P4491 [HAOI2018]染色

题目链接：洛谷

题目大意：$n$个位置染$m$种颜色，如果出现次数恰为$S$次的颜色有$k$种，则对答案有$W_k$的贡献，求所有染色方案的答案之和$mod 1004535809$。

数据范围：$nleq 10^7,mleq 10^5,Sleq 150,0leq W_ileq 1004535808$

首先是要推式子的。

首先我们知道，出现次数恰为$S$次的至多$up=min(m,frac{n}{S})$种。

设恰好出现$S$次的颜色至少$i$种，则

$$f_i=C_m^i*frac{n!}{(S!)^i(n-iS)!}*(m-i)^{n-iS}$$

这三项分别是选$i$种颜色，对现在这$i+1$种颜色（选定的$i$种和未染色的）做可重集排列，对剩下的$n-iS$个位置任意染色。

然后用容斥原理就可以得出，恰好出现$S$次的颜色正好$i$种的方案数

$$ans_i=sum_{j=i}^{up}(-1)^{j-i}C_j^if_j$$

所以

$$frac{ans_i}{i!}=sum_{j=i}^{up}frac{(-1)^{j-i}}{(j-i)!}*frac{f_j}{j!}$$

如果你不知道如何做，把第一个数组先反转再向右平移$up$位，再把计算后的$ans_i$左移$up$位就可以了。

#include<cstdio>
#include<algorithm>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 400003, G = 3, Gi = 334845270, mod = 1004535809;
int n, m, s, up, W[N], fac[10000003], inv[10000003], f[N], g[N], ans;
inline int kasumi(int a, int b){
    int res = 1;
    while(b){
        if(b & 1) res = (LL) res * a % mod;
        a = (LL) a * a % mod;
        b >>= 1;
    }
    return res;
}
inline void init(){
    fac[0] = 1;
    for(Rint i = 1;i <= 10000000;i ++) fac[i] = (LL) i * fac[i - 1] % mod;
    inv[0] = inv[1] = 1;
    for(Rint i = 2;i <= 10000000;i ++) inv[i] = (LL) inv[mod % i] * (mod - mod / i) % mod;
    for(Rint i = 1;i <= 10000000;i ++) inv[i] = (LL) inv[i] * inv[i - 1] % mod;
}
inline int C(int n, int m){
    if(n < m) return 0;
    return (LL) fac[n] * inv[m] % mod * inv[n - m] % mod;
}
int rev[N];
inline void NTT(int *A, int limit, int type){
    for(Rint i = 0;i < limit;i ++)
        if(i < rev[i]) swap(A[i], A[rev[i]]);
    for(Rint mid = 1;mid < limit;mid <<= 1){
        int Wn = kasumi(type == 1 ? G : Gi, (mod - 1) / (mid << 1));
        for(Rint j = 0;j < limit;j += mid << 1){
            int w = 1;
            for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % mod){
                int x = A[j + k], y = (LL) w * A[j + k + mid] % mod;
                A[j + k] = (x + y) % mod;
                A[j + k + mid] = (x - y + mod) % mod;
            }
        }
    }
    if(type == -1){
        int inv = kasumi(limit, mod - 2);
        for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * inv % mod;
    }
}
int main(){
    scanf("%d%d%d", &n, &m, &s);
    for(Rint i = 0;i <= m;i ++) scanf("%d", W + i);
    init(); up = min(m, n / s);
    for(Rint i = 0;i <= up;i ++)
        f[i] = (LL) C(m, i) * fac[i] % mod * fac[n] % mod * kasumi(inv[s], i) % mod * inv[n - i * s] % mod * kasumi(m - i, n - i * s) % mod;
    for(Rint i = 0;i <= up;i ++)
        g[i] = ((up - i) & 1) ? (mod - inv[up - i]) : inv[up - i];
    int limit = 1, L = -1;
    while(limit <= (up << 1)){limit <<= 1; ++ L;}
    for(Rint i = 0;i < limit;i ++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
    NTT(f, limit, 1); NTT(g, limit, 1);
    for(Rint i = 0;i < limit;i ++) f[i] = (LL) f[i] * g[i] % mod;
    NTT(f, limit, -1);
    for(Rint i = 0;i <= up;i ++)
        ans = (ans + (LL) W[i] * f[up + i] % mod * inv[i] % mod) % mod;
    printf("%d
", ans);
}