TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: 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?

Submission attempts (in order):
#coding: utf-8
import numpy as np

N, A, B, C = map(int, input().split())
bumboo = np.array([int(input()) for _ in range(N)])
magic = 0
for i in [A, B, C]:
    bumboo_diff = bumboo - i
    bumboo_diff = np.where(bumboo_diff >= 0)
    if np.sum(bumboo_diff) >= 2:
        magic += 10
    else:
        magic += np.min(bumboo_diff)
print(magic)

#coding: utf-8
import numpy as np

N, A, B, C = map(int, input().split())
bumboo = np.array([int(input()) for _ in range(N)])
magic = 0
for i in [A, B, C]:
    bumboo_diff = bumboo - i
    bumboo_diff = np.where(bumboo_diff >= 0)
    if np.sum(bumboo_diff) == 2:
        magic += 10
    else:
        magic += np.min(bumboo_diff)
print(magic)

# coding: utf-8
import sys
sys.setrecursionlimit(10**9)
def Base_10_to_n(X, n):
    X_dumy = X
    out = ''
    while X_dumy>0:
        out = str(X_dumy%n)+out
        X_dumy = int(X_dumy/n)
    return out
N, A, B, C = map(int, input().split())
L = []
for i in range(N):
    L.append(int(input()))
# L.sort()
def dfs(cnt, a, b, c):
    if cnt == N:
        return abs(A - a) + abs(B - b) + abs(C - c) - 30 if min(a,b,c) > 0 else 10**9
    r0 = dfs(cnt+1, a, b, c)
    r1 = dfs(cnt+1, a+L[cnt], b, c) + 10
    r2 = dfs(cnt+1, a, b+L[cnt], c) + 10
    r3 = dfs(cnt+1, a, b, c+L[cnt]) + 10
    return min(r0, r1, r2, r3)
    
print(dfs(0,0,0,0))

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
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)
    ret2 = dfs(cur + 1, a, b + l[cur], c)
    ret3 = dfs(cur + 1, a, b, c + l[cur])
    return min(ret0, ret1, ret2, ret3)

print(dfs(ret0, ret1, ret2, ret3))


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))


Verdicts: Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
from itertools import product, permutations

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

ans = 10 ** 9
for groups in product(range(4), repeat=N):
    items = []
    group_a = []
    group_b = []
    group_c = []

    for group, length in zip(groups, L):
        if group == 0:
            items.append((length, 0))
        elif group == 1:
            group_a.append(length)
        elif group == 2:
            group_b.append(length)
        elif group == 3:
            group_c.append(length)

    if len(items) < 3:
        continue

    if group_a:
        items.append((sum(group_a), (len(group_a) - 1) * 10))
    if group_b:
        items.append((sum(group_b), (len(group_b) - 1) * 10))
    if group_c:
        items.append((sum(group_c), (len(group_c) - 1) * 10))

    for one, two, three in permutations((A, B, C), r=3):
        cost = 0
        _items = list(items)

        _items = sorted(_items, key=lambda x: (abs(x[0] - one), x[1]))
        cost += abs(_items[0][0] - one) + _items[0][1]
        _items.pop(0)

        _items = sorted(_items, key=lambda x: (abs(x[0] - two), x[1]))
        cost += abs(_items[0][0] - two) + _items[0][1]
        _items.pop(0)

        _items = sorted(_items, key=lambda x: (abs(x[0] - three), x[1]))
        cost += abs(_items[0][0] - three) + _items[0][1]

        ans = min(ans, cost)

print(ans)


from itertools import product

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

ans = 10 ** 9
for groups in product(range(4), repeat=N):
    length_a = 0
    length_b = 0
    length_c = 0
    cost_a = -10
    cost_b = -10
    cost_c = -10

    for group, length in zip(groups, L):
        if group == 1:
            length_a += length
            cost_a += 10
        elif group == 2:
            length_b += length
            cost_b += 10
        elif group == 3:
            length_c += length
            cost_c += 10

    if 0 in {length_a, length_b, length_c}:
        continue

    ans = min(ans, sum([
        abs(A - length_a),
        abs(B - length_b),
        abs(C - length_c),
        cost_a,
        cost_b,
        cost_c,
    ]))

print(ans)


Verdicts: Wrong Answer

Accepted
Problem: 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?

Submission attempts (in order):
@n, @a, @b, @c = gets.chomp.split(" ").map(&:to_i)

@l = []
@n.times do
  @l << gets.chomp.to_i
end
INF = 10**9

def dfs(cur, a, b, c)
  if cur == @n
    if [a, b, c].min > 0
      return (a - @a).abs + (b - @b).abs + (c - @c).abs - 30
    else
      return INF
    end
  end
  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 [ret0, ret1, ret2, ret3].min
end

puts dfs(0, 0, 0, 0)

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))


Verdicts: Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
N, A, B, C = map(int, input().split())
l = list(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))

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

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))

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
N, A, B, C = map(int, input().split())
L = [int(input()) for i in range(N)]
print(L)


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) else 1000000000

    path_null = dfs(cur+1, a, b, c)
    path_a = dfs(cur+1, a+L[cur], b, c) + 10
    path_b = dfs(cur+1, a, b+L[cur], c) + 10
    path_c = dfs(cur+1, a, b, c+L[cur]) + 10

    return min(path_null, path_a, path_b, path_c)

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


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) else 1000000000

    path_null = dfs(cur+1, a, b, c)
    path_a = dfs(cur+1, a+L[cur], b, c) + 10
    path_b = dfs(cur+1, a, b+L[cur], c) + 10
    path_c = dfs(cur+1, a, b, c+L[cur]) + 10

    return min(path_null, path_a, path_b, path_c)

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


Verdicts: Wrong Answer

Accepted
Problem: 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?

Submission attempts (in order):
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))

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))

N,A,B,C = map(int,input().split())
l = [int(input()) for i in range(N)]
INF = 100000000
def rec(i,a,b,c):
    if(i==N):
        return(abs(a-A)+abs(b-B)+abs(c-C)-30 if min(a,b,c)>0 else INF)
    ret0 = rec(i+1,a,b,c)
    ret1 = rec(i+1,a+l[i],b,c)+10
    ret2 = rec(i+1,a,b+l[i],c)+10
    ret3 = rec(i+1,a,b,c+l[i])+10
    return min(ret0,ret1,ret2,ret3)
print(rec(0,0,0,0))


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: 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?

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

min_ans = 10 ** 10
for i in range(4 ** n):
	mm = [[], [], [],[]]
	tmp = i
	for i2 in range(n):
		g = tmp // (4 ** (n - i2 - 1))
		tmp = tmp % (4 ** (n - i2 - 1))
		mm[g%4].append(ls[i2])
	
	if len(mm[0]) * len(mm[1]) * len(mm[2]) == 0:
		continue

	print(mm)
	ans = abs(sum(mm[0]) - a) + abs(sum(mm[1]) - b) + abs(sum(mm[2]) - c) + (len(mm[0]) - 1) * 10 + (len(mm[1]) - 1) * 10 + (len(mm[2]) - 1) * 10
	min_ans = min(ans, min_ans)

print(min_ans)

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

min_ans = 10 ** 10
for i in range(4 ** n):
	mm = [[], [], [],[]]
	tmp = i
	for i2 in range(n):
		g = tmp // (4 ** (n - i2 - 1))
		tmp = tmp % (4 ** (n - i2 - 1))
		mm[g%4].append(ls[i2])
	
	if len(mm[0]) * len(mm[1]) * len(mm[2]) == 0:
		continue

	#print(mm)
	ans = abs(sum(mm[0]) - a) + abs(sum(mm[1]) - b) + abs(sum(mm[2]) - c) + (len(mm[0]) - 1) * 10 + (len(mm[1]) - 1) * 10 + (len(mm[2]) - 1) * 10
	min_ans = min(ans, min_ans)

print(min_ans)

Verdicts: Wrong Answer

Accepted
Problem: 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?

Submission attempts (in order):
import numpy as np

n, a, b, c = map(int, input().split(' '))
ls = []
for _ in range(n):
    ls.append(int(input()))
n = 0


def solve(obj):
    c = len(ls)
    dp = np.zeros((obj + 1, c + 1), dtype=int)
    dp[1:, 0] = 10000
    for j in range(1, c + 1):
        dp[0, j] = min(ls[:j])
    for i in range(1, obj + 1):
        for j in range(1, c + 1):
            dp[i, j] = min(dp[i - 1, j] + 1, dp[i, j - 1], abs(ls[j - 1] - i))
            if i >= ls[j - 1]:
                dp[i, j] = min(dp[i, j], dp[i - ls[j - 1], j - 1] + 10)
  
    return dp[obj, c]


for obj in [a, b, c]:
    if obj in ls:
        continue
    else:
        n += solve(obj)
print(n)

import numpy as np

n, a, b, c = map(int, input().split(' '))
ls = []
for _ in range(n):
    ls.append(int(input()))
n = 0
def solve(obj):
  c = len(ls)
  dp = np.zeros((obj+1, c+1), dtype=int)
  dp[1:, 0] = np.inf
  for j in range(1, c+1):
    dp[0, j] = min(ls[:j])
  for i in range(1, obj+1):
    for j in range(1, c+1):
      dp[i, j] = min(dp[i-1, j] + 1, dp[i, j-1], dp[i-ls[j-1], j-1] + 10, abs(ls[j-1]) - i)
  return dp[obj, c]

for obj in [a, b, c]:
  if obj in ls:
    continue
  else:
    n += solve(obj)
print(n)


import numpy as np

N, A, B, C = map(int, input().split(' '))
ls = []
for _ in range(N):
    ls.append(int(input()))
n = 0
INF = 10 ** 9

def dfs(n, a, b, c):
    if n == N:
        return INF if min(a, b, c) == 0 else abs(a-A) + abs(b-B) + abs(c-C) - 30
    res0 = dfs(n+1, a, b, c)
    res1 = dfs(n+1, a+ls[n], b, c) + 10
    res2 = dfs(n+1, a, b+ls[n], c) + 10
    res3 = dfs(n+1, a, b, c+ls[n]) + 10
    return min(res0, res1, res2, res3)

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

Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):

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))


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))



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))


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
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)


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(input())
    ret = solve(N, A, B, C, inputs(N))
    print(ret)


Verdicts: Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
from itertools import product 
n,a,b,c = map(int,input().split())
l = [int(input()) for _ in range(n)]
tar = [a,b,c]
ans = float("inf")
for p in product(range(4),repeat=8):
    mp = 0
    length = [0]*3
    for i,j in enumerate(p):
        if j == 3: continue
        if length[j] > 0: mp += 10
        length[j] += l[i]
    for i in range(3):
        if length[i] == 0: break
        mp += abs(length[i]-tar[i])
    else:
        ans = min(ans, mp)
print(ans)

from itertools import product 
n,a,b,c = map(int,input().split())
l = [int(input()) for _ in range(n)]
tar = [a,b,c]
ans = float("inf")
for p in product(range(4),repeat=n):
    mp = 0
    length = [0]*3
    for i,j in enumerate(p):
        if j == 3: continue
        if length[j] > 0: mp += 10
        length[j] += l[i]
    for j in range(3):
        if length[j] == 0: break
        mp += abs(length[j]-tar[j])
    else:
        ans = min(ans, mp)
print(ans)

Verdicts: Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
def dfs(a, b, c, k)
  
  if k == N
    nil if [a, b, c].min == 0
    return (A-a).abs + (B-b).abs + (C-c).abs - 10 * 3
  end
  
  x = dfs(a+L[k], b, c, k+1) + 10
  y = dfs(a, b+L[k], c, k+1) + 10
  z = dfs(a, b, c+L[k], k+1) + 10
  w = dfs(a, b, c, k+1)
  
  [x, y, z, w].compact.min
end

N, A, B, C = gets.split.map(&:to_i)

L = Array.new(N)
N.times{ |i| L[i] = gets.to_i }

p dfs(0, 0, 0, 0)



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))

Verdicts: Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
N,A,B,C=[int(i) for i in input().split()]
l=[int(input()) for _ in range(n)]
INF=10**18

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

N,A,B,C=[int(i) for i in input().split()]
l=[int(input()) for _ in range(N)]
INF=10**18

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))


Verdicts: Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
N,A,B,C = map(int,input().split())

l = [int(input()) for _ in range(N)]

ans = 2 ** 20 -1
for i in range(N):
    for j in range(N):
        if i == j: continue
        for k in range(N):
            if i == k or j == k : continue
        
            x,y,z = l[i],l[j],l[k]
            cnt = 0
            
            for t in range(N):
                if t == i or t == j or t == k : continue

                if l[t] > 10:
                    cnt += l[t]
                    x += l[t]
                if l[t] > 10:
                    cnt += l[t]
                    y += l[t]
                if l[t] > 10:
                    cnt += l[t]
                    z += l[t]
            cnt += abs(x-A) 
            cnt += abs(y-B) 
            cnt += abs(z-C) 
            ans = min(ans,cnt)
                
                
print(ans)
        

N,A,B,C = map(int,input().split())
l = [int(input()) for _ in range(N)]
lnum = [0] * N
Alist = [A,B,C]
ans = 2**30+1
for i in range(4**N):
    cnt = -30
    tmp = i
    for j in range(N):
        lnum[j] = tmp%4
        tmp //= 4
    
    if not(1 in lnum and 2 in lnum and 3 in lnum):
        continue
    x = [0] * 3
    for j in range(N):
        if lnum[j] != 0:
            x[lnum[j]-1] += l[j]
            cnt += 10
    for j in range(3):
        cnt += abs(Alist[j]-x[j])
    
    ans = min(ans,cnt)
    
print(ans)
    
        

Verdicts: Wrong Answer

Accepted
Problem: 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?

Submission attempts (in order):
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
    print(ret0)
    print(ret1)
    print(ret2)
    print(ret3)
    return min(ret0, ret1, ret2, ret3)

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

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))

Verdicts: Wrong Answer

Accepted
Problem: 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?

Submission attempts (in order):
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) 
    else:
      return inf
  else:  
  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))

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))


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
  else:  
    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))


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
import bisect

ans = 10**20

N, K = list(map(int, input().split()))
x = [-10**20] + list(map(int, input().split())) + [10**20]

zero = bisect.bisect_left(x, 0)

left = x[:zero] + [0] 
right = [0] + x[zero:]
left.reverse()
l = len(left)
r = len(right)

idxs = [[K-i, i] for i in range(K+1) if K-i < l and i < r]

for l, r in idxs:
    n = abs(left[l]) + right[r] + min(abs(left[l]), right[r])
    ans = min(ans, n)
print(ans)

def search(idx, a, b, c):
    if idx == N:
        if min(a, b, c) > 0:
            return abs(A-a) + abs(B-b) + abs(C-c) - 30
        else:
            return 10**20
            
    n0 = search(idx+1, a, b, c)
    n1 = search(idx+1, a+l[idx], b, c) + 10
    n2 = search(idx+1, a, b+l[idx], c) + 10
    n3 = search(idx+1, a, b, c+l[idx]) + 10
    
    return min(n0, n1, n2, n3)
    
N, A, B, C = list(map(int, input().split()))
l = [int(input()) for i in range(N)]

print(search(0, 0, 0, 0))

Verdicts: Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
N, A, B, C = map(int, input().split())
l = [int(input()) for i in range(N)]
def calc(take, a, b, c):
    if take == N:
        return abs(a-A)+abs(b-B)+abs(c-C) 
    ret0 = calc(take+1, a, b, c)
    ret1 = calc(take+1, a+l[take], b, c) + 10
    ret2 = calc(take+1, a, b+l[take], c) + 10
    ret3 = calc(take+1, a, b, c+l[take]) + 10
    return min(ret0, ret1, ret2, ret3)
    
print(calc(0,0,0,0))

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

Verdicts: Wrong Answer

Accepted
Problem: 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?

Submission attempts (in order):
8 1000 800 100
300
333
400
444
500
555
600
666

N, A, B, C = map(int, input().split())
l = []
for i in range(N):
    l.append(int(input()))
ans = 100000000


def rec(a, b, c, d):
    if d == N:
        if a[1] > 0 and b[1] > 0 and c[1] > 0:
            global ans
            ans = min(
                ans,
                abs(
                    a[0] - A) + abs(b[0] - B) + abs(c[0] - C)
                + 10 * (a[1] + b[1] + c[1] - 3)
            )
        return
    rec((a[0] + l[d], a[1] + 1), b, c, d + 1)
    rec(a, (b[0] + l[d], b[1] + 1), c, d + 1)
    rec(a, b, (c[0] + l[d], c[1] + 1), d + 1)
    rec(a, b, c, d + 1)


rec((0, 0), (0, 0), (0, 0), 0)
print(ans)


Verdicts: Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):
import sys
sys.setrecursionlimit("10**7")

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))


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))

import sys
sys.setrecursionlimit("10**7")

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))


import sys
sys.setrecursionlimit(10**7)

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))

Verdicts: Runtime Error

Runtime Error

Runtime Error

Accepted
Problem: 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?

Submission attempts (in order):


import numpy as np
from copy import deepcopy

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

L = np.array(L)
max_merge = N-3
#min_MP = float('Inf')



def bamboo2MP_nomerge(bamboo, A, B, C):
  arr = deepcopy(bamboo)
  n_bamboo = len(arr)
  bamboo_A = np.argmin(np.abs(arr-A))
  MP_A = np.abs(arr[bamboo_A]-A)
  np.delete(arr, MP_A)
  bamboo_B = np.argmin(np.abs(arr-B))
  MP_B = np.abs(arr[bamboo_B]-B)
  np.delete(arr, MP_B)
  bamboo_C = np.argmin(np.abs(arr-C))
  MP_C = np.abs(arr[bamboo_C]-C)
  #np.delete(arr, MP_C)
  return int(MP_A + MP_B + MP_C)
  


min_MP = bamboo2MP_nomerge(L, A, B, C)
  


  
print(min_MP)
  

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): 
    bamboo = [0]*4
    for i in range(len(state)): 
      if state[i] < 3:
        bamboo[state[i]] += L[i]
    if 0 in bamboo[0:3]: 
      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)
  

Verdicts: Wrong Answer

Accepted
Problem: 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?

Submission attempts (in order):
N,A,B,C = map(int, input().split())

bms = [int(input()) for _ in Range(N)]

def dfs(cur,a,b,c):
    if cur == N:
        s = abs(A-a)+abs(B-b)+abs(C-c)-30 if min(a,b,c) > 0 else 10**9
        return s
    s_0 = dfs(cur+1,a,b,c)
    s_a = dfs(cur+1,a+bms[cur],b,c) + 10
    s_b = dfs(cur+1,a,b+bms[cur],c) + 10
    s_c = dfs(cur+1,a,b,c+bms[cur]) + 10
    return min(s_0,s_a,s_b,s_c)

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

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

targets = [A,B,C]
bamboos.sort(reverse=True)
targets.sort(reverse=True)

for target in targets:
    if target in bamboos:
        targets.remove(target)
        bamboos.remove(target)

if not len(targets):
  print(0)

import numpy as np
print(np.nan)

N,A,B,C = map(int, input().split())
 
bms = [int(input()) for _ in Range(N)]
 
def dfs(cur,a,b,c):
    if cur == N:
        s = abs(A-a)+abs(B-b)+abs(C-c)-30 if min(a,b,c) > 0 else 10**9
        return s
    s_0 = dfs(cur+1,a,b,c)
    s_a = dfs(cur+1,a+bms[cur],b,c) + 10
    s_b = dfs(cur+1,a,b+bms[cur],c) + 10
    s_c = dfs(cur+1,a,b,c+bms[cur]) + 10
    return min(s_0,s_a,s_b,s_c)

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

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

bms = [int(input()) for _ in range(N)]

def dfs(cur,a,b,c):
    if cur == N:
        s = abs(A-a)+abs(B-b)+abs(C-c)-30 if min(a,b,c) > 0 else 10**9
        return s
    s_0 = dfs(cur+1,a,b,c)
    s_a = dfs(cur+1,a+bms[cur],b,c) + 10
    s_b = dfs(cur+1,a,b+bms[cur],c) + 10
    s_c = dfs(cur+1,a,b,c+bms[cur]) + 10
    return min(s_0,s_a,s_b,s_c)

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

Verdicts: Runtime Error

Wrong Answer

Wrong Answer

Runtime Error

Accepted