Problem Statement

There are N dishes, numbered ,2,…,N. Initially, for each i (1≤i≤N), Dish i has ai​ (1≤ai​≤3) pieces of sushi on it.

Taro will perform the following operation repeatedly until all the pieces of sushi are eaten:

• Roll a die that shows the numbers 1,2,…,N with equal probabilities, and let i be the outcome. If there are some pieces of sushi on Dish i, eat one of them; if there is none, do nothing.

Find the expected number of times the operation is performed before all the pieces of sushi are eaten.

Constraints

• All values in input are integers.
• 1≤N≤300
• 1≤ai​≤3

Input

Input is given from Standard Input in the following format:

N
a1​ a2​ ……aN​

Output

Print the expected number of times the operation is performed before all the pieces of sushi are eaten. The output is considered correct when the relative difference is not greater than 10−910−9.

Sample 1

3
1 1 1

5.5

The expected number of operations before the first piece of sushi is eaten, is 11. After that, the expected number of operations before the second sushi is eaten, is 1.51.5. After that, the expected number of operations before the third sushi is eaten, is 33. Thus, the expected total number of operations is 1+1.5+3=5.51+1.5+3=5.5.

Sample 2

1
3

3

Outputs such as 3.00, 3.000000003 and 2.999999997 will also be accepted.

Sample 3

2
1 2

4.5

Sample 4

10
1 3 2 3 3 2 3 2 1 3

54.48064457488221

Sponsor

解析：

题解 AT4531 Sushi - 触情离殇 的博客 - 洛谷博客 (luogu.com.cn)

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<math.h>
#include<map>
#define ll long long
using namespace std;
const int MAXN = 310;int n, a[5];
double dp[MAXN][MAXN][MAXN];int main()
{cin >> n;for (int i = 1; i <= n; i++){int tmp;cin >> tmp;a[tmp]++;}for (int k = 0; k <= n; k++){for (int j = 0; j <= n; j++){for (int i = 0; i <= n; i++){double s = i + j + k;if (s != 0)//防止 i,j,k 同时为零{if (i)dp[i][j][k] += dp[i - 1][j][k] * i / s;if (j)dp[i][j][k] += dp[i + 1][j - 1][k] * j / s;if (k)dp[i][j][k] += dp[i][j + 1][k - 1] * k / s;dp[i][j][k] += (double)n / s;}}}}printf("%.10lf\n", dp[a[1]][a[2]][a[3]]);return 0;
}