Luogu 4491 [HAOI2018]染色

BZOJ 5306

考虑计算恰好出现$s$次的颜色有$k$种的方案数。

首先可以设$lim = min(m, \left \lfloor \frac{n}{s} \right \rfloor)$，我们在计算的时候只要算到这个$lim$就可以了。

设$f(k)$表示出现$s$次的颜色至少有$k$种的方案数，则

$$f(k) = \binom{m}{k}\binom{n}{ks}\frac{(ks)!}{(s!)^k}(m - k)^{n - ks}$$

就是先选出$k$个颜色和$ks$个格子放这些颜色，这样子总的方案数是全排列除以限排列，剩下的颜色随便放。

处理一下组合数，整个$f$可以在$O(nlogn)$时间内算出来。

设$g(k)$表示出现$s$次的颜色刚好有$k$种的方案数，考虑到对于$\forall i < j$，$g(j)$在$f(i)$中被计算了$\binom{j}{i}$次，所以有

$$f(k) = \sum_{i = k}^{lim}\binom{i}{k}g(i)$$

直接二项式反演回来，

$$g(k) = \sum_{i = k}^{lim}(-1)^{i - k}\binom{i}{k}f(i)$$

拆开组合数，

$$g(k) = \sum_{i = k}^{lim}(-1)^{i - k}\frac{i!}{k!(i - k)!}f(i) = \frac{1}{k!}\sum_{i = k}^{lim}\frac{(-1)^{i - k}}{(i - k)!}(f(i) * (i!))$$

设$A(i) = f(i) * (i!)$，$B(i) = \frac{(-1)^{i}}{i!}$

$$g(k)* (k!) = \sum_{i = k}^{lim}A(i)B(i - k)$$

咕，并不是卷积。

把$B$翻转，再设$B'(i) = B(lim - i)$

$$g(k)* (k!) = \sum_{i = k}^{lim}A(i)B'(lim + k - i)$$

注意到$A*B'$的第$lim + k$项的系数是$\sum_{i = 0}^{lim + k}A(i)B'(lim + k - i)$，但是需要满足

$$ 0 \leq i \leq lim$$

$$ 0 \leq lim + k - i \leq lim $$

$$k \leq i \leq lim$$

刚好满足。

时间复杂度$O(n + mlogm)$。

Code：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
typedef vector <ll> poly;

const int N = 1e7 + ;
const int M = 1e5 + ;

int n, m, s;
ll w[M], fac[N], ifac[N], f[M], g[M], sum[M];

inline void deb(poly c) {
    for (int i = ; i < (int)c.size(); i++)
        printf("%lld%c", c[i], " \n"[i == (int)c.size() - ]);
}

namespace Poly {
    const int L =  << ;
    const ll P = 1004535809LL;

    int lim, pos[L];

    template <typename T>
    inline void inc(T &x, T y) {
        x += y;
        if (x >= P) x -= P;
    }

    template <typename T>
    inline void sub(T &x, T y) {
        x -= y;
        if (x < ) x += P;
    }

    inline void reverse(poly &c) {
        for (int i = , j = (int)c.size() - ; i < j; i++, j--) swap(c[i], c[j]);
    }

    inline ll fpow(ll x, ll y) {
        ll res = ;
        for (; y > ; y >>= ) {
            if (y & ) res = res * x % P;
            x = x * x % P;
        }
        return res;
    }

    inline void prework(int len) {
        int l = ;
        for (lim = ; lim < len; lim <<= , ++l);
        for (int i = ; i < lim; i++)
            pos[i] = (pos[i >> ] >> ) | ((i & ) << (l - ));
    }

    inline void ntt(poly &c, int opt) {
        c.resize(lim, );
        for (int i = ; i < lim; i++)
            if (i < pos[i]) swap(c[i], c[pos[i]]);
        for (int i = ; i < lim; i <<= ) {
            ll wn = fpow(, (P - ) / (i << ));
            if (opt == -) wn = fpow(wn, P - );
            for (int len = i << , j = ; j < lim; j += len) {
                ll w = ;
                for (int k = ; k < i; k++, w = w * wn % P) {
                    ll x = c[j + k], y = c[j + k + i] * w % P;
                    c[j + k] = (x + y) % P, c[j + k + i] = (x - y + P) % P;
                }
            }
        }

        if (opt == -) {
            ll inv = fpow(lim, P - );
            for (int i = ; i < lim; i++) c[i] = c[i] * inv % P;
        }
    }

    inline poly mul(const poly x, const poly y) {
        poly u = x, v = y, res;
        prework(x.size() + y.size() - );
        ntt(u, ), ntt(v, );
        for (int i = ; i < lim; i++) res.push_back(u[i] * v[i] % P);
        ntt(res, -);
        res.resize(x.size() + y.size() - );
        return res;
    }

}

using Poly :: P;
using Poly :: fpow;
using Poly :: mul;
using Poly :: inc;
using Poly :: reverse;

template <typename T>
inline void read(T &X) {
    X = ; char ch = ; T op = ;
    for (; ch > ''|| ch < ''; ch = getchar())
        if (ch == '-') op = -;
    for (; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}

inline void prework(int len) {
    fac[] = ;
    for (int i = ; i <= len; i++) fac[i] = fac[i - ] * i % P;
    ifac[len] = fpow(fac[len], P - );
    for (int i = len - ; i >= ; i--) ifac[i] = ifac[i + ] * (i + ) % P;
}

inline ll getC(int x, int y) {
    return fac[x] * ifac[y] % P * ifac[x - y] % P;
}

int main() {
    read(n), read(m), read(s);
    for (int i = ; i <= m; i++) read(w[i]);

    int rep = min(m, n / s);
    prework(max(n, m));

    for (int i = ; i <= rep; i++)
        f[i] = getC(m, i) * getC(n, i * s) % P * fac[i * s] % P * fpow(ifac[s], i) % P * fpow(m - i, n - i * s) % P;

/*    for (int i = 0; i <= rep; i++)
        printf("%lld%c", f[i], " \n"[i == rep]);     */

    poly a, b;
    a.resize(rep + , ), b.resize(rep + , );
    for (int i = ; i < rep + ; i++) {
        a[i] = f[i] * fac[i] % P;
        b[i] = ifac[i];
        if (i & ) b[i] = (P - b[i]) % P;
    }
    reverse(b);

//    deb(a), deb(b);

    poly c = mul(a, b);

//    deb(c);

/*    for (int i = rep; i >= 0; i--) {
        sum[i] = sum[i + 1];
        inc(sum[i], c[i]);
    }   

    for (int i = 0; i <= rep; i++)
        printf("%lld%c", sum[i], " \n"[i == rep]);    */ 

    ll ans = ;
    for (int i = ; i <= rep; i++) {
        g[i] = ifac[i] * c[rep + i] % P;
        inc(ans, w[i] * g[i] % P);
    }

    printf("%lld\n", ans);
    return ;
}