[BZOJ5306][Haoi2018]染色（容斥+组合数学+NTT）

Address

https://www.lydsy.com/JudgeOnline/problem.php?id=5306

Solution

先回顾一下「 $\{1,2,...,M\}$ 个中恰好 $K$ 个合法」的容斥求法：

$$\sum_{S\subset\{1,2,...,M\}}^{|S|\ge K}(-1)^{|S|-K}C_{|S|}^K\times(\{1,2,...,M\}的子集S合法)$$
接下来考虑一个问题：在 $M$ 种颜色中选出 $i$ 种，每种颜色染 $S$ 个格子，剩下的 $N-iS$ 个格子分别染剩下的 $M-i$ 种颜色之一，求方案数。
（1）选出 $i$ 种颜色，方案数：
$$C_M^i$$
（2）每种颜色分别染 $S$ 格：
$$\prod_{j=1}^iC_{N-(j-1)S}^S=\frac{N!}{(S!)^i(N-iS)!}$$
（3）染剩下 $N-iS$ 格：
$$(M-i)^{N-iS}$$
所以，方案数：
$$\frac{N!}{(S!)^i(N-iS)!}C_M^i(M-i)^{N-iS}$$
所以，答案为：
$$\sum_{k=0}^MW_k\sum_{i=k}^M\frac{N!}{(S!)^i(N-iS)!}(-1)^{i-k}C_i^kC_M^i(M-i)^{N-iS}$$
复杂度显然是平方的。
但根据式 (mo) 子 (shu) ，容易想到把原式转化为卷积形式。
考虑把 $\sum_{i=k}^M$ 后的式子改成枚举 $i$ 满足 $i+k$ 为合法颜色数：
$$\sum_{k=0}^MW_k\sum_{i=0}^{M-k}\frac{N!}{(S!)^{i+k}(N-(i+k)S)!}(-1)^iC_{i+k}^kC_M^{i+k}(M-i-k)^{N-(i+k)S}$$
$$=N!\sum_{k=0}^MW_k\sum_{i=0}^{M-k}\frac{(-1)^i\times\frac{(i+k)!}{i!k!}\times\frac{M!}{(i+k)!(M-i-k)!}(M-i-k)^{N-(i+k)S}}{(S!)^{i+k}(N-(i+k)S)!}$$
$$=N!M!\sum_{k=0}^M\frac{M_k}{k!(S!)^k}\sum_{i=0}^{M-k}\frac{(-1)^i}{i!(S!)^i}\times\frac{(M-i-k)^{N-(i+k)S}}{(M-i-k)!(N-(i+k)S)!}$$
$$=N!M!\sum_{k=0}^M\frac{M_k}{k!(S!)^k}\sum_{i=0}^{M-k}\frac{(-1)^i}{i!(S!)^i}\times\frac{(M-k-i)^{N+(M-k-i-M)S}}{(M-k-i)!(N+(M-k-i-M)S)!}$$
设：
$$F(i)=\frac{(-1)^i}{i!(S!)^i}$$
$$G(i)=\frac{i^{N+(i-M)S}}{i!(N+(i-M)S)!}$$
$$H(i)=\frac{M_i}{i!(S!)^i}$$
那么就变成了：
$$N!M!\sum_{k=0}^MH(k)(F\bigotimes G)(M-k)$$
是一个卷积的形式，可以使用 NTT 计算出。
复杂度 $O(n+m\log m)$ 。

Code

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define For(i, a, b) for (i = a; i <= b; i++)
#define Step(i, a, b, x) for (i = a; i <= b; i += x)
#define Pow(k, n) for (k = 1; k < n; k <<= 1)
using namespace std;
inline int read() {
    int res = 0; bool bo = 0; char c;
    while (((c = getchar()) < '0' || c > '9') && c != '-');
    if (c == '-') bo = 1; else res = c - 48;
    while ((c = getchar()) >= '0' && c <= '9')
        res = (res << 3) + (res << 1) + (c - 48);
    return bo ? ~res + 1 : res;
}
const int N = 3e5 + 5, M = 1e7 + 5, ZZQ = 1004535809;
int n, m, s, w[N], f[N], g[N], h[N], fac[M], inv[M],
spw[N], rev[N], ff = 1, gg, tot, gp[N], res[N], ans;
int qpow(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) res = 1ll * res * a % ZZQ;
        a = 1ll * a * a % ZZQ;
        b >>= 1;
    }
    return res;
}
void FFT(int n, int *a, int op) {
    int i, j, k, sp = n;
    gp[n] = qpow(op == 1 ? 3 : 334845270, (ZZQ - 1) / n);
    For (i, 0, n - 1) if (i < rev[i]) swap(a[i], a[rev[i]]);
    For (i, 1, tot) sp >>= 1,
        gp[sp] = 1ll * gp[sp << 1] * gp[sp << 1] % ZZQ;
    Pow(k, n) {
        int x = gp[k << 1];
        Step (i, 0, n - 1, k << 1) {
            int w = 1;
            For (j, 0, k - 1) {
                int u = a[i + j], v = 1ll * w * a[i + j + k] % ZZQ;
                a[i + j] = (u + v) % ZZQ;
                a[i + j + k] = (u - v + ZZQ) % ZZQ;
                w = 1ll * w * x % ZZQ;
            }
        }
    }
}
int main() {
    int i; fac[0] = inv[0] = inv[1] = spw[0] = 1;
    n = read(); m = read(); s = read();
    For (i, 0, m) w[i] = read();
    For (i, 1, max(n, m)) fac[i] = 1ll * fac[i - 1] * i % ZZQ;
    For (i, 2, max(n, m))
        inv[i] = 1ll * (ZZQ - ZZQ / i) * inv[ZZQ % i] % ZZQ;
    For (i, 2, max(n, m)) inv[i] = 1ll * inv[i] * inv[i - 1] % ZZQ;
    For (i, 1, m) spw[i] = 1ll * spw[i - 1] * inv[s] % ZZQ;
    For (i, 0, m) {
        f[i] = i & 1 ? ZZQ - 1 : 1;
        f[i] = 1ll * f[i] * inv[i] % ZZQ * spw[i] % ZZQ;
    }
    For (i, 0, m) {
        if (n + (i - m) * s < 0) continue;
        g[i] = qpow(i, n + (i - m) * s);
        g[i] = 1ll * g[i] * inv[i] % ZZQ * inv[n + (i - m) * s] % ZZQ;
    }
    For (i, 0, m) h[i] = 1ll * w[i] * inv[i] % ZZQ * spw[i] % ZZQ;
    while (ff <= (m << 1)) ff <<= 1, tot++;
    gg = qpow(ff, ZZQ - 2);
    For (i, 0, ff - 1)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << tot - 1);
    FFT(ff, f, 1); FFT(ff, g, 1);
    For (i, 0, ff - 1) res[i] = 1ll * f[i] * g[i] % ZZQ;
    FFT(ff, res, -1);
    For (i, 0, ff - 1) res[i] = 1ll * res[i] * gg % ZZQ;
    For (i, 0, m) ans = (ans + 1ll * h[i] * res[m - i] % ZZQ) % ZZQ;
    cout << 1ll * fac[m] * fac[n] % ZZQ * ans % ZZQ << endl;
    return 0;
}