模板 - 数学 - 数值分析 - 拉格朗日插值

int lagrange(int n, int *x, int *y, int xi) {
    int ans = 0;
    for (int i = 0; i <= n; i++) {
        int s1 = 1, s2 = 1;
        for (int j = 0; j <= n; j++)
            if (i != j) {
                s1 = 1ll * s1 * (xi - x[j]) % mod;
                s2 = 1ll * s2 * (x[i] - x[j]) % mod;
            }
        ans = (1ll * ans + 1ll * y[i] * s1 % mod * quick_pow(s2, mod - 2) % mod) % mod;
    }
    return (ans + mod) % mod;
}

int lagrange(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;
}