  for (int i = 0; i < len; i++) {
    h[i] = (long long)1 * (2 - (long long)1 * gg[i] * h[i] % mod + mod) % mod *
           h[i] % mod;
  }

  NTT(h, len, -1);
  fill(h + deg, h + len, 0);
}

int n, t[maxn];

// deg:次数
// f:被开根数组
// h:答案数组
inline void sqrt(int deg, int *f, int *h) {
  if (deg == 1) {
    h[0] = 1;
    return;
  }

  sqrt(deg + 1 >> 1, f, h);

  int len = 1;
  while (len < deg * 2) {  //倍增
    len *= 2;
  }
  fill(g, g + len, 0);
  inv(deg, h, g);
  copy(f, f + deg, t);
  fill(t + deg, t + len, 0);
  NTT(t, len, 1);
  NTT(g, len, 1);
  NTT(h, len, 1);

  for (int i = 0; i < len; i++) {
    h[i] = (long long)1 * inv2 *
           ((long long)1 * h[i] % mod + (long long)1 * g[i] * t[i] % mod) % mod;
  }
  NTT(h, len, -1);
  fill(h + deg, h + len, 0);
}

int main() {
  cin >> n;

  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sqrt(n, a, b);

  for (int i = 0; i < n; i++) {
    printf("%d ", b[i]);
  }

  return 0;
}

Newton's Method

参见 Newton's Method.

例题

  1. 「Codeforces Round #250」E. The Child and Binary Tree
build本页面最近更新：，更新历史
edit发现错误？想一起完善？ 在 GitHub 上编辑此页！
people本页面贡献者：OI-wiki
copyright本页面的全部内容在 CC BY-SA 4.0 和 SATA 协议之条款下提供，附加条款亦可能应用

评论