[模板] 多项式求逆

提交至 【模板】多项式求逆

(F(x)G(x)-1 equiv 0 pmod {x^{lceil frac n 2 rceil}})

(F(x)^2G(x)^2-2F(x)G(x)+1equiv 0 pmod {x^n})

(F(x)[2G(x)-F(x)G(x)^2] equiv 1 pmod {x^n})

(F(x)^{-1} equiv G(x)[2-F(x)G(x)] pmod {x^n})

代码丑。

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#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int mod = 998244353;
const int g = 3;
const int maxn = 100010;

int (int x, int y) {
int ret = 1;
while (y) {
if (y & 1) {
ret = 1LL*ret*x%mod;
}
x = 1LL*x*x%mod;
y >>= 1;
}
return ret;
}

int w_a[maxn*4], w_b[maxn*4], w_c[maxn*4], rev[maxn*4];

int n = 0;

struct poly {
int *a, len;
poly (int l = 0) {
a = new int[l];
len = l;
for (int i = 0; i < l; i++) {
a[i] = 0;
}
}
};

void ntt(int *a, int t, int ty) {
int len = (1 << t);
for (int i = 1; i < len; i++) rev[i] = (rev[i>>1]>>1)|((i&1)<<(t-1));
for (int i = 0; i < len; i++) if (rev[i] > i) swap(a[rev[i]], a[i]);
for (int l = 2; l <= len; l <<= 1) {
int wn = qpow(g, (mod-1)/l);
for (int s = 0; s < len; s += l) {
int w = 1;
for (int i = s; i < (s + (l >> 1)); i++) {
int v1 = a[i], v2 = 1LL*a[i+(l>>1)]*w%mod;
a[i] = (v1+v2) % mod;
a[i+(l>>1)] = (v1-v2+mod) % mod;
w = 1LL*w*wn%mod;
}
}
}
if (ty == -1) {
for (int i = 1; i < len/2; i++) swap(a[i], a[len-i]);
int r = qpow(len, mod-2);
for (int i = 0; i < len; i++) a[i] = 1LL*a[i]*r%mod;
}
}

poly operator*(poly A, poly B) {
poly ret(A.len + B.len - 1);
int t = 0;
while ((1<<t) < ret.len) t ++;
for (int i = 0; i < (1<<t); i++) w_a[i] = w_b[i] = 0;
for (int i = 0; i < A.len; i++) w_a[i] = A.a[i];
for (int i = 0; i < B.len; i++) w_b[i] = B.a[i];
ntt(w_a, t, 1); ntt(w_b, t, 1);
for (int i = 0; i < (1<<t); i++) w_c[i] = 1LL*w_a[i]*w_b[i]%mod;
ntt(w_c, t, -1);
for (int i = 0; i < ret.len; i++) ret.a[i] = w_c[i];
return ret;
}

poly inverse(poly A) {
if (A.len == 1) {
poly ret(1);
ret.a[0] = qpow(A.a[0], mod-2);
return ret;
}
int nlen = (A.len+1)/2;
poly nA(nlen);
for (int i = 0; i < nlen; i++) nA.a[i] = A.a[i];
poly r = inverse(nA);
poly tmp = A*r;
tmp.len = A.len;
for (int i = 0; i < tmp.len; i++) tmp.a[i] = (-tmp.a[i]+mod)%mod;
tmp.a[0] = (tmp.a[0]+2) % mod;
poly ret = r*tmp;
ret.len = A.len;
return ret;
}

int main() {
scanf("%d", &n);
poly F(n);
for (int i = 0; i < n; i++) scanf("%d", &F.a[i]);
poly G = inverse(F);
for (int i = 0; i < n; i++) printf("%d ", G.a[i]);
printf("n");
return 0;
}