1.冪級數的復合

對於冪級數\(F(x)\)和\(G(x)\)，我們稱\(F(G(x))\)為冪級數F和G的復合

2.復合逆：

如果\(F(x)\)和\(G(x)\)滿足\(F(G(x))=G(F(x))=x\)則稱它們互為復合逆

3.拉格朗日反演：

如果\(F(x)\)和\(G(x)\)互為復合逆，則有\([x^n]G(x)=\frac1n[x^{n-1}](\frac{1}{F(x)/x})^n\)

可以通過這個在\(O(n\log n)\)(多項式exp和ln求快速冪，巨大常數)或\(O(n\log^2n)\)(倍增快速冪，小常數)來求復合逆的某一項（如果求整個復合逆的最優復雜度為\(O(n^2)\)

一下為多項式倍增快速冪取模和拉格朗日反演的板子，可以配合ghj1222的多項式板子食用

void poly_qpow(int *a, int len, int n)
{
    int *tmp = new int[len * 2];
    for (int i = 0; i < len * 2; i++) tmp[i] = i >= len ? 0 : a[i], a[i] = (i == 0);
    while (n > 0)
    {
        ntt(tmp, len * 2, 1);
        if (n & 1)
        {
            ntt(a, len * 2, 1);
            for (int i = 0; i < len * 2; i++) a[i] = a[i] * (long long)tmp[i] % p;
            ntt(a, len * 2, -1);
            for (int i = len; i < len * 2; i++) a[i] = 0;
        }
        for (int i = 0; i < len * 2; i++) tmp[i] = tmp[i] * (long long)tmp[i] % p;
        ntt(tmp, len * 2, -1);
        for (int i = len; i < len * 2; i++) tmp[i] = 0;
        n >>= 1;
    }
    delete []tmp;
}

int lagrange_inversion(int *aa, int len, int n)
{
    int *a = new int[len * 2];
    for (int i = 0; i < len; i++) a[i] = aa[i];
    for (int i = len; i < len * 2; i++) a[i] = 0;
    for (int i = 1; i < len; i++) a[i - 1] = a[i];
    poly_inv(a, len);
    poly_qpow(a, len, n);
    int ans = a[n - 1] * (long long)qpow(n, p - 2) % p;
    delete []a;
    return ans;
}
 

拉格朗日反演