Luo Gu 4781: Lagrange polynomial

Luo Gu 4781: Lagrange

• We know that \ (n + 1 \) points can determine a \ (n \) order polynomial.

• How this polynomial seek it? This is the Lagrange polynomial, Lagrange interpolation polynomial can find.

• For example a first direct, for example, a quadratic polynomial, of course, need three points.

• For example, \ (F (. 3) = 2, F (. 4) =. 3, F (. 6) =. 1 \) .

• First, write three polynomial:

  • \(l_0(x)=\frac{(x-4)(x-6)}{(3-4)(3-6)}\).
  • \(l_1(x)=\frac{(x-3)(x-6)}{(4-3)(4-6)}\).
  • \(l_2(x)=\frac{(x-3)(x-4)}{(6-3)(6-4)}\).

• \(P(x)=f(3)l_0(x)+f(4)l_1(x)+f(6)l_2(x)\).

• \(=2\frac{(x-4)(x-6)}{(3-4)(3-6)}+3\frac{(x-3)(x-6)}{(4-3)(4-6)}+1\frac{(x-3)(x-4)}{(6-3)(6-4)}\).

• \(=-\frac{2}{3}x^2+\frac{34}{6}x-9\).

• You can be substituted into the verify:

  • \(f(3)=-6+17-9=2\).
  • \(f(4)=3\).
  • \(f(6)=-24+34-9=1\).

• This is the Lagrange polynomial.

• Formula summarized as follows:

  • \(F(x)=\sum_{i=1}^ny_i*\prod_{j\neq i}\frac{x-x_j}{x_i-x_j}\)。

• This allows the (O (n ^ 2) \ ) \ determined interpolation polynomial time.

• Of course, this problem requires that \ (f (k) \) values, so the \ (X \) band into \ (K \) on the line.

  • \(ans=\sum_{i=1}^ny_i*\prod_{j\neq i}\frac{k-x_j}{x_i-x_j}\).

• #include<bits/stdc++.h>
  using namespace std;
  typedef long long ll;
  const int maxn = 2000 + 10;
  const int mod = 998244353;
  ll n, k, x[maxn], y[maxn];
  ll qmi(ll a, ll b)
  {
      ll res = 1;
      while(b)
      {
          if(b&1) res = (res*a)%mod;
          a = (a*a)%mod;
          b >>= 1;
      } return res%mod;
  }
  
  ll ans, s1, s2;
  void Largrange()
  {
      for(int i = 1; i <= n; i++)
      {
          s1 = y[i] % mod;
          s2 = 1ll;
          for(int j = 1; j <= n; j++)
          {
              if(i != j)
              {
                  s1 = s1*(k-x[j])%mod;
                  s2 = s2*((x[i] - x[j]%mod)%mod)%mod;
              }
          }
          ans += s1 * qmi(s2, mod-2)%mod;
          ans = (ans+mod)%mod;
      }
      printf("%lld\n", ans);
  }
  
  int main()
  {
      scanf("%lld%lld", &n, &k);
      for(int i = 1; i <= n; i++)
          scanf("%lld%lld", &x[i], &y[i]);
      Largrange();
      return 0;
  }