（CCPC 2020 网络选拔赛）HDU 6900 Residual Polynomial（分治 + NTT）

Residual Polynomial

写出所有的fi(x)f_i(x)fi(x)出来，fi,jf_{i, j}fi,j表示fi(x)f_i(x)fi(x)的第jjj项系数

{f1,0f1,1f1,2…f1,n−1f1,nf2,0f2,1f2,2…f2,n−1f2,nf3,0f3,1f3,2…f3,n−1f3,n⋮⋮⋮⋱⋮⋮fn−1,0fn−1,1fn−1,2…fn−1,n−1fn−1,nfn,0fn,1fn,2…fn,n−1fn,n}\left\{ \begin{matrix} f_{1, 0} & f_{1, 1} & f_{1, 2} & \dots & f_{1, n - 1} & f_{1, n}\\ f_{2, 0} & f_{2, 1} & f_{2, 2} & \dots & f_{2, n - 1} & f_{2, n}\\ f_{3, 0} & f_{3, 1} & f_{3, 2} & \dots & f_{3, n - 1} & f_{3, n}\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ f_{n - 1, 0} & f_{n - 1, 1} & f_{n - 1, 2} & \dots & f_{n - 1, n - 1} & f_{n - 1, n}\\ f_{n, 0} & f_{n, 1} & f_{n, 2} & \dots & f_{n, n - 1} & f_{n, n}\\ \end{matrix} \right\} ⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧f1,0f2,0f3,0⋮fn−1,0fn,0f1,1f2,1f3,1⋮fn−1,1fn,1f1,2f2,2f3,2⋮fn−1,2fn,2………⋱……f1,n−1f2,n−1f3,n−1⋮fn−1,n−1fn,n−1f1,nf2,nf3,n⋮fn−1,nfn,n⎭⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎫
第一行是a0,a1,a2,…,an−1,ana_0, a_1, a_2, \dots, a_{n - 1}, a_{n}a0,a1,a2,…,an−1,an。

考虑下一行如何从上一行转移，分两种：

一、fi,j×ci+1=fi+1,jf_{i, j} \times c_{i + 1} = f_{i + 1, j}fi,j×ci+1=fi+1,j

二、fi,j×j×bi+1=fi+1,j−1f_{i, j} \times j \times b_{i + 1} = f_{i + 1, j - 1}fi,j×j×bi+1=fi+1,j−1

可以想想成每个点有一条向下的连边value=cvalue = cvalue=c，同时有一条向左下的连边value=b×jvalue = b \times jvalue=b×j。

我们从f1,if_{1, i}f1,i出发，走向fn,jf_{n, j}fn,j，显然选择了i−ji- ji−j个b×jb \times jb×j，所以还有n−1−(i−j)n - 1 - (i - j)n−1−(i−j)个ccc。

先不考虑jjj，只算b,cb, cb,c的贡献，设选bbb的生成函数为F(x)F(x)F(x)，对于某个点有F(x)=c+bxF(x) = c + bxF(x)=c+bx，因为上面的转移是具有区间性的。

所以F(l,r,x)=F(l,mid,x)∗F(mid,r,x)F(l, r, x) = F(l, mid, x) * F(mid, r, x)F(l,r,x)=F(l,mid,x)∗F(mid,r,x)，也就是在左区间选几个，右区间选几个，然后组合一下，这个组合刚好满足卷积的形式。

这一步就可以递归O(nlog⁡nlog⁡n)O(n \log n \log n)O(nlognlogn)求解了。

接下来考虑jjj的影响：

fn,k=∑i−k=jF(j)f1,ii!k!f_{n, k} = \sum\limits_{i - k = j} F(j)f_{1, i} \frac{i!}{k!}fn,k=i−k=j∑F(j)f1,ik!i!

这里可以简单理解一下，每次只能向左向下移动，k是我们最后在的列，只能从初始列i≥ki \geq ki≥k的列转移过来。

(k!×fn,k)=∑i−k=jF(j)(f1,i×i!)\left(k! \times f_{n, k}\right) = \sum\limits_{i - k = j} F(j) \left(f_{1, i} \times i!\right)(k!×fn,k)=i−k=j∑F(j)(f1,i×i!)

设H(k)=k!×fn,k,G(i)=f1,i×i!设H(k) = k! \times f_{n, k}, G(i) = f_{1, i} \times i !设H(k)=k!×fn,k,G(i)=f1,i×i!，考虑把G翻转，G(n−i)=G(i)考虑把G翻转，G(n - i) = G(i)考虑把G翻转，G(n−i)=G(i)，

H(k)=∑i+j=n−kF(j)G(i)H(k) = \sum_{i + j = n - k} F(j) G(i)H(k)=∑i+j=n−kF(j)G(i)，于是再做一次卷积即可求得答案。

#include <bits/stdc++.h>using namespace std;const int mod = 998244353, inv2 = mod + 1 >> 1;namespace Quadratic_residue {struct Complex {int r, i;Complex(int _r = 0, int _i = 0) : r(_r), i(_i) {}};int I2;Complex operator * (const Complex &a, Complex &b) {return Complex((1ll * a.r * b.r % mod  + 1ll * a.i * b.i % mod * I2 % mod) % mod, (1ll * a.r * b.i % mod + 1ll * a.i * b.r % mod) % mod);}Complex quick_pow(Complex a, int n) {Complex ans = Complex(1, 0);while (n) {if (n & 1) {ans = ans * a;}a = a * a;n >>= 1;}return ans;}int get_residue(int n) {mt19937 e(233);if (n == 0) {return 0;}if(quick_pow(n, (mod - 1) >> 1).r == mod - 1) {return -1;}uniform_int_distribution<int> r(0, mod - 1);int a = r(e);while(quick_pow((1ll * a * a % mod - n + mod) % mod, (mod - 1) >> 1).r == 1) {a = r(e);}I2 = (1ll * a * a % mod - n + mod) % mod;int x = quick_pow(Complex(a, 1), (mod + 1) >> 1).r, y = mod - x;if(x > y) swap(x, y);return x;}
}const int N = 1e6 + 10;int r[N], inv[N], b[N], c[N], d[N], e[N], t[N];int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * a * ans % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans;
}void get_r(int lim) {for (int i = 0; i < lim; i++) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);}
}void get_inv(int n) {inv[1] = 1;for (int i = 2; i <= n; i++) {inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;}
}void NTT(int *f, int lim, int rev) {for (int i = 0; i < lim; i++) {if (i < r[i]) {swap(f[i], f[r[i]]);}}for (int mid = 1; mid < lim; mid <<= 1) {int wn = quick_pow(3, (mod - 1) / (mid << 1));for (int len = mid << 1, cur = 0; cur < lim; cur += len) {int w = 1;for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;}}}if (rev == -1) {int inv = quick_pow(lim, mod - 2);reverse(f + 1, f + lim);for (int i = 0; i < lim; i++) {f[i] = 1ll * f[i] * inv % mod;}}
}void polyinv(int *f, int *g, int n) {if (n == 1) {g[0] = quick_pow(f[0], mod - 2);return ;}polyinv(f, g, n + 1 >> 1);for (int i = 0; i < n; i++) {t[i] = f[i];}int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(t, lim, 1);NTT(g, lim, 1);for (int i = 0; i < lim; i++) {int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;g[i] = 1ll * g[i] * cur % mod;t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}
}void polysqrt(int *f, int *g, int n) {if (n == 1) {g[0] = Quadratic_residue::get_residue(f[0]);return ;}polysqrt(f, g, n + 1 >> 1);polyinv(g, b, n);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);for (int i = 0; i < n; i++) {t[i] = f[i];}NTT(g, lim, 1);NTT(b, lim, 1);NTT(t, lim, 1);for (int i = 0; i < lim; i++) {g[i] = (1ll * inv2 * g[i] % mod + 1ll * inv2 * b[i] % mod * t[i] % mod) % mod;b[i] = t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}
}void derivative(int *a, int *b, int n) {for (int i = 0; i < n; i++) {b[i] = 1ll * a[i + 1] * (i + 1) % mod;}
}void integrate(int *a, int n) {for (int i = n - 1; i >= 1; i--) {a[i] = 1ll * a[i - 1] * inv[i] % mod;}a[0] = 0;
}void polyln(int *f, int *g, int n) {polyinv(f, b, n);derivative(f, g, n);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(g, lim, 1);NTT(b, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * b[i] % mod;b[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}integrate(g, n);
}void polyexp(int *f, int *g, int n) {if (n == 1) {g[0] = 1;return ;}polyexp(f, g, n + 1 >> 1);int lim = 1;while (lim < 2 * n) {lim <<= 1;}polyln(g, d, n);for (int i = 0; i < n; i++) {t[i] = (f[i] - d[i] + mod) % mod;}t[0] = (t[0] + 1) % mod;get_r(lim);NTT(g, lim, 1);NTT(t, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * t[i] % mod;t[i] = d[i] =  0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}
}/*b存放多项式逆，c存放多项式开根，d存放多项式对数ln，e存放多项式指数exp，t作为中间转移数组,如果要用到polyln，得提前调用get_inv(n)先预先得到我们想要得到的逆元范围。
*/int B[N], C[N], f[N], fac[N], inv1[N], n;vector<int> F[N];int A1[N], B1[N];void init() {fac[0] = 1;for (int i = 1; i < N; i++) {fac[i] = 1ll * fac[i - 1] * i % mod;}inv1[N - 1] = quick_pow(fac[N - 1], mod - 2);for (int i = N - 2; i >= 0; i--) {inv1[i] = 1ll * inv1[i + 1] * (i + 1) % mod;}
}void merge(int rt) {int ls = rt << 1, rs = rt << 1 | 1;int n = F[ls].size(), m = F[rs].size();F[rt].resize(n + m);for (int i = 0; i < n + m; i++) {F[rt][i] = 0;}if (n <= 50 && m <= 50) {for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {F[rt][i + j] = (F[rt][i + j] + 1ll * F[ls][i] * F[rs][j] % mod) % mod;}}return ;}for (int i = 0; i < n; i++) {A1[i] = F[ls][i];}for (int i = 0; i < m; i++) {B1[i] = F[rs][i];}int lim = 1;while (lim < n + m) {lim <<= 1;}get_r(lim);NTT(A1, lim, 1);NTT(B1, lim, 1);for (int i = 0; i < lim; i++) {A1[i] = 1ll * A1[i] * B1[i] % mod;}NTT(A1, lim, -1);for (int i = 0; i < n + m; i++) {F[rt][i] = A1[i];}for (int i = 0; i < lim; i++) {A1[i] = B1[i] = 0;}
}void divide(int rt, int l, int r) {if (l == r) {F[rt].push_back(C[l]);F[rt].push_back(B[l]);return ;}int mid = l + r >> 1;divide(rt << 1, l, mid), divide(rt << 1 | 1, mid + 1, r);merge(rt);F[rt << 1].clear(), F[rt << 1 | 1].clear();
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;init();scanf("%d", &T);while (T--) {scanf("%d", &n);for (int i = 0; i <= n; i++) {scanf("%d", &f[i]);f[i] = 1ll * f[i] * fac[i] % mod;}for (int i = 0; i <= n - 2; i++) {scanf("%d", &B[i]);}for (int i = 0; i <= n - 2; i++) {scanf("%d", &C[i]);}divide(1, 0, n - 2);int lim = 1;while (lim <= 2 * n) {lim <<= 1;}get_r(lim);for (int i = 0; i <= n; i++) {A1[i] = f[n - i];}for (int i = 0; i < F[1].size(); i++) {B1[i] = F[1][i];}F[1].clear();NTT(A1, lim, 1);NTT(B1, lim, 1);for (int i = 0; i < lim; i++) {A1[i] = 1ll * A1[i] * B1[i] % mod;}NTT(A1, lim, -1);for (int i = 0; i <= n; i++) {printf("%lld%c", 1ll * inv1[i] * A1[n - i] % mod, i == n ? '\n' : ' ');}for (int i = 0; i < lim; i++) {A1[i] = B1[i] = 0;}}return 0;
}

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（CCPC 2020 网络选拔赛）HDU 6900 Residual Polynomial（分治 + NTT）

Residual Polynomial

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