拉格朗日插值法板子

int solve(int n, int *x, int *y, int xi) {
  int ans = 0;
  s1[0] = (xi - x[0]) % mod, s2[n + 1] = 1;
  for (int i = 1; i <= n; i++) 
    s1[i] = 1ll * s1[i - 1] * (xi - x[i]) % mod;
  for (int i = n; i >= 0; i--) 
    s2[i] = 1ll * s2[i + 1] * (xi - x[i]) % mod;
  ifac[0] = ifac[1] = 1;
  for (int i = 2; i <= n; i++) 
    ifac[i] = -1ll * mod / i * ifac[mod % i] % mod;
  for (int i = 2; i <= n; i++) 
    ifac[i] = 1ll * ifac[i] * ifac[i - 1] % mod;
  for (int i = 0; i <= n; i++) {
    (ans += 1ll * y[i] * (i == 0 ? 1 : s1[i - 1]) % mod * s2[i + 1] % mod
            * ifac[i] % mod * (((n - i) & 1) ? -1 : 1) * ifac[n - i] % mod) %= mod;
  }
  return (ans + mod) % mod;
}