题意：
有 (n) 个人，(m) 间浴室，每间浴室有(a[ i ])个浴缸，每个人要洗澡的话都要排队，假如一群人进入同一个浴室，他们总倾向于使得最长的队伍最短，现在问你所有队伍中最长的期望？
中文题解：
用状态 (dp[i][j][k]) 表示还剩 (i) 间浴室，还剩 (j) 个人，之前最长队伍的长度为 (k) 的期望最长队伍长度。
那么状态转移方程为:
$dp[i][j][k] = sum_{i=1}{m}sum_{j=0}{n}sum_{k=1}{n}sum_{c=1}{j}(dp[i-1][j-c][max(k, frac{c+a[i]-1}{a[i]})] * frac{(i-1)^{j-c}}{ i^j } * C(j, c)) $
其中 (c) 是枚举当前去第 (j) 间浴室的人数。
那么答案就是 (dp[m][n][0]) 。
时间复杂度：(O(n^{3}*m))
英文题解：
This problem is solved by dynamic programming
Consider the following dynamics: (dp[i][j][k]).
(i) --- number of not yet processed students,
(j) --- number of not yet processed rooms,
(k) --- maximum queue in the previous rooms.
The value we need is in state (dp[n][m][0]). Let's consider some state ((i, j, k)) and search through all (c) from 0 to (i). If (c) students will go to (j)th room, than a probability of such event consists of factors: (C_{i}^{c}) --- which students will go to (j)th room.
((1 / j)^c· ((j - 1) / j)^{i-c}) --- probability, that (c) students will go to (j)th room,and the rest of them will go to the rooms from first to (j - 1)th.
Sum for all (ñ) from 0 to (i) values of
((1 / j)^c· ((j - 1) / j)^{i-c}·C_{i}^{c}· dp[i-c][j-1][mx]) . Do not forget to update maximum queue value and get the accepted.
代码：
#include<bits/stdc++.h>
#pragma GCC optimize ("O3")
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
int n,m;
int a[56];
double dp[56][56][56];
double C[56][56];
//题解：
//http://www.cnblogs.com/LzyRapx/p/7692702.html
int main()
{
cin>>n>>m;
C[0][0] = 1.0;
for(int i=1;i<=55;i++)
{
C[i][0] = 1.0;
for(int j=1;j<=i;j++)
{
C[i][j] = C[i-1][j-1] + C[i-1][j];
}
}
for(int i=1;i<=m;i++) cin>>a[i];
for(int i=0;i<=n;i++) dp[0][0][i] = i;
for(int i=1;i<=m;i++)
{
for(int j=0;j<=n;j++)
{
for(int k=0;k<=n;k++)
{
for(int c=0;c<=j;c++)
{
int Max = max(k,(c+a[i]-1)/a[i]);
dp[i][j][k] += dp[i-1][j-c][Max] * pow(i-1,j-c) / pow(i,j) * C[j][c];
}
}
}
}
printf("%.10f
",dp[m][n][0]);
return 0;
}