【UOJ449】[集训队作业2018]喂鸽子
【UOJ449】[集训队作业2018]喂鸽子
题面
题解
考虑\(\text{min-max}\)容斥，那么答案为
\[\sum_{i=1}^n(-1)^{i+1}{n\choose i}f_i
\]
其中\(f_i\)表示选出\(i\)只鸽子然后其中第一只鸽子被喂饱的期望时间。
那么\(f_i=\sum_{j}p_j\times j\)，其中\(p_j\)表示上述过程发生在时刻\(j\)的概率，
这样子不好统计，我们换一种统计方法：
\(f_i=\sum_jp'_j\)，\(p'_j\)表示在\(j\)时刻选出的鸽子没有一只喂饱的概率，也就是上述过程发生时刻大于等于\(j+1\)的概率。
再定义一个数组\(g_{c,s}\)表示\(c\)只鸽子单独拿出来，时刻\(s\)没有喂饱的的概率。
那么
\[\begin{aligned}
p'_i&=\sum_{j=0}^{i}g_{c,j} {i\choose j}\big(\frac {c}{n}\big)^j\big(\frac {n-c}{n}\big)^{i-j}
\end{aligned}
\]
也就是说
\[\begin{aligned}
f_c&=\sum_{i\geq 1}p'_i\\
&=\sum_{i\geq 1}\sum_{j=0}^ig_{c,j} {i\choose j}\big(\frac {c}{n}\big)^j\big(\frac {n-c}{n}\big)^{i-j}\\
&=\sum_{j=0}^{c(k-1)}g_{c,j}\big(\frac cn\big)^j\sum_{i\geq 0}{i+j\choose i}\big(\frac{n-c}{n}\big)^i
\end{aligned}
\]
注意到后面那一坨如果令\(x=\frac {n-c}c\)的话，就是\(\sum_{i\geq 0}{i+j\choose j}x^i=(\frac {1}{1-x})^{j+1}=(\frac nc)^{j+1}\)
那么\(f_c=\frac nc\sum_{j=0}^{c(k-1)} g_{c,j}\)
如果暴力的话可以考虑加入一个鸽子然后dp：
\(g_{c,s}=\sum_{i}g_{c-1,s-i}(\frac 1c)^i(\frac {c-1}{c})^{s-i}{s\choose i}\)
NTT 优化做到总复杂度\(O(n^2k(\log n+\log k))\)
考虑怎么把这个\(\log\)去掉，令 EGF \(g_c(x)=({\sum_{i=0}^{k} \frac {x^i}{i!}})^c\)（实际上是\(k-1\)这里为了方便所以写成\(k\)，下同），那么
\[g(x)=1+x+\frac {x^2}{2!}+...+\frac {x^k}{k!}\\
g'(x)=g(x)-\frac {x^k}{k!}\\
\begin{aligned}
g'_c(x)&=cg_{c-1}(x)\big(g(x)-\frac {x^k}{k!}\big)\\
&=c\big(g_{c}(x)-\frac {x^k}{k!}g_{c-1}(x)\big)
\end{aligned}\\
g_{c,s+1}=\frac{c}{s+1}(g_{c,s}-\frac {1}{k!}g_{c-1,s-k})
\]
于是可以递推了，最后\(g_{c,s}\)乘上一个\(\frac {1}{c^s}\)就是概率了。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int Mod = 998244353;
int fpow(int x, int y) {
int res = 1;
while (y) {
if (y & 1) res = 1ll * res * x % Mod;
x = 1ll * x * x % Mod;
y >>= 1;
}
return res;
}
const int MAX_N = 55, MAX_M = MAX_N * 1005;
int g[MAX_N][MAX_M], ifc[MAX_M], fac[MAX_M];
int f[MAX_N];
int N, K;
int C(int n, int m) { return 1ll * fac[n] * ifc[m] % Mod * ifc[n - m] % Mod; }
int main () {
#ifndef ONLINE_JUDGE
freopen("cpp.in", "r", stdin);
#endif
cin >> N >> K;
for (int i = fac[0] = 1; i <= N * K; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
ifc[N * K] = fpow(fac[N * K], Mod - 2);
for (int i = N * K - 1; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % Mod;
for (int i = 0; i < K; i++) g[1][i] = ifc[i];
for (int i = 2; i <= N; i++) {
g[i][0] = 1;
for (int j = 1; j <= i * (K - 1); j++) {
g[i][j] = g[i][j - 1];
if (j - K >= 0) g[i][j] = (g[i][j] - 1ll * ifc[K - 1] * g[i - 1][j - K]) % Mod;
g[i][j] = (g[i][j] + Mod) % Mod;
g[i][j] = 1ll * i * ifc[j] % Mod * fac[j - 1] % Mod * g[i][j] % Mod;
}
}
for (int i = 1; i <= N; i++) {
int t = 1ll * ifc[i] * fac[i - 1] % Mod;
for (int pw = 1, j = 0; j <= i * (K - 1); j++, pw = 1ll * pw * t % Mod)
g[i][j] = 1ll * g[i][j] * fac[j] % Mod * pw % Mod;
}
for (int i = 1; i <= N; i++) {
for (int j = 0; j <= i * (K - 1); j++) f[i] = (f[i] + g[i][j]) % Mod;
f[i] = 1ll * f[i] * N % Mod * ifc[i] % Mod * fac[i - 1] % Mod;
}
int ans = 0;
for (int i = 1; i <= N; i++) {
if (i & 1) ans = (ans + 1ll * C(N, i) * f[i]) % Mod;
else ans = (ans - 1ll * C(N, i) * f[i]) % Mod, ans = (ans + Mod) % Mod;
}
printf("%d\n", ans);
return 0;
}
