Problem:
 Score : 100 points    Problem Statement There are N dishes, numbered 1, 2, \ldots, N .
Initially, for each i ( 1 \leq i \leq N ), Dish i has a_i ( 1 \leq a_i \leq 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, \ldots, 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 \leq N \leq 300  1 \leq a_i \leq 3         Input Input is given from Standard Input in the following format:  N  a_1  a_2  \ldots  a_N       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^{-9} .        Sample Input 1 3
1 1 1      Sample Output 1 5.5  The expected number of operations before the first piece of sushi is eaten, is 1 .
After that, the expected number of operations before the second sushi is eaten, is 1.5 .
After that, the expected number of operations before the third sushi is eaten, is 3 .
Thus, the expected total number of operations is 1 + 1.5 + 3 = 5.5 .       Sample Input 2 1
3      Sample Output 2 3  Outputs such as 3.00 , 3.000000003 and 2.999999997 will also be accepted.       Sample Input 3 2
1 2      Sample Output 3 4.5       Sample Input 4 10
1 3 2 3 3 2 3 2 1 3      Sample Output 4 54.48064457488221

Solution (Python):
from collections import Counter, defaultdict
N = int(input())
A = list(map(int, input().split()))
C = Counter(A)

U = N + 1
inf = 1e18

dp = [[[inf] * U for _ in range(U)] for _ in range(U)]
dp[0][0][0] = 0.0

for c in range(N + 1):
    for b in range(N + 1):
        if c + b > N:
            break
        for a in range(N + 1):
            if a + b + c == 0:
                continue
            if a + b + c > N:
                break
            prob = 0
            E = N
            if a > 0:
                E += a * dp[a - 1][b][c]
                prob += a
            if b > 0 and a < N:
                E += b * dp[a + 1][b - 1][c]
                prob += b
            if c > 0 and b < N:
                E += c * dp[a][b + 1][c - 1]
                prob += c
            dp[a][b][c] = E / prob

x = C[1]
y = C[2]
z = C[3]
print(dp[x][y][z])

Solution (Python):
import bisect
import copy
import heapq
import math
import sys
from collections import *
from itertools import accumulate, combinations, permutations, product

from functools import lru_cache
def input():
    return sys.stdin.readline()[:-1]
def ruiseki(lst):
    return [0]+list(accumulate(lst))
sys.setrecursionlimit(5000000)
mod=pow(10,9)+7
al=[chr(ord('a') + i) for i in range(26)]
direction=[[1,0],[0,1],[-1,0],[0,-1]]

n=int(input())
a=list(map(int,input().split()))

lst=[0]*3
for i in range(n):
    lst[a[i]-1]+=1

dp=[[[0.0]*310 for i in range(310)] for j in range(310)]

for k in range(lst[2]+1):
    for j in range(lst[1]+lst[2]+1-k):
        for i in range(lst[0]+lst[1]+lst[2]+1-k-j):
            if i==j==k==0:
                continue
            tmp=1.0
            if i>0:
                tmp+=dp[i-1][j][k]*i/n
            if j>0:
                tmp+=dp[i+1][j-1][k]*j/n
            if k>0:
                tmp+=dp[i][j+1][k-1]*k/n
            
            tmp*=n/(i+j+k)

            dp[i][j][k]=tmp

print(dp[lst[0]][lst[1]][lst[2]])

Solution (Python):
N=int(input())
C=[0]*4
for i in map(int, input().split()):
    if i==1:
        C[1]+=1
    elif i==2:
        C[2]+=1
    else:
        C[3]+=1
dp = [[[.0] * (N + 2) for _ in range(N + 2)] for _ in range(N + 2)]
for i in range(C[3] + 1):
    for j in range(N - C[1] - i + 1):
        for k in range(N - i - j + 1):
            if i + j + k != 0:
                dp[i][j][k] = (
                    N +
                    i * dp[i - 1][j + 1][k] +
                    j * dp[i][j - 1][k + 1] +
                    k * dp[i][j][k - 1]
                ) / (i + j + k)
print(dp[C[3]][C[2]][C[1]])

Solution (Python):
def main():
    N = int(input())
    a = map(int, input().split())
    single = double = triple = 0
    for sushi in a:
        if sushi == 1:
            single += 1
        elif sushi == 2:
            double += 1
        elif sushi == 3:
            triple += 1
    dp = [[[0 for i in range(1+single+double+triple-j-k)]
           for j in range(double + triple + 1 - k)] for k in range(triple + 1)]
    for k in range(triple+1):
        for j in range(double + triple + 1-k):
            for i in range(1+single+double+triple-j-k):
                if i == 0 and j == 0 and k == 0:
                    continue
                first = second = third = 0
                if i-1 >= 0 and j >= 0 and k >= 0:
                    first = dp[k][j][i-1]
                if i+1 >= 0 and j-1 >= 0 and k >= 0:
                    second = dp[k][j - 1][i+1]
                if i >= 0 and j+1 >= 0 and k-1 >= 0:
                    third = dp[k-1][j + 1][i]
                dp[k][j][i] = (first*i + second * j +
                               third * k + N) / (i + j + k)
    print(dp[triple][double][single])


if __name__ == "__main__":
    main()

Solution (Python):
import bisect
import copy
import heapq
import math
import sys
from collections import *
from itertools import accumulate, combinations, permutations, product

from functools import lru_cache
def input():
    return sys.stdin.readline()[:-1]
def ruiseki(lst):
    return [0]+list(accumulate(lst))
sys.setrecursionlimit(5000000)
mod=pow(10,9)+7
al=[chr(ord('a') + i) for i in range(26)]
direction=[[1,0],[0,1],[-1,0],[0,-1]]

n=int(input())
a=list(map(int,input().split()))

lst=[0]*3
for i in range(n):
    lst[a[i]-1]+=1

dp=[[[0.0]*310 for i in range(310)] for j in range(310)]

for k in range(lst[2]+1):
    for j in range(lst[1]+lst[2]+1-k):
        for i in range(lst[0]+lst[1]+lst[2]+1-k-j):
            if i==j==k==0:
                continue
            tmp=1.0
            if i>0:
                tmp+=dp[i-1][j][k]*i/n
            if j>0:
                tmp+=dp[i+1][j-1][k]*j/n
            if k>0:
                tmp+=dp[i][j+1][k-1]*k/n
            
            tmp*=n/(i+j+k)

            dp[i][j][k]=tmp

print(dp[lst[0]][lst[1]][lst[2]])

Solution (Python):
import sys
import math
from decimal import Decimal, ROUND_HALF_UP, ROUND_HALF_EVEN
from collections import deque
from bisect import bisect_left
from itertools import product
def I(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI2(N): return [list(map(int, sys.stdin.readline().split())) for i in range(N)]

def LSI(): return list(map(int, list(sys.stdin.readline().rstrip())))
def LSI2(N): return [list(map(int, list(sys.stdin.readline().rstrip()))) for i in range(N)]

def ST(): return sys.stdin.readline().rstrip()
def LST(): return sys.stdin.readline().rstrip().split()
def LST2(N): return [sys.stdin.readline().rstrip().split() for i in range(N)]
def FILL(i,h): return [i for j in range(h)]
def FILL2(i,h,w): return [FILL(i,w) for j in range(h)]
def FILL3(i,h,w,d): return [FILL2(i,w,d) for j in range(h)]
def FILL4(i,h,w,d,d2): return [FILL3(i,w,d,d2) for j in range(h)]
def sisha(num,digit): return Decimal(str(num)).quantize(Decimal(digit),rounding=ROUND_HALF_UP)

MOD = 1000000007
INF = float("inf")
sys.setrecursionlimit(10**6+10)

N = I()
A = LI()
one = 0
two = 0
three = 0
for a in A:
    if a==1:
        one += 1
    elif a==2:
        two += 1
    elif a==3:
        three += 1

dp = FILL3(0.0,N+2,N+2,N+2)
dp[0][0][0] = 0
for  k in range(0,three+1):
    for j in range(0,two+three-k+1):
        for i in range(0,one+two+three-k-j+1):
            if i==j==k==0:
                continue
            zero = N-i-j-k
            p_0 = zero/N
            p_1 = i/N
            p_2 = j/N
            p_3 = k/N
            ret = 1
            if i>0:
                ret += p_1*dp[i-1][j][k]
            if j>0:
                ret += p_2*dp[i+1][j-1][k]
            if k>0:
                ret += p_3*dp[i][j+1][k-1]
            ret /= 1-p_0
            dp[i][j][k] = ret

print(dp[one][two][three])

Solution (Python):
import sys
import math
from decimal import Decimal, ROUND_HALF_UP, ROUND_HALF_EVEN
from collections import deque
from bisect import bisect_left
from itertools import product
def I(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI2(N): return [list(map(int, sys.stdin.readline().split())) for i in range(N)]

def LSI(): return list(map(int, list(sys.stdin.readline().rstrip())))
def LSI2(N): return [list(map(int, list(sys.stdin.readline().rstrip()))) for i in range(N)]

def ST(): return sys.stdin.readline().rstrip()
def LST(): return sys.stdin.readline().rstrip().split()
def LST2(N): return [sys.stdin.readline().rstrip().split() for i in range(N)]
def FILL(i,h): return [i for j in range(h)]
def FILL2(i,h,w): return [FILL(i,w) for j in range(h)]
def FILL3(i,h,w,d): return [FILL2(i,w,d) for j in range(h)]
def FILL4(i,h,w,d,d2): return [FILL3(i,w,d,d2) for j in range(h)]
def sisha(num,digit): return Decimal(str(num)).quantize(Decimal(digit),rounding=ROUND_HALF_UP)

MOD = 1000000007
INF = float("inf")
sys.setrecursionlimit(10**6+10)

N = I()
A = LI()
one = 0
two = 0
three = 0
for a in A:
    if a==1:
        one += 1
    elif a==2:
        two += 1
    elif a==3:
        three += 1

dp = FILL3(0.0,N+2,N+2,N+2)
dp[0][0][0] = 0.0
for  k in range(0,three+1):
    for j in range(0,two+three-k+1):
        for i in range(0,one+two+three-k-j+1):
            if i==j==k==0:
                continue
            zero = N-i-j-k
            p_0 = zero/N
            p_1 = i/N
            p_2 = j/N
            p_3 = k/N
            ret = 1.0
            if i>0:
                ret += p_1*dp[i-1][j][k]
            if j>0:
                ret += p_2*dp[i+1][j-1][k]
            if k>0:
                ret += p_3*dp[i][j+1][k-1]
            ret /= 1-p_0
            dp[i][j][k] = ret

print(dp[one][two][three])

Solution (Python):
import sys
def input(): return sys.stdin.readline().rstrip()
def main():
    n=int(input())
    A=list(map(int,input().split()))
    
    dp=[[[0.0]*(n+1) for j in range(n+1)]for i in range(n+1)]
    max_i,max_j=A.count(1),A.count(2)
    max_k=n-max_i-max_j
    for k in range(max_k+1):
        for j in range(max_j+max_k+1):
            for i in range(min(max_i+max_j+max_k+1,n-(j+k)+1)):
                if i+j+k==0:continue
                res=n
                if i>0:res+=dp[i-1][j][k]*i
                if j>0:res+=dp[i+1][j-1][k]*j
                if k>0:res+=dp[i][j+1][k-1]*k
                res/=i+j+k
                dp[i][j][k]=res
    print(dp[max_i][max_j][max_k])

if __name__=='__main__':
    main()

Solution (Python):
import sys
def input(): return sys.stdin.readline().rstrip()
from collections import Counter
def main():
    n=int(input())
    A=list(map(int,input().split()))
    ac=Counter(A)
    
    dp=[[[0.0]*(n+1) for j in range(n+1)]for i in range(n+1)]
    max_i,max_j,max_k=ac[1],ac[2],ac[3]
    for k in range(max_k+1):
        for j in range(max_j+max_k+1):
            for i in range(min(max_i+max_j+max_k+1,n-(j+k)+1)):
                if i+j+k==0:continue
                res=n
                if i>0:res+=dp[i-1][j][k]*i
                if j>0:res+=dp[i+1][j-1][k]*j
                if k>0:res+=dp[i][j+1][k-1]*k
                res/=i+j+k
                dp[i][j][k]=res
    print(dp[max_i][max_j][max_k])

if __name__=='__main__':
    main()

Solution (Python):
from collections import Counter
n = int(input())
a = list(map(int,input().split()))
c = Counter(a)
c1,c2,c3 = c[1],c[2],c[3]
dp = [[[0 for i in range(c3+1)] for j in range(c2+c3+1)] for k in range(c1+c2+c3+1)]

for k in range(c3+1):
  for j in range(c2+c3+1):
    for i in range(n+1):
      if i+j+k > n or i == j == k == 0:
        continue
      x = (n-i-j-k)/n
      dp[i][j][k] = 1
      if 0<i :
        dp[i][j][k] += dp[i-1][j][k]*(i/n)
      if 0<j and i+1<=n:
        dp[i][j][k] += dp[i+1][j-1][k]*(j/n)
      if 0<k and j+1<=c2+c3:
        dp[i][j][k] += dp[i][j+1][k-1]*(k/n)
      dp[i][j][k] /= 1-x
print(dp[c1][c2][c3])

Solution (Python):

n=int(input())
a=list(map(int,input().split()))
dp=[[[-1]*(301) for i in range(301)] for j in range(301)]











def memo(i,j,k):
    if dp[i][j][k]+1:
        return dp[i][j][k]
    ans=n
    if i>0:
        ans+=memo(i-1,j,k)*i
    if j>0:
        ans+=memo(i+1,j-1,k)*j
    if k>0:
        ans+=memo(i,j+1,k-1)*k
    dp[i][j][k]=ans/(i+j+k)
    return ans/(i+j+k)
dp[0][0][0]=0
x,y,z=0,0,0
for i in a:
    if i==1:
        x+=1
    elif i==2:
        y+=1
    else:
        z+=1
print(memo(x,y,z))

Solution (Python):
'''
N枚の皿それぞれにa_i個の寿司があり、等確率で皿を選んだ時その皿の寿司を一つ食べる。
その皿に寿司が無い場合は何も行わない。
すべての皿がなくなるまでの操作回数の期待値を求めよ。

terms:
    1<=N<=300
    1<=a_i<=3

input:
3
1 1 1

output:
    5.5

N<=300: O(N^3)が通る

期待値DP。
同じ寿司の量の皿の数にさえ注目すればいいところが鍵。
※PythonもPyPyも遅くて通らない...
再帰をfor文に書き直せばPyPyで通るはず
'''

from collections import defaultdict,deque
import sys, bisect, math, itertools
from functools import lru_cache

sys.setrecursionlimit(10**8)
input = sys.stdin.readline
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp():
    '''
    一つの整数
    '''
    return int(input())
def inpl():
    '''
    一行に複数の整数
    '''
    return list(map(int, input().split()))


n = inp()
a = inpl()
'''
dp[i][j][k] : 
残り1個の皿がi枚、残り2個の皿がj枚、残り3個の皿がk枚の状態から
すべての寿司をなくすのに必要な回数の期待値
'''
one, two, three = 0, 0, 0
for i in a:
    if i == 1:
        one += 1
    elif i == 2:
        two += 1
    else:
        three+=1
MAX = max(one, two, three)
dp = [[[0.0] * (n + 1) for i in range(n + 1)] for j in range(n + 1)]

for k in range(three+1):
    for j in range(two+three+1):
        for i in range(one+two+three+1):
            if i + j + k == 0 or i + j + k > n:
                continue
            res = 0.0
            if i != 0:
                res += dp[i - 1][j][k] * i
            if j != 0:
                res += dp[i + 1][j - 1][k] * j
            if k != 0:
                res += dp[i][j + 1][k - 1] * k
            res += n
            res *= 1.0 / (i + j + k)
            dp[i][j][k] = res
print(dp[one][two][three])






    














'''
dp遷移は
次に引く皿の寿司 = 次の状態 × 確率 として
zero = dp[i][j][k] * (N-i-j-k)/N
one =  dp[i-1][j][k] * i/N
two =  dp[i+1][j-1][k] * j/N
three= dp[i][j+1][k-1] * k/N
dp[i][j][k]= zero + one + two + three + 1
(最後の1で追加しているイメージ)
となる。しかし両辺にdp[i][j][k]があると更新できない。(自己ループ)
よってこれをdp[i][j][k]について解くと
dp[i][j][k] = (one + two + three + 1)*N/(i+j+k)

dp[i][j][k] = (dp[i-1][j][k]*i + dp[i+1][j-1][k]*j + dp[i][j+1][k-1]*k + N)/(i+j+k)

ただし、それぞれが0の時に例外処理が必要。
'''

Solution (Python):
'''
N枚の皿それぞれにa_i個の寿司があり、等確率で皿を選んだ時その皿の寿司を一つ食べる。
その皿に寿司が無い場合は何も行わない。
すべての皿がなくなるまでの操作回数の期待値を求めよ。

terms:
    1<=N<=300
    1<=a_i<=3

input:
3
1 1 1

output:
    5.5

N<=300: O(N^3)が通る

期待値DP。
同じ寿司の量の皿の数にさえ注目すればいいところが鍵。
※PythonもPyPyも遅くて通らない...
再帰をfor文に書き直せばPyPyで通るはず
'''

from collections import defaultdict,deque
import sys, bisect, math, itertools
from functools import lru_cache

sys.setrecursionlimit(10**8)
input = sys.stdin.readline
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp():
    '''
    一つの整数
    '''
    return int(input())
def inpl():
    '''
    一行に複数の整数
    '''
    return list(map(int, input().split()))


n = inp()
a = inpl()
'''
dp[i][j][k] : 
残り1個の皿がi枚、残り2個の皿がj枚、残り3個の皿がk枚の状態から
すべての寿司をなくすのに必要な回数の期待値
'''
one, two, three = 0, 0, 0
for i in a:
    if i == 1:
        one += 1
    elif i == 2:
        two += 1
    else:
        three+=1
MAX = max(one, two, three)
dp = [[[0.0] * (n + 1) for i in range(n + 1)] for j in range(n + 1)]

dp[0][0][0] = 1.0
for k in range(three+1):
    for j in range(two+three+1):
        for i in range(one+two+three+1):
            if i + j + k == 0 or i + j + k > n:
                continue
            res = 0.0
            if i != 0:
                res += dp[i - 1][j][k] * i
            if j != 0:
                res += dp[i + 1][j - 1][k] * j
            if k != 0:
                res += dp[i][j + 1][k - 1] * k
            res += n
            res *= 1.0 / (i + j + k)
            dp[i][j][k] = res
print(dp[one][two][three]-1.0)






    














'''
dp遷移は
次に引く皿の寿司 = 次の状態 × 確率 として
zero = dp[i][j][k] * (N-i-j-k)/N
one =  dp[i-1][j][k] * i/N
two =  dp[i+1][j-1][k] * j/N
three= dp[i][j+1][k-1] * k/N
dp[i][j][k]= zero + one + two + three + 1
(最後の1で追加しているイメージ)
となる。しかし両辺にdp[i][j][k]があると更新できない。(自己ループ)
よってこれをdp[i][j][k]について解くと
dp[i][j][k] = (one + two + three + 1)*N/(i+j+k)

dp[i][j][k] = (dp[i-1][j][k]*i + dp[i+1][j-1][k]*j + dp[i][j+1][k-1]*k + N)/(i+j+k)

ただし、それぞれが0の時に例外処理が必要。
'''

Solution (Python):
from collections import Counter

n = int(input())
a = list(map(int, input().split()))
cnt = Counter(a)
p, q, r = cnt[1], cnt[2], cnt[3]

dp = [[[0.0] * (n + 1) for _ in range(n + 1)] for _ in range(n + 1)]

for k in range(r + 1):
    for j in range(r + q + 1):
        for i in range(n + 1):
            if not (0 < i + j + k <= n):
                continue
            if i > 0:
                dp[i][j][k] += dp[i - 1][j][k] * i
            if j > 0:
                dp[i][j][k] += dp[i + 1][j - 1][k] * j
            if k > 0:
                dp[i][j][k] += dp[i][j + 1][k - 1] * k
            dp[i][j][k] += n
            dp[i][j][k] /= (i + j + k)

print(dp[p][q][r])

Solution (Python):
from collections import Counter

n = int(input())
a = list(map(int, input().split()))
cnt = Counter(a)
p, q, r = cnt[1], cnt[2], cnt[3]

dp = [[[0.0] * (n + 1) for _ in range(n + 1)] for _ in range(n + 1)]

for k in range(r + 1):
    for j in range(r + q + 1):
        for i in range(n + 1):
            if not (0 < i + j + k <= n):
                continue
            tmp = 0.0
            if i > 0:
                tmp += dp[i - 1][j][k] * i
            if j > 0:
                tmp += dp[i + 1][j - 1][k] * j
            if k > 0:
                tmp += dp[i][j + 1][k - 1] * k
            tmp += n
            dp[i][j][k] = tmp / (i + j + k)

print(dp[p][q][r])

Solution (Python):
import sys
input = sys.stdin.readline

n = int(input())
a = list(map(int, input().split()))
cnt = [0, 0, 0]
for ai in a:
    cnt[ai-1] += 1
p, q, r = cnt[0], cnt[1], cnt[2]

dp = [[[0.0] * (n + 1) for _ in range(n + 1)] for _ in range(n + 1)]

for k in range(r + 1):
    for j in range(r + q + 1):
        for i in range(n + 1):
            if not (0 < i + j + k <= n):
                continue
            tmp = 0.0
            if i > 0:
                tmp += dp[i - 1][j][k] * i
            if j > 0:
                tmp += dp[i + 1][j - 1][k] * j
            if k > 0:
                tmp += dp[i][j + 1][k - 1] * k
            tmp += n
            dp[i][j][k] = tmp / (i + j + k)

print(dp[p][q][r])

Solution (Python):
import sys
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def NI(): return int(sys.stdin.readline())


N = NI()
a = [0, 0, 0]
for x in LI():
    a[x-1] += 1
dp = [[[0.0] * (N+1) for _ in range(N+1)] for _ in range(N+1)]

for k in range(a[2]+1):
    for j in range(a[1]+a[2]+1):
        for i in range(N+1):
            if not (0 < i+j+k <= N):
                continue
            x = 0.0
            if i > 0:
                x += dp[i-1][j][k] * (i/N)
            if j > 0:
                x += dp[i+1][j-1][k] * (j/N)
            if k > 0:
                x += dp[i][j+1][k-1] * (k/N)
            x += 1.0
            x *= N / (i+j+k)
            dp[i][j][k] = x
print(dp[a[0]][a[1]][a[2]])

Solution (Python):
import sys,queue,math,copy,itertools,bisect,collections,heapq
sys.setrecursionlimit(10**7)
INF = 10**18
MOD = 10**9 + 7
LI = lambda : [int(x) for x in sys.stdin.readline().split()]
NI = lambda : int(sys.stdin.readline())

N = NI()
a = [0,0,0]
for x in LI():
    a[x-1] += 1
dp = [[[0.0] * (N+1) for _ in range(N+1)] for _ in range(N+1)]

for k in range(a[2]+1):
    for j in range(a[1]+a[2]+1):
        for i in range(N+1):
            if not (0 < i+j+k <= N): continue
            x = 0.0
            if i > 0: x += dp[i-1][j][k] * (i/N)
            if j > 0: x += dp[i+1][j-1][k] * (j/N)
            if k > 0: x += dp[i][j+1][k-1] * (k/N)
            x += 1.0
            x *= N / (i+j+k)
            dp[i][j][k] = x
print(dp[a[0]][a[1]][a[2]])

Solution (Python):
import sys
sys.setrecursionlimit(2147483647)
INF = float("inf")
MOD = 10**9 + 7 
input = lambda:sys.stdin.readline().rstrip()
def resolve():
    n = int(input())
    counter = [0] * 3
    for a in map(int, input().split()):
        counter[a - 1] += 1

    
    
    dp = [[[None] * (n + 1) for _ in range(n + 1)] for _ in range(n + 1)]
    dp[0][0][0] = 0
    for s in range(1, n + 1):
        for k in range(n + 1):
            for j in range(n + 1):
                i = s - j - k
                if i < 0:
                    break
                res = n
                if i: res += i * dp[i - 1][j][k]
                if j: res += j * dp[i + 1][j - 1][k]
                if k: res += k * dp[i][j + 1][k - 1]

                res /= s
                dp[i][j][k] = res

    i, j, k = counter
    print(dp[i][j][k])
resolve()

Solution (Python):
N=int(input())
A=list(map(int,input().split()))
d={1:0,2:0,3:0}
for a in A:
    d[a]+=1
dp=[[[0]*(N+1) for j in range(d[2]+d[3]+1)] for i in range(d[3]+1)]
for i in range(d[3]+1):
    for j in range(d[2]+d[3]-i+1):
        for k in range(N-i-j+1):
            if i+j+k==0:continue
            dp[i][j][k]=N
            if i>0:
                dp[i][j][k]+=dp[i-1][j+1][k]*i
            if j>0:
                dp[i][j][k]+=dp[i][j-1][k+1]*j
            if k>0:
                dp[i][j][k]+=dp[i][j][k-1]*k
            dp[i][j][k]/=i+j+k
print(dp[d[3]][d[2]][d[1]])

Solution (Python):
import sys,queue,math,copy,itertools,bisect,collections,heapq
sys.setrecursionlimit(10**7)
INF = 10**18
MOD = 10**9 + 7
LI = lambda : [int(x) for x in sys.stdin.readline().split()]
NI = lambda : int(sys.stdin.readline())

N = NI()
a = [0,0,0]
for x in LI():
    a[x-1] += 1
dp = [[[0.0] * (N+1) for _ in range(N+1)] for _ in range(N+1)]

for k in range(a[2]+1):
    for j in range(a[1]+a[2]+1):
        for i in range(N+1):
            if not (0 < i+j+k <= N): continue
            x = 0.0
            if i > 0: x += dp[i-1][j][k] * (i/N)
            if j > 0: x += dp[i+1][j-1][k] * (j/N)
            if k > 0: x += dp[i][j+1][k-1] * (k/N)
            x += 1.0
            x *= N / (i+j+k)
            dp[i][j][k] = x
print(dp[a[0]][a[1]][a[2]])

Solution (Python):
n = int(input())
a = list(map(int, input().split()))
dp = [[[float(0)]*(n+1) for _ in range(n+1)] for _ in range(n+1)]
I,J,K = 0,0,0
for i in a:
  if i == 1:
    I += 1
  elif i == 2:
    J += 1
  else:
    K += 1
for k in range(n+1):
  for j in range(n+1):
    for i in range(n+1):
      if (i+j+k) <= n:
        if (i+j+k) != 0:
          dp[i][j][k] = float(n)/float(i+j+k)
        if i > 0:
          dp[i][j][k] += dp[i-1][j][k]*float(i) / float(i+j+k)
        if j > 0:
          dp[i][j][k] += dp[i+1][j-1][k]*float(j) / float(i+j+k)
        if k > 0:
          dp[i][j][k] += dp[i][j+1][k-1]*float(k) / float(i+j+k)
print(dp[I][J][K])

Solution (Python):
n = int(input())
ns = list(map(int, input().split()))
one = ns.count(1)
two = ns.count(2)
three = n-one-two
dp = [[[0 for _ in range(n+1)] for _ in range(two+three+1)] for _ in range(three+1)]
for k in range(three+1):
    for j in range(two+three-k+1):
        for i in range(n-j-k+1):
            v = k+i+j
            if v==0:
                continue
            res = n
            if i:
                res += i * dp[k][j][i-1]
            if j:
                res += j * dp[k][j-1][i+1]
            if k:
                res += k * dp[k-1][j+1][i]
            dp[k][j][i] = res / v
print(dp[three][two][one])

Solution (Python):
n=int(input())
a=tuple(map(int,input().split()))
c1 = a.count(1)
c2 = a.count(2)
c3 = a.count(3)

dp = [[[0.0]*(n+1) for i in range(n+1)] for j in range(n+1)] 
dp[0][0][0] = 0
for i in range(c3+1):
    for j in range(c2+c3-i+1):
        for k in range(n-i-j+1):
            if i == j == k == 0:
                continue
            tmp = n
            if k != 0:
                tmp += k*dp[k-1][j][i]
            if j != 0:
                tmp += j*dp[k+1][j-1][i]
            if i != 0:
                tmp += i*dp[k][j+1][i-1]
            dp[k][j][i] = tmp/(i+j+k)

print(dp[c1][c2][c3])