传送门

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调了1h竟然是因为1004535809写成了998244353

“恰好有\(K\)种颜色出现了\(S\)次”的限制似乎并不容易达到，考虑容斥计算。

令\(c_j\)表示强制\(j\)种颜色恰好出现\(S\)次，其他颜色随意染的方案数。可以通过生成函数知道

\(\begin{align*} c_j &= \binom{m}{j} n! [x^n] (\frac{x^k}{k!})^j (\sum\limits_{i=0}^\infty \frac{x^i}{i!})^{m-j} \\ &= \binom{m}{j} n! [x^n] (\frac{x^k}{k!})^j e^{(m-j)x} \\ &= \binom{m}{j} n! [x^n] (\frac{x^k}{k!})^j \sum\limits_{i=0}^\infty \frac{x^i (m-j)^i}{i!} \\ &= \binom{m}{j}n! \frac{(m-j)^{n-jk}}{(k!)^j (n-jk)!} \end{align*}\)

显然是可以预处理阶乘之后\(O(1)\)计算的。注意不要漏掉了指数型生成函数前面要乘的\(n!\)。

又设\(h_j\)表示恰好有\(j\)种颜色出现了\(S\)次的方案总数。有\(h_j = c_j - \sum\limits_{i=j+1}^{Max} h_{i}A(i,j)\)，\(A(i,j)\)是一个与\(i,j\)相关的容斥系数，表示\(h_i\)在\(c_j\)中的出现次数。

既然\(h_i\)中恰好有\(i\)种颜色出现了\(S\)次，那么对于任意一个对\(h_i\)产生贡献的状态，只要枚举到当前状态中\(i\)种恰好出现了\(S\)次的颜色构成的集合的任意一个大小为\(j\)的子集时都会对\(c_j\)产生\(1\)的贡献。所以\(A(i,j) = \binom{i}{j}\)

所以可以得到\(h_j = c_j - \sum\limits_{i=j+1}^{Max}h_i \binom{i}{j}\)，两边同乘\(j!\)得到\(h_jj! = c_jj! - \sum\limits_{i=j+1}^{Max}\frac{h_ii!}{(i-j)!}\)

设多项式\(H = \sum\limits_{i=0}^{Max}h_ii!x^i , C = \sum\limits_{i=0}^{Max}c_ii!x^i\)，记\(rev(H)\)为多项式\(H\)所有系数翻转过来之后的多项式，那么不难得到\(rev(H) = rev(C) - W * rev(H)\)，其中\(W = \sum\limits_{i=0}^{Max} \frac{1}{i!} - 1\)。多项式求逆即可。

#include<iostream>
#include<cstdio>
#include<random>
#include<cstring>
#include<algorithm>
//This code is written by Itst
using namespace std;

const int mod = 998244353;
inline int read(bool flg = 0){
    int a = 0;
    char c = getchar();
    bool f = 0;
    while(!isdigit(c) && c != EOF){
        if(c == '-')
            f = 1;
        c = getchar();
    }
    if(c == EOF)
        exit(0);
    while(isdigit(c)){
        if(flg)
            a = (a * 10ll + c - 48) % mod;
        else
            a = a * 10 + c - 48;
        c = getchar();
    }
    if(flg) a += mod;
    return f ? -a : a;
}

const int MAXN = (1 << 19) + 7 , MAXM = 1e7 + 7 , MOD = 1004535809;
#define PII pair < int , int >
#define st first
#define nd second

inline int poww(long long a , int b){
    int times = 1;
    while(b){
        if(b & 1)
            times = times * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return times;
}

namespace poly{
    const int G = 3 , INV = (MOD + 1) / G;
    int A[MAXN] , B[MAXN] , C[MAXN] , D[MAXN] , E[MAXN];
    int a[MAXN] , b[MAXN] , c[MAXN] , d[MAXN];
    int need , inv , dir[MAXN] , _inv[MAXN];
#define clear(x) memset(x , 0 , sizeof(int) * need)

    void init(int len){
        need = 1;
        while(need < len)
            need <<= 1;
        inv = poww(need , MOD - 2);
        for(int i = 1 ; i < need ; ++i)
            dir[i] = (dir[i >> 1] >> 1) | (i & 1 ? need >> 1 : 0);
    }

    void init_inv(){
        _inv[1] = 1;
        for(int i = 2 ; i < MAXN ; ++i)
            _inv[i] = MOD - 1ll * (MOD / i) * _inv[MOD % i] % MOD;
    }

    void NTT(int *arr , int type){
        for(int i = 1 ; i < need ; ++i)
            if(i < dir[i])
                arr[i] ^= arr[dir[i]] ^= arr[i] ^= arr[dir[i]];
        for(int i = 1 ; i < need ; i <<= 1){
            int wn = poww(type == 1 ? G : INV , (MOD - 1) / i / 2);
            for(int j = 0 ; j < need ; j += i << 1){
                long long w = 1;
                for(int k = 0 ; k < i ; ++k , w = w * wn % MOD){
                    int x = arr[j + k] , y = arr[i + j + k] * w % MOD;
                    arr[j + k] = x + y >= MOD ? x + y - MOD : x + y;
                    arr[i + j + k] = x < y ? x + MOD - y : x - y;
                }
            }
        }
    }

    void mul(int *a , int *b){
        NTT(a , 1);NTT(b , 1);
        for(int i = 0 ; i < need ; ++i)
            a[i] = 1ll * a[i] * b[i] % MOD;
        NTT(a , -1);
    }

    void getInv(int *a , int *b , int len){
        if(len == 1){
            b[0] = poww(a[0] , MOD - 2);
            return;
        }
        getInv(a , b , (len + 1) >> 1);
        memcpy(A , a , sizeof(int) * len);
        memcpy(B , b , sizeof(int) * len);
        init(len * 3);
        NTT(A , 1);NTT(B , 1);
        for(int i = 0 ; i < need ; ++i)
            A[i] = 1ll * A[i] * B[i] % MOD * B[i] % MOD;
        NTT(A , -1);
        for(int i = 0 ; i < len ; ++i)
            b[i] = (2 * b[i] - 1ll * A[i] * inv % MOD + MOD) % MOD;
        clear(A);clear(B);
    }
}
using namespace poly;
int F[MAXN] , H[MAXN] , jc[MAXM] , Inv[MAXM] , W[MAXN];
int N , M , K , Len;

void init(){
    jc[0] = 1;
    for(int i = 1 ; i <= N || i <= M ; ++i)
        jc[i] = 1ll * jc[i - 1] * i % MOD;
    Inv[max(N , M)] = poww(jc[max(N , M)] , MOD - 2);
    for(int i = max(N , M) - 1 ; i >= 0 ; --i)
        Inv[i] = Inv[i + 1] * (i + 1ll) % MOD;
}

int binom(int b , int a){
    return b < a ? 0 : 1ll * jc[b] * Inv[a] % MOD * Inv[b - a] % MOD;
}

int calc(int j){
    return 1ll * poww(Inv[K] , j) * Inv[N - j * K] % MOD * poww(M - j , N - j * K) % MOD * binom(M , j) % MOD * jc[N] % MOD;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
    //freopen("out","w",stdout);
#endif
    init_inv();
    N = read(); M = read(); K = read();
    for(int i = 0 ; i <= M ; ++i) W[i] = read();
    Len = min(N / K , M);
    init();
    for(int i = 0 ; i <= Len ; ++i)
        F[i] = 1ll * calc(i) * jc[i] % MOD;
    reverse(F , F + Len + 1);
    for(int i = 0 ; i <= Len ; ++i)
        H[i] = Inv[i];
    getInv(H , a , Len + 1);
    init((Len + 1) * 2);
    mul(F , a);
    reverse(F , F + Len + 1);
    int ans = 0;
    for(int i = 0 ; i <= Len ; ++i)
        ans = (ans + 1ll * F[i] * inv % MOD * Inv[i] % MOD * W[i]) % MOD;
    cout << ans;
    return 0;
}