Problem:


Score : 300 points

Problem Statement

You have N bamboos. The lengths (in centimeters) of these are l_1, l_2, ..., l_N, respectively.

Your objective is to use some of these bamboos (possibly all) to obtain three bamboos of length A, B, C. For that, you can use the following three kinds of magics any number:

Extension Magic: Consumes 1 MP (magic point). Choose one bamboo and increase its length by 1.

Shortening Magic: Consumes 1 MP. Choose one bamboo of length at least 2 and decrease its length by 1.

Composition Magic: Consumes 10 MP. Choose two bamboos and combine them into one bamboo. The length of this new bamboo is equal to the sum of the lengths of the two bamboos combined. (Afterwards, further magics can be used on this bamboo.)

At least how much MP is needed to achieve the objective?

Constraints

3 \leq N \leq 8

1 \leq C < B < A \leq 1000

1 \leq l_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B C
l_1
l_2
:
l_N

Output

Print the minimum amount of MP needed to achieve the objective.

Sample Input 1

5 100 90 80
98
40
30
21
80

Sample Output 1

23

We are obtaining three bamboos of lengths 100, 90, 80 from five bamboos 98, 40, 30, 21, 80. We already have a bamboo of length 80, and we can obtain bamboos of lengths 100, 90 by using the magics as follows at the total cost of 23 MP, which is optimal.

Use Extension Magic twice on the bamboo of length 98 to obtain a bamboo of length 100. (MP consumed: 2)

Use Composition Magic on the bamboos of lengths 40, 30 to obtain a bamboo of length 70. (MP consumed: 10)

Use Shortening Magic once on the bamboo of length 21 to obtain a bamboo of length 20. (MP consumed: 1)

Use Composition Magic on the bamboo of length 70 obtained in step 2 and the bamboo of length 20 obtained in step 3 to obtain a bamboo of length 90. (MP consumed: 10)

Sample Input 2

8 100 90 80
100
100
90
90
90
80
80
80

Sample Output 2

0

If we already have all bamboos of the desired lengths, the amount of MP needed is 0. As seen here, we do not necessarily need to use all the bamboos.

Sample Input 3

8 1000 800 100
300
333
400
444
500
555
600
666

Sample Output 3

243



Sample input:
5 100 90 80
98
40
30
21
80


Sample output:
23

Submission status: Runtime Error

Solution code:
N, A, B, C = map(int, input().split())
ls = [int(input) for i in range(N)]

def dfs(cur, a, b, c):
  if cur==N:
    return abs(A-a)+abs(B-b)+abs(C-c)-30 if min(a,b,c)>0 else 10**9
  ret0 = dfs(cur+1, a+ls[cur], b, c)
  ret1 = dfs(cur+1, a, b+ls[cur], c)
  ret2 = dfs(cur+1, a, b, c+ls[cur])
  ret3 = dfs(cur+1, a, b, c)
  return min(ret0, ret1, ret2, ret3)
print(dfs(0,0,0,0))

Submission status: Runtime Error

Solution code:
    N, A, B, C = map(int, input().split())
    l = [int(input()) for _ in range(N)]

    def dfs(i, a, b, c, x):
        if i == N:
            return (abs(A - a) + abs(B - b) + abs(C - c) + 10 * (x - 3)) if a * b * c != 0 else 10 ** 9
        ret1 = dfs(i + 1, a, b, c, x)
        ret2 = dfs(i + 1, a + l[i], b, c, x + 1)
        ret3 = dfs(i + 1, a, b + l[i], c, x + 1)
        ret4 = dfs(i + 1, a, b, c + l[i], x + 1)
        return (min(ret1, ret2, ret3, ret4))

Submission status: Runtime Error

Solution code:
#コピペAC
N, A, B, C = map(int, input().split())
l = [int(input()) for i in range(N)]
INF = 10 ** 9
def dfs(cur, a, b, c):
  if cur == N:
  return abs(a - A) + abs(b - B) + abs(c - C) - 30 if min(a, b, c) > 0 else INF
ret0 = dfs(cur + 1, a, b, c)
ret1 = dfs(cur + 1, a + l[cur], b, c) + 10
ret2 = dfs(cur + 1, a, b + l[cur], c) + 10
ret3 = dfs(cur + 1, a, b, c + l[cur]) + 10
return min(ret0, ret1, ret2, ret3)

print(dfs(0, 0, 0, 0))

Submission status: Runtime Error

Solution code:
#コピペAC
N, A, B, C = map(int, input().split())
l = [int(input()) for i in range(N)]
INF = 10 ** 9
def dfs(cur, a, b, c):
if cur == N:
return abs(a - A) + abs(b - B) + abs(c - C) - 30 if min(a, b, c) > 0 else INF
ret0 = dfs(cur + 1, a, b, c)
ret1 = dfs(cur + 1, a + l[cur], b, c) + 10
ret2 = dfs(cur + 1, a, b + l[cur], c) + 10
ret3 = dfs(cur + 1, a, b, c + l[cur]) + 10
return min(ret0, ret1, ret2, ret3)

print(dfs(0, 0, 0, 0))

Submission status: Runtime Error

Solution code:
N, A, B, C = map(int, input().split())
l = [int(input()) for _ in range(N)]
INF = 10 ** 9

def dfs(cur, a, b, c):
  if cur == N:
    return abs(a-A) + abs(b-B) + abs(c-C) - 30 if min(a, b, c) > 0 else INF
  
  # 使わない場合
  ret0 = dfs(cur + 1, A, B, C)
  # A に使う場合
  ret1 = dfs(cur + 1, A + l[cur], B, C) + 10
  # B に使う場合
  ret2 = dfs(cur + 1, A, B + l[cur], C) + 10
  # C に使う場合
  ret3 = dfs(cur + 1, A, B, C + l[cur]) + 10
  
  return min(ret0, ret1, ret2, ret3)
  
print(defs(0, 0, 0, 0))

Submission status: Runtime Error

Solution code:
import sys

n,A,B,C=map(int,input().split())
x=list(set(map(int,sys.stdin)))

inf=10**9

def dfs(cur,a,b,c):
    if cur==n:
        #print("abs",abs(a-A),abs(b-B),abs(c-C))
        #print("abs...",abs(a-A)+abs(b-B)+abs(c-C)-30)
        return abs(a-A)+abs(b-B)+abs(c-C)-30 if min(a,b,c)>0 else inf
    ret0 = dfs(cur+1,a,b,c)
    ret1 = dfs(cur+1,a+x[cur],b,c) + 10
    ret2 = dfs(cur+1,a,b+x[cur],c) + 10
    ret3 = dfs(cur+1,a,b,c+x[cur]) + 10
    #print(ret0,ret1,ret2,ret3)
    return min(ret0, ret1, ret2, ret3)

print(dfs(0,0,0,0))

Submission status: Runtime Error

Solution code:
N,A,B,C = map(int, input().split())
l = list(map(int, input().split()))
def dfs(cur,a,b,c):
    if cur == N:
        if min(a, b, c) > 0:
                return abs(a-A) + abs(b-B) + abs(c-C) - 30
        else:
                return 10**9
    ret0 = dfs(cur+1, a, b, c)
    ret1 = dfs(cur+1, a + l[cur], b, c) + 10
    ret2 = dfs(cur+1, a, b + l[cur], c) + 10
    ret3 = dfs(cur+1, a, b, c + l[cur]) + 10
    return min(ret0,ret1,ret2,ret3)

print(dfs(0, 0, 0, 0))

Submission status: Runtime Error

Solution code:
n,A,B,C = map(int,input().split(()))
l = [int(input()) for i in range(n)]
inf = 10 ** 9

def dfs(i,a,b,c):
    if i == n:
        if min(a,b,c) > 0:
            return abs(a-A) + abs(b-B) + abs(c-C) - 30
        else:
            return inf
    ret0 = dfs(i+1, a, b, c)
    ret1 = dfs(i+1, a + l[i], b, c) + 10
    ret2 = dfs(i+1, a, b + l[i], c) + 10
    ret3 = dfs(i+1, a, b, c + l[i]) + 10
    return min(ret0, ret1, ret2, ret3)

print(dfs(0,0,0,0))

Submission status: Runtime Error

Solution code:
tgt_b = int(str_in[2])
tgt_c = int(str_in[3])

source = [int(input()) for i in range(num)]

mp = float('inf')

for i in range(1,2**num):

  sum_a = 0
  len_a = 0
  
  bin_i = format(i,'0256b')  
  for t in range(1,num+1):
    sum_a += int(bin_i[-t])
    len_a += int(bin_i[-t]) * source[t-1]
    
  for j in range(1,2**num):
 
    sum_b = 0
    len_b = 0

    if (i & j) > 0:
      continue
    
    bin_j = format(j,'0256b')
    for t in range(1,num+1):
      sum_b += int(bin_j[-t])
      len_b += int(bin_j[-t]) * source[t-1]

    for k in range(1,2**num):

      sum_c = 0
      len_c = 0
      
      if (i & k) > 0:
        continue
      elif(j & k) > 0:
        continue
        
      bin_k = format(k,'08b')        
      
      for t in range(1,num+1):
        sum_c += int(bin_k[-t])
        len_c += int(bin_k[-t]) * source[t-1]
        
      mp_try = abs(len_a - tgt_a) + abs(len_b - tgt_b) + abs(len_c - tgt_c) + (sum_a + sum_b + sum_c - 3) * 10 
      if(mp > mp_try):
        mp = mp_try

print (mp)

Submission status: Runtime Error

Solution code:
def main():
    import itertools as it
    N,A,B,C = map(int,input().split())
    pre = [int(input()) for _ in range(N)]
    if N == 3:print(A-pre[0]+B-pre[1]+C-pre[2])
    else:
        m = [0,1,2,3,4,5,6,7]
        cpm = it.permutations(pre,N)
        ccb = list(it.combinations(m[:N], 3))
        minc = 99999999999999999
        for l in cpm:
            for d in ccb:
                cmp = (d[2] - 2) * 10
                asum = abs(A - sum(l[:d[0]+1]))
                if cmp+asum >= minc: continue
                bsum = abs(B - sum(l[d[0]+1:d[1]+1]))
                if cmp+asum+bsum >= minc: continue
                csum = abs(C - sum(l[d[1]+1:d[2]+1]))
                tc = cmp+asum+bsum+csum
                minc = min(minc,tc)
    print(minc)
main()

Submission status: Runtime Error

Solution code:
N, A, B, C = map(int, input().split())
l = [int(input() for i in range(n))]
inf = 10**4

def bamboo_connect(i, a, b, c):
    if i == N:
        if all(a, b, c):
            return abs(a-A) + abs(b-B) + abs(c-C) - 30
        else:
            return inf
    no_connect = bamboo_connect(i+1, a, b, c)
    choice_a = bamboo_connect(i+1, a+l[i], b, c) + 10
    choice_b = bamboo_connect(i+1, a, b+l[i], c) + 10
    choice_c = bamboo_connect(i+1, a, b, c+l[i]) + 10
    return min(no_connect, choice_a, choice_b, choice_c)

print(bamboo_connect(0, 0, 0, 0))

Submission status: Runtime Error

Solution code:
import sys
import math
from collections import deque

sys.setrecursionlimit(4100000)


def inputs(num_of_input):
    ins = [input() for i in range(num_of_input)]
    return ins


def solve(N, A, B, C, inputs):
    bamboos = [int(i) for i in inputs]

    def cost(cursor, a, b, c):
        if cursor == N:
            if min(a, b, c) > 0:
                return abs(A-a) + abs(B-b) + abs(C-c) - 30
            else:
                return 1000000000

        t = bamboos[cursor]
        results = []
        results.append(cost(cursor+1, a, b, c))
        results.append(cost(cursor+1, a + t, b, c) + 10)
        results.append(cost(cursor+1, a, b + t, c) + 10)
        results.append(cost(cursor+1, a, b, c + t) + 10)

        return min(results)

    return cost(0, 0, 0, 0)


def string_to_int(string):
    return list(map(lambda x: int(x), string.split()))


if __name__ == "__main__":
    [N, A, B, C] = string_to_int(inputs(1))
    ret = solve(N, A, B, C, inputs(N))
    print(ret)

Submission status: Runtime Error

Solution code:
n,a,b,c=map(int,input().split())
l=[]
for _ in range(n):
  x=int(input())
  l.append(x)

def dfs(cur,x,y,z):
  if cur==n and min(x,y,z)>0:
    return abs(a-x)+abs(b-y)+abs(c-z)-30
  elif cur==n:
    return inf
  ret0=dfs(cur+1,x,y,z)
  ret1=dfs(cur+1,x+l[cur],y,z)+10
  ret2=dfs(cur+1,x,y+l[cur],z)+10
  ret3=dfs(cur+1,x,y,z+l[cur])+10
  return min(ret0,ret1,ret2,ret3)

print(dfs(0,0,0,0))

Submission status: Runtime Error

Solution code:
def main():

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)

    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # open(0).read() is a convenient method:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions used
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    if __name__ != '__main__':
        # for test
        sys.stdin = open('sample.txt')

    # ----------------------------------
    # main

    N, A, B, C = R()
    L = [int(r()) for _ in range(N)]
    G = [[0,1,2,3] for _ in range(N)]
    ans = INF
    for  i in product(*G):
        group_sum = [0] * N
        group_count = [0] * N
        for j, k in enumerate(i):
            group_sum[k] += L[j]
            group_count[k] += 1
        ans = min(ans, (abs(A - group_sum[0]) + abs(B - group_sum[1]) + abs(C - group_sum[2]) + (max(group_count[0]-1,0)+max(group_count[1]-1,0)+max(group_count[2]-1,0))*10))
    print(ans)
    
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # for
    for k, v in Dic.items():
    """

if __name__ == '__main__':
    main()

Submission status: Runtime Error

Solution code:
def main():

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)

    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # open(0).read() is a convenient method:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions used
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    if __name__ != '__main__':
        # for test
        sys.stdin = open('sample.txt')

    # ----------------------------------
    # main

    N, A, B, C = Rmap()
    L = [r_int() for _ in range(N)]
    G = [[0,1,2,3] for _ in range(N)]
    ans = INF
    for  i in product(*G):
        group_sum = [0] * N
        group_count = [0] * N
        for j, k in enumerate(i):
            group_sum[k] += L[j]
            group_count[k] += 1
        ans = min(ans, 1)
    print(ans)
    
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # for
    for k, v in Dic.items():
    """

if __name__ == '__main__':