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: Accepted

Solution code:
import itertools

N, A, B, C = map(int,input().split())
L = [int(input()) for _ in range(N)]

A, B, C = sorted([A,B,C], reverse=True)

ans = 100000000 ## =Inf
for state in itertools.product(range(4), repeat=N): ## 4**N 通りの全探索, 3:not used
    bamboo = [0]*4
    for i in range(len(state)): ## stateに応じてbambooをグループ化する, 同グループは合成
      if state[i] < 3:
        bamboo[state[i]] += L[i]
    if 0 in bamboo[0:3]: ## 竹が3種類あることを確認
      continue
    cost_0 = max(state.count(0) - 1, 0) * 10
    cost_1 = max(state.count(1) - 1, 0) * 10
    cost_2 = max(state.count(2) - 1, 0) * 10
    cost = cost_0 + cost_1 + cost_2
    bamboo.sort(reverse=True)
    cost += abs(bamboo[0]-A) + abs(bamboo[1]-B) + abs(bamboo[2]-C)
    ans = min(ans, cost)
    
print(ans)

Submission status: Accepted

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

def base2n(Num,n):
    if Num<n:
        return str(Num)
    else:
        return base2n(Num//n,n)+str(Num%n)

ans=10**6
    
for i in range(4**N):
    i_4=("0"*N+base2n(i,4))[-N:]
    Group=[[] for i in range(4)]
    for j in range(N):
        Group[int(i_4[j])-1].append(l[j])
    if len(Group[0])*len(Group[1])*len(Group[2])!=0:
        ans=min(ans,(N-3-len(Group[3]))*10+sum([abs(ABC[j] - sum(Group[j])) for j in range(3)]))
        
print(ans)

Submission status: Accepted

Solution code:
from collections import deque
from itertools import combinations
N, A, B, C = map(int, input().split())
L = [int(input()) for _ in range(N)]

def dfs(d, a, b, c):
    if d == N:
        # A,B,Cの竹それぞれで最初の１回は合成していないため30引く
        return abs(a - A) + abs(b - B) + abs(c - C) - 30 if min(a, b, c)> 0 else 10**9
    
    p1 = dfs(d+1, a, b, c)
    p2 = dfs(d+1, a+L[d], b, c) + 10
    p3 = dfs(d+1, a, b+L[d], c) + 10
    p4 = dfs(d+1, a, b, c+L[d]) + 10
    
    return min(p1, p2, p3, p4)
    
print(dfs(0, 0, 0, 0))

Submission status: Accepted

Solution code:
from itertools import*
from math import*
from collections import*
from heapq import*
from bisect import bisect_left,bisect_right
from copy import deepcopy
inf = float("inf")
mod = 10**9+7
from functools import reduce
import sys
sys.setrecursionlimit(10**7)

#それぞれの竹はA,B,Cのどれかのために使うか使わないかしかない
N,A,B,C = map(int,input().split())
l = [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 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: Accepted

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

def dfs(idx, a, b, c):
    if idx == N:
        if a == 0 or b == 0 or c == 0:
            return 1e9
        return abs(A-a) + abs(B-b) + abs(C-c)
    
    x = dfs(idx+1,a,b,c)
    y = dfs(idx+1,a+L[idx],b,c) + 10*(a!=0)
    z = dfs(idx+1,a,b+L[idx],c) + 10*(b!=0)
    w = dfs(idx+1,a,b,c+L[idx]) + 10*(c!=0)

    return min(x,y,z,w)

ans = dfs(0,0,0,0)

print(ans)

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline

def inpl(): return list(map(int, input().split()))

N, A, B, C = inpl()
L = [0]*N
for i in range(N):
    L[i] = int(input())

ans = 10**9
for k in range(4**N):
    ABC = [0]*3
    CNT = [0]*3
    j = k*1
    for i in range(N):
        d, m = divmod(j, 4)
        if m <= 2:
            ABC[m] += L[i]
            CNT[m] += 1
        j = d
    if min(CNT) == 0:
        continue
    tmp = abs(ABC[0] - A) + abs(ABC[1] - B) + abs(ABC[2] - C) + sum([10*(cnt-1) for cnt in CNT])
    ans = min(ans, tmp)
print(ans)

Submission status: Accepted

Solution code:
n,x,y,z=map(int,input().split())
l=[]
cmin=1000000000
for i in range(n):
  l.append(int(input()))
for i in range(4**n):
  l1=[]
  for j in range(n):
    if i>=4:
      l1.append(i%4)
      i=(i//4)
    elif i!=0:
      l1.append(i)
      i=0
    else:
      l1.append(i)
  c1=0
  c2=0
  c3=0
  d1=0
  d2=0
  d3=0
  l1=l1[::-1]
  for j in range(n):
    if l1[j]==0:
      pass
    elif l1[j]==1:
      c1+=l[j]
      d1+=1
    elif l1[j]==2:
      c2+=l[j]
      d2+=1
    else:
      c3+=l[j]
      d3+=1
  if c1*c2*c3!=0:
    c=abs(c1-x)+abs(c2-y)+abs(c3-z)+(d1+d2+d3-3)*10
    if c<=cmin:
      cmin=c
print(cmin)

Submission status: Accepted

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

for i in range(N):
    l.append(int(input()))
    
def dfs(syn, n, a, b, c):
    if n == N:
        if a*b*c != 0:
            mp = abs(A-a) + abs(B-b) + abs(C-c)
            mp += syn * 10 - 30
            return mp
        else:
            return 10**9
    else:
        branch1 = dfs(syn+1, n+1, a + l[n], b, c)
        branch2 = dfs(syn+1, n+1, a, b + l[n], c)
        branch3 = dfs(syn+1, n+1, a, b, c + l[n])
        branch4 = dfs(syn, n+1, a, b, c)
        return min(branch1, branch2, branch3, branch4)
        
print(dfs(0, 0, 0, 0, 0))

Submission status: Accepted

Solution code:
import itertools

N, A, B, C = map(int, input().split())
l = []

for i in range(N):
    l.append(int(input()))
    
prod = list(itertools.product(['A','B','C','X'], repeat=N))

mps = []

for prod_e in prod:
    if prod_e.count('A') * prod_e.count('B') * prod_e.count('C') != 0:
        mp = 0
        length = {'A':0, 'B':0, 'C':0, 'X':0}
        for index, prod_e_e in enumerate(prod_e):
            length[prod_e_e] += l[index]
        mp += abs(length['A'] - A) + abs(length['B'] - B) + abs(length['C'] - C)
        mp += (prod_e.count('A')-1) * 10 + (prod_e.count('B')-1) * 10 + (prod_e.count('C')-1) * 10
        mps.append(mp)

print(min(mps))

Submission status: Accepted

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
    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: Accepted

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


def select(next_idx, a=0, b=0, c=0, point=0):
  if next_idx == N:
    if a and b and c:
      tmp = abs(A - a) + abs(B - b) + abs(C - c)
      yield abs(A - a) + abs(B - b) + abs(C - c) + point - 30
  else:
    length = ls[next_idx]
    flag = length > 10
    next_idx += 1

    yield from select(next_idx, a=a, b=b, c=c, point=point)
    point += 10
    if a == 0 or flag: 
      yield from select(next_idx, a=a + length, b=b, c=c, point=point)
    if b == 0 or flag: 
      yield from select(next_idx, a=a, b=b + length, c=c, point=point)
    if c == 0 or flag: 
      yield from select(next_idx, a=a, b=b, c=c + length, point=point)



print(min(select(0)))

Submission status: Accepted

Solution code:
from itertools import product


def main():
    n, *target = map(int, input().split())
    arr = [int(input()) for i in range(n)]

    tests = product("0123", repeat=n)
    test_arr = [[] for _ in range(3)]
    min_cost = float("inf")

    for test in tests:
        cost = 0

        if "1" not in test or "2" not in test or "3" not in test:
            continue

        for i in range(3):
            test_arr[i] = [arr[x] for x in range(n) if test[x] == str(i + 1)]
            cost += (len(test_arr[i]) - 1) * 10
            cost += abs(sum(test_arr[i]) - target[i])

        if min_cost > cost:
            min_cost = cost

    print(min_cost)


main()

Submission status: Accepted

Solution code:
N, A, B, C = map(int, input().split())
L = [int(input()) for _ in range(N)]
INF = float('inf')

def dfs(cursor=0, a=0, b=0, c=0):
    if cursor == N:
        return abs(a-A) + abs(b-B) + abs(c-C) - 30 if a*b*c > 0 else INF

    no_compound = dfs(cursor+1, a, b, c)
    A_compound = dfs(cursor+1, a+L[cursor], b, c) + 10
    B_compound = dfs(cursor+1, a, b+L[cursor], c) + 10
    C_compound = dfs(cursor+1, a, b, c+L[cursor]) + 10
    return min(no_compound, A_compound, B_compound, C_compound)

print(dfs())

Submission status: Accepted

Solution code:
from collections import deque

N, A, B, C = map(int, input().split())
L = [int(input()) for _ in range(N)]
minMP = 10**5

p = deque()
p.append([-1] * len(L))
while p:
    stats = p.pop()
    stats = list(stats)
    a, b, c = 0, 0, 0
    MP = 0
    cntA = 0; cntB = 0; cntC = 0
    for i in range(N):
        if stats[i] == 1:
            a += L[i]
            cntA += 1
        elif stats[i] == 2:
            b += L[i]
            cntB += 1
        elif stats[i] == 3:
            c += L[i]
            cntC += 1

    if cntA >= 1 and cntB >= 1 and cntC >= 1:
        MP += abs(A - a) + abs(B - b) + abs(C - c) + 10 * (cntA+cntB+cntC-3)
        minMP = min(minMP, MP)

    if not -1 in stats:
        continue
    j = stats.index(-1)
    for k in range(4):
        stats[j] = k
        newStats = tuple(stats)
        p.append(newStats)

print(minMP)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

import sys

input = sys.stdin.readline

def main():
    n, A, B, C = map(int, input().split())
    bumb = []
    for _ in range(n):
        bumb.append(int(input()))

    min_cost = 100000000
    for i in range(1<<n):
        for j in range(1<<n):
            if i & j:
                continue
            for k in range(1<<n):
                if i & k or j & k:
                    continue
                a = 0
                b = 0
                c = 0
                a_cost = 0
                b_cost = 0
                c_cost = 0
                for idx in range(n):
                    if i & 1<<idx:
                        a_cost += 10 if a != 0 else 0
                        a += bumb[idx]
                    if j & 1<<idx:
                        b_cost += 10 if b != 0 else 0
                        b += bumb[idx]
                    if k & 1<<idx:
                        c_cost += 10 if c != 0 else 0
                        c += bumb[idx]
                if a == 0 or b == 0 or c == 0:
                    continue
                m = abs(a - A) + abs(b - B) + abs(c - C) + a_cost + b_cost + c_cost
                min_cost = min(min_cost, m)
    print(min_cost)
main()

Submission status: Accepted

Solution code:
inf = 10**9

def dfs(cur, a, b, c):
    if cur==N:
        return abs(a-A) + abs(b-B) + abs(c-C) - 30 if a > 0 and b>0 and 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)


N, A, B, C = map(int, input().split())
l = [0] * N
for i in range(N):
    l[i] = int(input())
print(dfs(0, 0, 0, 0))

Submission status: Accepted

Solution code:
N,A,B,C=map(int, input().split())
L=[int(input()) for i in range(N)]
from itertools import product
ans=10**9
for p in product(range(4),repeat=N):
    l=[0]*3
    cost=0
    for i,t in enumerate(p):
        if t<3:
            if l[t]>0:
                cost+=10
            l[t]+=L[i]
    if min(l)>0:
        ans=min(ans,cost+abs(A-l[0])+abs(B-l[1])+abs(C-l[2]))
print(ans)

Submission status: Accepted

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

inf=3000

def mp(g,a,b,c):
    if g==N:
        return abs(A-a)+abs(B-b)+abs(C-c)-30 if min(a,b,c)>0 else inf

    l0=mp(g+1,a+l[g],b,c)+10
    l1=mp(g+1,a,b+l[g],c)+10
    l2=mp(g+1,a,b,c+l[g])+10
    l3=mp(g+1,a,b,c)
    
    return min(l0,l1,l2,l3)

print(mp(0,0,0,0))

Submission status: Accepted

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

bit = [0, 0, 0]
ans = 100000000
for bit[0] in range(1, 1<<N):
  for bit[1] in range(1, 1<<N):
    if bit[1] & bit[0]: continue
    for bit[2] in range(1, 1<<N):
      if bit[2] & bit[0]: continue
      if bit[2] & bit[1]: continue

      sums = [0, 0, 0]
      nums = [0, 0, 0]
      for i in range(N):
        for j in range(3):
          if bit[j] & (1<<i):
            sums[j] += l[i]
            nums[j] += 1

      score = 0
      for i in range(3):
        score += abs(sums[i] - As[i])
        score += (nums[i] - 1) * 10

      ans = min(ans, score)

print(ans)

Submission status: Accepted

Solution code:
n, *goals = list(map(int, input().split()))
bamboos = list(map(int, [input().rstrip() for _ in range(n)]))


def get_best(li, goal):
    bam_mp_lis = {0: (-10, tuple())}
    li.sort()
    for i, v in enumerate(li):
        tmp = bam_mp_lis.copy()
        for bam, (mp, lis) in tmp.items():
            b0 = bam + v
            m0 = mp + 10
            l0 = lis + (i,)
            if b0 not in bam_mp_lis or m0 < bam_mp_lis[b0][0]:
                bam_mp_lis[b0] = (m0, l0)
    del bam_mp_lis[0]
    best = 10 ** 9
    for bam, (mp, lis) in bam_mp_lis.items():
        cost = mp + abs(goal - bam)
        if cost < best:
            best = cost
            beslis = lis
    return best, beslis


ans = 10 ** 9
from itertools import permutations

for junjo in permutations(goals):
    tmp = 0
    bambooo = bamboos.copy()
    for ju in junjo:
        if len(bambooo) == 0:
            tmp = 10 ** 9
            break
        be, beli = get_best(bambooo, ju)
        beli = sorted(beli, reverse=True)
        for beli0 in beli:
            del bambooo[beli0]
        tmp += be
    if tmp < ans:
        ans = tmp

print(ans)

Submission status: Accepted

Solution code:
# ABC119C - Synthetic Kadomatsu
N, A, B, C = map(int, input().split())
L = [int(input()) for i in range(N)]

INF = float('inf')

def dfs(cursor=0, a=0, b=0, c=0):  # cursor:カーソル a,b,c:現在の竹の長さ
    if cursor == N:  # cursorが最後まで行ったら終了する。
        return abs(a - A) + abs(b - B) + abs(c - C) - 30 if min(a, b, c) > 0 else INF
    # abs(a - A) + abs(b - B) + abs(c - C) - 30でなぜ 30を減じているのかというと、
    # dfs(0,0,0,0)で始まる以上、最初に選ばれるa,b,cを決定する際にもコストが 10増加してしまうからである。
    # また、全探索を行う中で、a,b,cの初期値が 0,0,0である以上、a,b,cのどれかが 0のまま終了する場合が存在する。
    # この場合はa,b,cに竹が対応してないと言えるため、解にはならない。
    # そこで三項演算子を利用して、その場合についてコストをINFとしている

    # 以下は4^Nで展開される再起処理となる。
    # カーソルの当たっている竹に対して、(A or B or Cに合成する) or (合成しない)の場合に分ける
    no_compound_pattern = dfs(cursor + 1, a, b, c)
    compound_a_pattern = dfs(cursor + 1, a + L[cursor], b, c) + 10
    compound_b_pattern = dfs(cursor + 1, a, b + L[cursor], c) + 10
    compound_c_pattern = dfs(cursor + 1, a, b, c + L[cursor]) + 10

    # 結果的に以下の値が返るのはそれぞれのパターンのコストが決定されてからなので
    # 以下のコードは最終的なコストの最小値
    return min(
        no_compound_pattern, compound_a_pattern, compound_b_pattern, compound_c_pattern
    )


print(dfs())

Submission status: Accepted

Solution code:
n,a,b,c = map(int, input().split())
l = [int(input()) for _ in range(n)]
def dfs(k,p,q,r):
  if k == n:
    if min(p,q,r) > 0:
      return abs(p-a)+abs(q-b)+abs(r-c)-30
    else:
      return 10**7
  t1 = dfs(k+1,p,q,r)
  t2 = dfs(k+1,p+l[k],q,r)+10
  t3 = dfs(k+1,p,q+l[k],r)+10
  t4 = dfs(k+1,p,q,r+l[k])+10
  return min(t1,t2,t3,t4)

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

Submission status: Accepted

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

def toq(N):
    ret = ""
    t = 8
    while 4**t > 1:
        ret += str(N//(4**t))
        N -= (4**t)*(N//(4**t))
        t -= 1
    ret += str(N%4)
    return ret[::-1]

ans = float("inf")
for k in range(4**N):
    a, b, c = 0, 0, 0
    e = 0
    q = toq(k)
    for d in range(N):
        if q[d] == "1":
            a += L[d]
            e += 1
        if q[d] == "2":
            b += L[d]
            e += 1
        if q[d] == "3":
            c += L[d]
            e += 1
    if a*b*c > 0:
        ans = min(ans, abs(A-a)+abs(B-b)+abs(C-c) + 10*(e-3))
print(ans)

Submission status: Accepted

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

    list_mp = []
    for i in range(4 ** N):
        materials = [0] * N
        tmp = i
        zero_cnt = 0

        for j in range(N):
            materials[(N-1)-j] = tmp % 4
            tmp //= 4
            if materials[(N-1)-j] == 0:
                zero_cnt += 1

        if (1 in materials and 2 in materials and 3 in materials) == False:
            continue

        a_mat, b_mat, c_mat = 0, 0, 0
        cost_mp = 0
        for k in range(len(len_banboo)):
            if materials[k] == 0:
                continue
            elif materials[k] == 1:
                if a_mat > 0:
                    cost_mp += 10
                a_mat += len_banboo[k]
            elif materials[k] == 2:
                if b_mat > 0:
                    cost_mp += 10
                b_mat += len_banboo[k]
            elif materials[k] == 3:
                if c_mat > 0:
                    cost_mp += 10
                c_mat += len_banboo[k]

        cost_mp += (abs(A - a_mat) + abs(B - b_mat) + abs(C - c_mat))
        list_mp.append(cost_mp)

    print(min(list_mp))


if __name__ == "__main__":
    main()

Submission status: Accepted

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

#n本目の竹をどう扱うかを考える

def dfs(n,a,b,c):
  if n == N:
    if min(a,b,c) > 0:
      return abs(a-A)+abs(b-B)+abs(c-C) - 30
    else:
      return INF

  rate_add_Not = dfs(n+1,a,b,c)
  rate_add_A   = dfs(n+1,a+l[n],b,c) + 10
  rate_add_B   = dfs(n+1,a,b+l[n],c) + 10
  rate_add_C   = dfs(n+1,a,b,c+l[n]) + 10
  return min(rate_add_Not,rate_add_A,rate_add_B,rate_add_C)

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

Submission status: Accepted

Solution code:
n,A,B,C=map(int,input().split())
l=[]
for _ in range(n):
    l.append(int(input()))
ans=10**9+7
for i in range(4**n):
    k=i
    a=0
    b=0
    c=0
    cnt=[0,0,0]
    for j in range(n):
        if k%4==1:
            a+=l[j]
            cnt[0]+=1
        elif k%4==2:
            b+=l[j]
            cnt[1]+=1
        elif k%4==3:
            c+=l[j]
            cnt[2]+=1
        k=k//4
    if cnt[0]>0 and cnt[1]>0 and cnt[2]:
        mp=abs(A-a)+abs(B-b)+abs(C-c)+max(cnt[0]*10-10,0)+max(cnt[1]*10-10,0)+max(cnt[2]*10-10,0)
        if mp<ans:
            ans=mp
print(ans)

Submission status: Accepted

Solution code:
from itertools import product
N,*ABC=map(int,input().split())
L=list(int(input()) for _ in range(N))
mp=10**9
for i in product([0,1,2,3] ,repeat=N):
    if {0,1,2}<=set(i):
        m=[0,0,0]
        for x,y in zip(i,L):
            if x!=3:
                m[x]+=y
        t=sum(abs(p-q) for p,q in zip(ABC,m))
        r=sum(1 if k!=3 else 0 for k in i)-3
        mp=min(mp,t+r*10)
print(mp)

Submission status: Accepted

Solution code:
import sys
from collections import deque
input = sys.stdin.readline

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:
        if min(a, b, c) > 0:
            return abs(a-A) + abs(b-B) + abs(c-C) - 30 #最初の三回
        else:
            return INF #長さ0になった場合
    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: Accepted

Solution code:
import sys
from collections import deque
input = sys.stdin.readline

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:
        if min(a, b, c) > 0:
            return abs(a-A) + abs(b-B) + abs(c-C) - 30
        else:
            return INF #長さ0になった場合
    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: Accepted

Solution code:
N ,A, B, C = [int(i) for i in input().split()]
L = []
for _ in range(N):
    L.append(int(input()))
minMP = 100000
for Ai in range(2**N):
    bin_Ai = format(Ai, "08b")
    bin_Ai = list(bin_Ai)
    bin_Ai.reverse()
    bambooA = []
    bambooBC = []
    for ni in range(N):
        if bin_Ai[ni] == "1":
            bambooA.append(L[ni])
        else:
            bambooBC.append(L[ni])
    if bambooA == []:
        continue
    k = len(bambooBC)
    for Bi in range(2**k):
        bin_Bi = format(Bi,"08b")
        bin_Bi = list(bin_Bi)
        bin_Bi.reverse()
        bambooB = []
        bambooCD = []
        for ki in range(k):
            if bin_Bi[ki] == "1":
                bambooB.append(bambooBC[ki])
            else:
                bambooCD.append(bambooBC[ki])
        if bambooB == []:
            continue
        m = len(bambooCD)
        for Ci in range(2**m):
            bin_Ci = format(Ci, "08b")
            bin_Ci = list(bin_Ci)
            bin_Ci.reverse()
            bambooC = []
            for mi in range(m):
                if bin_Ci[mi] == "1":
                    bambooC.append(bambooCD[mi])
            if bambooC == []:
                continue
            MP = 10*(len(bambooA)-1)+abs(sum(bambooA)-A)+10*(len(bambooB)-1)+abs(sum(bambooB)-B)+10*(len(bambooC)-1)+abs(sum(bambooC)-C)
            if MP < minMP:
                minMP = MP
print(minMP)

Submission status: Accepted

Solution code:
N, A, B, C = map(int, input().split())
lLst = [0] * N
for i in range(N):
    lLst[i] = int(input())
maxM = 10000

def dfs(nCnt, atake, btake, ctake):
    if nCnt == N:
        if min(atake, btake, ctake) > 0:
            return abs(A - atake) + abs(B - btake) + abs(C - ctake) - 30
        else:
            return maxM

    ret0 = dfs(nCnt + 1, atake, btake, ctake)
    ret1 = dfs(nCnt + 1, atake + lLst[nCnt], btake, ctake) + 10
    ret2 = dfs(nCnt + 1, atake, btake + lLst[nCnt], ctake) + 10
    ret3 = dfs(nCnt + 1, atake, btake, ctake + lLst[nCnt]) + 10
    return min(maxM, ret0, ret1, ret2, ret3)

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

Submission status: Accepted

Solution code:
N, A, B, C, *l = map(int, open(0).read().split())

INF = 1e9

def dfs(cur, a, b, c):
    if cur == N:
        return abs(A - a) + abs(B - b) + abs(C - c) - 30 if a * b * c != 0 else INF
    
    v1 = dfs(cur + 1, a, b, c)
    v2 = dfs(cur + 1, a + l[cur], b, c) + 10
    v3 = dfs(cur + 1, a, b + l[cur], c) + 10
    v4 = dfs(cur + 1, a, b, c + l[cur]) + 10
    
    return min(v1, v2, v3, v4)

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

Submission status: Accepted

Solution code:
n,a,b,c=map(int,input().split())
l=sorted([int(input()) for i in range(n)])
def kad(w,x,y,z):
    if w==n:
        if min(x,y,z)>0:
            return abs(x-a)+abs(y-b)+abs(z-c)-30
        else:
            return 10e4
        
    c0=kad(w+1,x,y,z)
    c1=kad(w+1,x+l[w],y,z)+10
    c2=kad(w+1,x,y+l[w],z)+10
    c3=kad(w+1,x,y,z+l[w])+10
    return min(c0,c1,c2,c3)

print(kad(0,0,0,0))

Submission status: Accepted

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

def solve(w,x,y,z):
    if w==n:
        if min(x,y,z)>0:
            return abs(x-a)+abs(y-b)+abs(z-c)-30
        else:
            return 10e4
    can0=solve(w+1,x,y,z)
    can1=solve(w+1,x+l[w],y,z)+10
    can2=solve(w+1,x,y+l[w],z)+10
    can3=solve(w+1,x,y,z+l[w])+10
    return min(can0,can1,can2,can3)

print(solve(0,0,0,0))

Submission status: Accepted

Solution code:
N, *P = map(int, input().split())
L = [int(input()) for _ in range(N)]

mp = sum(P) - 30

minmp = 1e10
Pcnt = [0] * 3
Ks = [-1] * N

while True:

    i = N-1
    while i >= 0:
        pre = cur = Ks[i]
        cur += 1
        if cur >= 3:
            cur = -1
        Ks[i] = cur

        for d, x in ((-1, pre), (1, cur)):

            if x >= 0:
                Pcnt[x] += d
                mp -= abs(P[x])
                P[x] -= d * L[i]
                mp += abs(P[x]) + d * 10

        if cur >= 0:
            break
        i -= 1
    else:
        break

    if min(Pcnt) > 0:
        minmp = min(minmp, mp)

print(minmp)

Submission status: Accepted

Solution code:
from itertools import product
n, a, b, c = map(int, input().split())
parts = [int(input()) for _ in range(n)]

usage = [['a', 'b', 'c', '_'] for _ in range(n)]

ans = 1e9
for cmb in product(*usage):
    if not ('a' in cmb and 'b' in cmb and 'c' in cmb):
        continue
    cost_a, cost_b, cost_c = 0, 0, 0
    len_a, len_b, len_c = 0, 0, 0
    for i in range(n):
        if cmb[i] == 'a':
            cost_a += 10 if len_a > 0 else 0
            len_a += parts[i]
        elif cmb[i] == 'b':
            cost_b += 10 if len_b > 0 else 0
            len_b += parts[i]
        elif cmb[i] == 'c':
            cost_c += 10 if len_c > 0 else 0
            len_c += parts[i]
    cost_a += abs(a - len_a)
    cost_b += abs(b - len_b)
    cost_c += abs(c - len_c)
    ans = min(ans, sum((cost_a, cost_b, cost_c)))

print(ans)

Submission status: Accepted

Solution code:
N, A, B, C = map(int, raw_input().split())
target = [A, B, C]
memo = [[[0, 0, 0], 0]]
for i in range(0, N):
	t = int(raw_input())
	m = len(memo)
	for j in range(0, m):
		now = memo[j]
		trees = now[0]
		mp = now[1]
		for k in range(0, 3):
			addmp = 0
			if trees[k] != 0: addmp = 10
			newmemo = [[trees[0], trees[1], trees[2]], mp+addmp]
			newmemo[0][k] += t
			memo.append(newmemo)
min = 1000000007
for i in range(0, len(memo)):
	if 0 not in memo[i][0]:
		sum = memo[i][1]
		for j in range(0, 3):
			sum += abs(target[j]-memo[i][0][j])
		if sum < min: min = sum
print min

Submission status: Accepted

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

K = 4

minmp = 1e8
for x in range(pow(4, N)):
    v = x
    c = [0] * K
    s = [0] * K
    i = 0
    while v > 0:
        m = v % K
        c[m] += 1
        s[m] += L[i]
        i += 1
        v //= K

    if c[1] > 0 and c[2] > 0 and c[3] > 0:
        minmp = min(minmp, (c[1] + c[2] + c[3] - 3) * 10 + abs(s[1] - A) + abs(s[2] - B) + abs(s[3] - C))

print(minmp)

Submission status: Accepted

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

INF = 10 **9
def optimal_value(a_step, a_cumA, a_cumB, a_cumC):
    if a_step == N:
        if min(a_cumA, a_cumB, a_cumC) == 0:
            return INF
        return abs(A - a_cumA) + abs(B - a_cumB) + abs(C-a_cumC)
     
    dA = optimal_value(a_step + 1, a_cumA + Ls[a_step], a_cumB, a_cumC)
    if a_cumA > 0:
        dA += 10
    
    dB = optimal_value(a_step + 1, a_cumA, a_cumB + Ls[a_step], a_cumC)
    if a_cumB > 0:
        dB += 10
   
    dC = optimal_value(a_step + 1, a_cumA, a_cumB, a_cumC + Ls[a_step])
    if a_cumC > 0:
        dC += 10
    
    dE = optimal_value(a_step + 1, a_cumA, a_cumB, a_cumC)
    
    return min(dA, dB, dC, dE)

print(optimal_value(0,0,0,0))

Submission status: Accepted

Solution code:
n, A, B, C = map(int, input().split())

l = [0 for _ in range(n)]

for i in range(n):
    l[i] = int(input())


def dfs(count, a, b, c):
    if count == n:
        if a == 0 or b == 0 or c == 0:
            return 10000000
        return abs(A - a) + abs(B - b) + abs(C - c) - 30
    else:
        return min(dfs(count + 1, a, b, c),
                   dfs(count + 1, a + l[count], b, c) + 10,
                   dfs(count + 1, a, b + l[count], c) + 10,
                   dfs(count + 1, a, b, c + l[count]) + 10,
                   )


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

Submission status: Accepted

Solution code:
from collections import defaultdict, Counter
from itertools import product, groupby, count, permutations, combinations
from math import pi, sqrt
from collections import deque
from bisect import bisect, bisect_left, bisect_right
from string import ascii_lowercase
from functools import lru_cache
import sys
sys.setrecursionlimit(100000)
INF = float("inf")
YES, Yes, yes, NO, No, no = "YES", "Yes", "yes", "NO", "No", "no"
dy4, dx4 = [0, 1, 0, -1], [1, 0, -1, 0]
dy8, dx8 = [0, -1, 0, 1, 1, -1, -1, 1], [1, 0, -1, 0, 1, 1, -1, -1]


def inside(y, x, H, W):
    return 0 <= y < H and 0 <= x < W


def ceil(a, b):
    return (a + b - 1) // b


# aとbの最大公約数
def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)


# aとbの最小公倍数
def lcm(a, b):
    g = gcd(a, b)
    return a / g * b


def dfs(i, a, b, c, N, A, B, C, L):
    if i >= N:
        if a == 0 or b == 0 or c == 0:
            return INF
        return abs(A - a) + abs(B - b) + abs(C - c)

    ret = INF
    ret = min(ret, dfs(i + 1, a + L[i], b, c, N, A, B, C, L) + (10 if a > 0 else 0))
    ret = min(ret, dfs(i + 1, a, b + L[i], c, N, A, B, C, L) + (10 if b > 0 else 0))
    ret = min(ret, dfs(i + 1, a, b, c + L[i], N, A, B, C, L) + (10 if c > 0 else 0))
    ret = min(ret, dfs(i + 1, a, b, c, N, A, B, C, L))

    return ret


def main():
    N, A, B, C = map(int, input().split())
    L = [int(input()) for _ in range(N)]

    print(dfs(0, 0, 0, 0, N, A, B, C, L))


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import itertools


n, a, b, c = map(int, input().split())
ls = [int(input()) for _ in range(n)]

mint = 1e9
for comb in itertools.product((0, 1, 2, 3), repeat=len(ls)):
    fa = [ls[i] for i in range(len(ls)) if comb[i] == 1]
    fb = [ls[i] for i in range(len(ls)) if comb[i] == 2]
    fc = [ls[i] for i in range(len(ls)) if comb[i] == 3]

    if len(fa) == 0 or len(fb) == 0 or len(fc) == 0:
        continue

    mint = min(mint, (len(fa) + len(fb) + len(fc) - 3) * 10 +
               abs(sum(fa) - a) + abs(sum(fb) - b) + abs(sum(fc) - c))


print(mint)