题目
为了报答小 C 的苹果, 小 G 打算送给热爱美术的小 C 一块画布, 这块画布可 以抽象为一个长度为 (N) 的序列, 每个位置都可以被染成 (M) 种颜色中的某一种.
然而小 C 只关心序列的 (N) 个位置中出现次数恰好为 (S) 的颜色种数, 如果恰 好出现了 (S) 次的颜色有 (K) 种, 则小 C 会产生 (W_k) 的愉悦度.
小 C 希望知道对于所有可能的染色方案, 他能获得的愉悦度的和对 1004535809 取模的结果是多少.
输入格式
从标准输入读入数据. 第一行三个整数 (N, M, S).
接下来一行 (M + 1) 个整数, 第 (i) 个数表示 (W_{i-1})
.
输出格式
输出到标准输出中. 输出一个整数表示答案．
输入样例
8 8 3
3999 8477 9694 8454 3308 8961 3018 2255 4910
输出样例
524070430
提示
特殊性质: (forall 1 le i le m, W_i = 0)
对于 (100\%) 的数据, 满足 (0 le W_i < 10045358090)
题解
令(E = min(lfloor frac{N}{S}
floor,M))
我们枚举有几种颜色有(K)次，那么剩余的就不能有(K)次，利用容斥我们可以得出式子：
[ans = sumlimits_{i = 0}^{E} w[i]{M choose i} {N choose iS} frac{(iS)!}{(S!)^{i}} sumlimits_{j = 0}^{E - i} (-1)^{j} {M - i choose j} {N - iS choose jS} frac{(jS)!}{(S!)^{j}} (M - i - j)^{N - iS - jS}
]
那个((M - i - j)^{N - iS - jS})非常不好处理，我们考虑转化一下：
[ans = sumlimits_{i = 0}^{E} w[i]{M choose i} {N choose iS} frac{(iS)!}{(S!)^{i}} sumlimits_{j = i}^{E} (-1)^{j - i} {M - i choose j - i} {N - iS choose jS - iS} frac{(jS - iS)!}{(S!)^{j - i}} (M - j)^{N - jS}
]
然后展开组合数，分子分母对消，剩余如下：
[ans = sumlimits_{i = 0}^{E} w[i]frac{M!N!}{i!} sumlimits_{j = i}^{E} frac{(-1)^{j - i}(M - j)^{N -jS}}{(j - i)!(M - j)!(N - jS)!(S!)^{j}}
]
我们交换一下(i,j)的位置，经整理得：
[ans = sumlimits_{j = 0}^{E} frac{M!N!(M-j)^{N - jS}}{(M - j)!(N - jS)!(S!)^{j}} sumlimits_{i = 0}^{j} frac{w[i]}{i!} * frac{(-1)^{j - i}}{(j - i)!}
]
左边是只与(j)有关的式子，右边是(f(x) = frac{w[x]}{x!})与(g(x) = frac{(-1)^{x}}{x!})的卷积
NTT即可
(1004535809)的原根是(3)
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 100005,maxm = 10000005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
const int G = 3,P = 1004535809;
int qpow(int a,int b){
int ans = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) ans = 1ll * ans * a % P;
return ans;
}
int fac[maxm],fv[maxm],inv[maxm];
int L,R[maxn << 2],f[maxn << 2],g[maxn << 2],n,m;
int N,M,S,W[maxn];
void NTT(int* a,int F){
for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
for (int i = 1; i < n; i <<= 1){
int gn = qpow(G,(P - 1) / (i << 1));
for (int j = 0; j < n; j += (i << 1)){
LL g = 1,x,y;
for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
x = a[j + k]; y = 1ll * g * a[j + k + i] % P;
a[j + k] = (x + y) % P; a[j + k + i] = (x - y + P) % P;
}
}
}
if (F == 1) return;
int nv = qpow(n,P - 2); reverse(a + 1,a + n);
for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
void init(){
int E = max(N,M);
fac[0] = 1;
for (int i = 1; i <= E; i++) fac[i] = 1ll * fac[i - 1] * i % P;
inv[0] = inv[1] = 1;
for (int i = 2; i <= E; i++) inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
fv[0] = 1;
for (int i = 1; i <= E; i++) fv[i] = 1ll * fv[i - 1] * inv[i] % P;
}
void solve(){
int t,E = min(M,N / S);
for (int i = 0; i <= E; i++){
t = (i & 1) ? -1 : 1;
f[i] = 1ll * W[i] * fv[i] % P;
g[i] = 1ll * t * fv[i] % P;
}
L = 0; m = E + E;
for (n = 1; n <= m; n <<= 1) L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
NTT(f,1); NTT(g,1);
for (int i = 0; i < n; i++) f[i] = 1ll * f[i] * g[i] % P;
NTT(f,-1);
int ans = 0,tmp;
for (int i = 0; i <= E; i++){
tmp = 1ll * fac[M] * fac[N] % P * qpow(M - i,N - i * S) % P;
tmp = 1ll * tmp * fv[N - i * S] % P * fv[M - i] % P * qpow(fv[S],i) % P;
ans = (ans + 1ll * tmp * f[i] % P) % P;
}
printf("%d
",(ans % P + P) % P);
}
int main(){
N = read(); M = read(); S = read();
for (int i = 0; i <= M; i++) W[i] = read();
init();
solve();
return 0;
}