多项式求逆学习笔记

用法及推导

这个主要用于在满足多项式(~A(x) * B(x) = C(x)~)且已知(~A(x)， C(x)~)时来求多项式(~B(x)~)。可知(~B(x) = C(x) * A ^{-1}(x)~)，其中(~A ^ {-1}(x))是(~A(x)~)在模(~x ^ n~)意义下的逆元。
考虑如何来求这个(~A^ {-1}(x)~).显然当(~n = 1~)时，(~A(x)~)中只有一个元素，根据费马小定理，(~A ^ {-1}(x)~)就是(~A(0) ^ {mod - 2}~)，现在知道了这一点，我们再来推导，设(~B(x) = A ^ {-1}(x)~​)，则有

[A(x) * B(x) equiv 1 ~(mod ~x ^ n) ]

而如果我们知道了(~A(x)~)在模(~x ^ {frac{n}{2}}~)意义下的逆元为(~B'(x)~)，那么(B(x)~)在此意义下也成立，有

[A(x) * B'(x) equiv 1 ~(mod~{x ^ {frac{n}{2}}}) ]

[A(x) * B(x) equiv 1 ~(mod~{x ^ {frac{n}{2}}}) ]

两式相减得

[A(x) * (B(x) - B'(x)) equiv 0 ~(mod~{x ^ {frac{n}{2}}}) Leftrightarrow B(x) - B'(x) equiv 0 ~(mod~ x ^ {frac{n}{2}}) ]

两边同乘平方，得

[B^2(x) - (2B*B')(x) + B' ^ 2(x) equiv 0~(mod~ x ^ {frac{n}{2}}) ]

再同乘个(~A(x)~)，可以消掉所有的(~B(x)~)，得

[B(x) - 2B'(x) + (A * B'^2)(x)equiv 0 ~(mod~ x ^ {frac{n}{2}}) ]

移项可得

[B(x) equiv 2B'(x) - (A * B'^2)(x)~(mod~ x ^ {frac{n}{2}}) ]

到这里式子就推完了。其实很好推也很好记。观察这个式子，可以发现(~B(x)~)是由(~A(x)~)和(~B'(x)~)计算而来的。那么考虑递归，在回溯的时候往前代计算答案就行了，代码这样写。

inline void Inv(int *a, int *b, int len) { // b is the inv of a

    if (len == 1) { b[0] = qpow(a[0], mod - 2); return; }

    Inv(a, b, len >> 1);
    
    bit = 0; for (siz = 1; siz <= len; siz <<= 1) ++ bit;
    For(i, 0, siz - 1) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));

    For(i, 0, len - 1) p[i] = a[i], q[i] = b[i];
    
    NTT(p, 1), NTT(q, 1);
    For(i, 0, siz - 1) p[i] = 1ll * q[i] * q[i] % mod * p[i] % mod;
    NTT(p, -1);
    
    For(i, 0, len - 1) b[i] = add(2 * b[i] % mod, mod - p[i]);
}

完整代码洛谷模板题

#include<bits/stdc++.h>
#define For(i, j, k) for (int i = j; i <= k; ++i)
#define Forr(i, j, k) for (int i = j; i >= k; --i)
using namespace std;

inline int read() {
    int x = 0, p = 1; char c = getchar();
    while(!isdigit(c)) { if(c == '-') p = -1; c = getchar(); }
    while(isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x *= p;
}

inline void File() {
    freopen("luogu4238.in", "r", stdin);
    freopen("luogu4238.out", "w", stdout);
}

const int N = 4e5 + 10, mod = 998244353;
int powg[N], invg[N], a[N], b[N], rev[N], bit, siz, n;
int p[N], q[N];

inline int qpow(int a, int b) {
    static int res;
    for (res = 1; b; b >>= 1, a = 1ll * a * a % mod)
        if (b & 1) res = 1ll * res * a % mod;
    return res;
}

inline int add(int x, int y) { return (x += y) >= mod ? x -= mod : x; }

inline void NTT(int *a, int flag) {
    For(i, 0, siz - 1) if (rev[i] > i) swap(a[rev[i]], a[i]);	

    for (int i = 2; i <= siz; i <<= 1) {
        int wn = flag > 0 ? powg[i] : invg[i];

        for (int j = 0; j < siz; j += i) {
            int w = 1;
            for (int k = 0; k < i >> 1; w = 1ll * w * wn % mod, ++ k) {
                int x = a[j + k], y = 1ll * w * a[j + k + (i >> 1)] % mod;
                a[k + j] = add(x, y), a[k + j + (i >> 1)] = add(x, mod - y);
            }
        }
    }

    if (flag == -1) {
        int g = qpow(siz, mod - 2);
        For(i, 0, siz - 1) a[i] = 1ll * a[i] * g % mod;
    }
}

inline void Inv(int *a, int *b, int len) {

    if (len == 1) { b[0] = qpow(a[0], mod - 2); return; }

    Inv(a, b, len >> 1);
    
    bit = 0; for (siz = 1; siz <= len; siz <<= 1) ++ bit;
    For(i, 0, siz - 1) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));

    For(i, 0, len - 1) p[i] = a[i], q[i] = b[i];
    
    NTT(p, 1), NTT(q, 1);
    For(i, 0, siz - 1) p[i] = 1ll * q[i] * q[i] % mod * p[i] % mod;
    NTT(p, -1);
    
    For(i, 0, len - 1) b[i] = add(2 * b[i] % mod, mod - p[i]);
}

int main() {
    n = read() - 1;
    For(i, 0, n) a[i] = read();
    
    for (siz = 1; siz <= n << 1; siz <<= 1) ++ bit;

    int g = qpow(3, mod - 2);
    for (int i = 1; i <= siz; i <<= 1) {
        powg[i] = qpow(3, (mod - 1) / i);
        invg[i] = qpow(g, (mod - 1) / i);
    }

    int len = siz >> 1;
    Inv(a, b, len);	

    For(i, 0, n) printf("%d ", b[i]);

    return 0;
}