Problem:


Score : 500 points

Problem Statement

Let N be a positive integer.

There is a numerical sequence of length 3N, a = (a_1, a_2, ..., a_{3N}).
Snuke is constructing a new sequence of length 2N, a', by removing exactly N elements from a without changing the order of the remaining elements.
Here, the score of a' is defined as follows: (the sum of the elements in the first half of a') - (the sum of the elements in the second half of a').

Find the maximum possible score of a'.

Constraints

1 ≤ N ≤ 10^5

a_i is an integer.

1 ≤ a_i ≤ 10^9

Partial Score

In the test set worth 300 points, N ≤ 1000.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{3N}

Output

Print the maximum possible score of a'.

Sample Input 1

2
3 1 4 1 5 9

Sample Output 1

1

When a_2 and a_6 are removed, a' will be (3, 4, 1, 5), which has a score of (3 + 4) - (1 + 5) = 1.

Sample Input 2

1
1 2 3

Sample Output 2

-1

For example, when a_1 are removed, a' will be (2, 3), which has a score of 2 - 3 = -1.

Sample Input 3

3
8 2 2 7 4 6 5 3 8

Sample Output 3

5

For example, when a_2, a_3 and a_9 are removed, a' will be (8, 7, 4, 6, 5, 3), which has a score of (8 + 7 + 4) - (6 + 5 + 3) = 5.



Sample input:
2
3 1 4 1 5 9


Sample output:
1

Submission status: Wrong Answer

Solution code:
import heapq
#N = int(input())
#A = list(map(int,input().split()))
N = 1000
S = "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 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1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000"
A = list(map(int,S.split()))
# 左側
l = A[0:N]
lsum = [sum(l)]
heapq.heapify(l)
for i in range(N,2*N):
    x = heapq.heappushpop(l,A[i])
    lsum.append(lsum[-1] + A[i] - x)

# 右側
r = list(map(lambda x: -x, A[-N:N*3]))
rsum = [sum(r)]
heapq.heapify(r)
for i in range(N,2*N):
    x = heapq.heappushpop(r, -A[3*N-1-i])
    rsum.append(rsum[-1] - A[3*N-1-i] - x)

ans = -9999999999999999
for i in range(len(rsum)):
    ans = max(ans, lsum[i]+rsum[len(rsum)-1-i])

print(ans)

Submission status: Wrong Answer

Solution code:
import heapq
N = int(input())
A = list(map(int,input().split()))

# 左側
l = A[0:N]
lsum = [sum(l)]
heapq.heapify(l)
for i in range(N,2*N):
    x = heapq.heappushpop(l,A[i])
    lsum.append(lsum[-1] + A[i] - x)

# 右側
r = list(map(lambda x: -x, A[-N:N*3]))
rsum = [sum(r)]
heapq.heapify(r)
for i in range(N,2*N):
    x = heapq.heappushpop(r, -A[3*N-1-i])
    rsum.append(rsum[-1] - A[3*N-1-i] - x)

ans = -99999999999
for i in range(len(rsum)):
    ans = max(ans, lsum[i]+rsum[len(rsum)-1-i])

print(ans)

Submission status: Wrong Answer

Solution code:
import heapq

N = int(input())
a_list = list(map(int, input().split()))

max_score = - float("inf")

red_list = a_list[:N]
blue_list = [-a for a in a_list[2 * N:]]
heapq.heapify(red_list)
heapq.heapify(blue_list)
red_sum = [sum(red_list)]
blue_sum = [sum(blue_list)]

for i in range(N, 2 * N):
    heapq.heappush(red_list, a_list[i])
    item = heapq.heappop(red_list)
    red_sum.append(max(red_sum[-1], red_sum[-1] + a_list[i] - item))

for i in range(2 * N - 1, N - 1, -1):
    heapq.heappush(blue_list, -a_list[i])
    item = heapq.heappop(blue_list)
    blue_sum.append(max(blue_sum[-1], blue_sum[-1] - a_list[i] - item))
for i in range(N):
    score = red_sum[i] + blue_sum[-(i + 1)]
    if max_score < score:
        max_score = score
print(max_score)

Submission status: Wrong Answer

Solution code:
n = int(input())
a = input().split()
a = [int(m) for m in a]
t1 = []
t2 = []
t3 = []
for i in range(3 * n):
    if i < (3 * n) // 3:
        t1.append(a[i])
    elif (3 * n) // 3 <= i < (3 * n) // 3 * 2:
        t2.append(a[i])
    else:
        t3.append(a[i])
t1.sort()
t3.sort(reverse=True)
sumt1 = sum(t1)
sumt3 = sum(t3)
y1 = []
y3 = []
w3 = []
for i in range(n + 1):
    if i == 0:
        sumt1 = sumt1
        y1.append(sumt1)
        sumt3 = sumt3
        y3.append(sumt3)
        w3.append([sumt3, n + 1])
    else:
        sumt1 = sumt1 + t2[i - 1] - t1[i - 1]
        y1.append(sumt1)
        sumt3 = sumt3 + t2[n - i] - t3[i - 1]
        y3.append(sumt3)
        w3.append([sumt3, n - i + 1])

w3.sort()

v3 = []
for i in range(n + 1):#index取得
    if len(v3) != 0:
        if v3[-1] < w3[i][1]:
            v3.append(w3[i][1])
    else:
        v3.append(w3[i][1])

ry3 = []
for i in range(len(y3)):
    ry3.append(y3[-i - 1])

j = 0
ansli = []
for i in range(n):
    if i == v3[j]:
        j += 1
    ansli.append(y1[i] - ry3[v3[j] - 1])

print(max(ansli))

Submission status: Wrong Answer

Solution code:
import heapq
N = int(input())
A = [int(_) for _ in input().split()]

X = A[0:N]
B = A[N:2*N]
Y = [-_ for _ in A[2*N:3*N]]

heapq.heapify(X)
heapq.heapify(Y)

x = [sum(X)]
y = [sum(Y)]

for i in range(N):
    x += [B[i] - heapq.heappushpop(X, B[i])]
    y += [-B[-1-i] - heapq.heappushpop(Y, -B[-1-i])]

for i in range(N):
    x[i+1] = x[i]+x[i+1]
    y[i+1] = y[i]+y[i+1]

print(max([x[i] + y[-1-i] for _ in range(N+1)]))

Submission status: Wrong Answer

Solution code:
import heapq
N=int(input())
A=[int(i) for i in input().split()]
score=[0]*(N+1)
former_score=[0]*(N+1)
latter_score=[0]*(N+1)
former=A[0:N]
latter=list(map(lambda x:-x,A[2*N:3*N]))
heapq.heapify(former)
heapq.heapify(latter)
former_score[0]=sum(former)
latter_score[N]=sum(latter)
for i in range(1,N+1):
    heapq.heappush(former,A[N+i-1])
    former_score[i]=former_score[i-1]+A[N+i-1]-heapq.heappop(former)
    heapq.heappush(latter,A[2*N-i])
    latter_score[N-i]=latter_score[N-i+1]-A[2*N-i]-heapq.heappop(latter)
ans=-10**20
for i in range(N+1):
  ans=max(ans,former_score[i]+latter_score[i])
print(ans)

Submission status: Wrong Answer

Solution code:
import heapq
N=int(input())
A=[int(i) for i in input().split()]
score=[0]*(N+1)
former_score=[0]*(N+1)
latter_score=[0]*(N+1)
former=A[0:N]
latter=list(map(lambda x:-x,A[2*N:3*N]))
heapq.heapify(former)
heapq.heapify(latter)
former_score[0]=sum(A[0:N])
latter_score[N]=-sum(A[2*N:3*N])
for i in range(1,N+1):
  if former!=[]:
    if former[0]<A[N+i-1]:
      former_score[i]=former_score[i-1]+A[N+i-1]-heapq.heappop(former)
      heapq.heappush(former,A[N+i-1])
    else:
      former_score[i]=former_score[i-1]
  if latter!=[]:
    if latter[0]<(-1)*A[2*N-i]:
      latter_score[N-i]=latter_score[N-i+1]-A[2*N-i]-heapq.heappop(latter)
      heapq.heappush(latter,A[2*N-i])
    else:
      latter_score[N-i]=latter_score[N-i+1]
for i in range(N+1):
  score[i]=former_score[i]+latter_score[i]
print(max(score))

Submission status: Wrong Answer

Solution code:
import heapq
N=int(input())
A=[int(i) for i in input().split()]
score=[0]*(N+1)
former_score=[0]*(N+1)
latter_score=[0]*(N+1)
former=A[0:N]
latter=list(map(lambda x:-x,A[2*N:3*N]))
heapq.heapify(former)
heapq.heapify(latter)
former_score[0]=sum(A[0:N])
latter_score[N]=-sum(A[2*N:3*N])
for i in range(1,N+1):
  if former!=[]:
    if former[0]<A[N+i-1]:
      former_score[i]=former_score[i-1]+A[N+i-1]-former[0]
      heapq.heappush(former,A[N+i-1])
      heapq.heappop(former)
    else:
      former_score[i]=former_score[i-1]
  if latter!=[]:
    if latter[0]<(-1)*A[2*N-i]:
      latter_score[N-i]=latter_score[N-i+1]-A[2*N-i]-latter[0]
      heapq.heappush(latter,A[2*N-i])
      heapq.heappop(latter)
    else:
      latter_score[N-i]=latter_score[N-i+1]
for i in range(N+1):
  score[i]=former_score[i]+latter_score[i]
print(max(score))

Submission status: Wrong Answer

Solution code:
N=int(input())
A=[int(i) for i in input().split()]
score=[0]*(N+1)
score[0]=sum(A[0:N])-sum(sorted(A[N:3*N])[0:N])
former=sorted(A[0:N-1])
latter=A[N:3*N]
sorted_latter=sorted(latter)
for i in range(1,N+1):
  score[i]=score[i-1]+A[N+i-1]-A[N+i-2]
  if former!=[]:
    if former[0]<A[N+i-2]:
      score[i]+=A[N+i-2]-former[0]
      del former[0]
      former.append(A[N+i-2])
      former.sort(key=lambda x:x)
  if sorted_latter[N-1]>=A[N+i-1]:
    score[i]+=A[N+i-1]-sorted_latter[N]
  del latter[0]
  sorted_latter=sorted(latter)
print(max(score)-1)

Submission status: Wrong Answer

Solution code:
#062-D
import heapq
import copy
n=int(input())
a=list(map(int,input().split()))
ca=copy.deepcopy(a)
f=ca[:n]
heapq.heapify(f)
fs=sum(f)
fsl=[0 for i in range(n+1)]
fsl[0]=fs
for i in range(n,2*n):
    heapq.heappush(f,a[i])
    fs=fs+a[i]-heapq.heappop(f)
    fsl[i-n+1]=fs
l=list(map(lambda x:-x,ca[2*n:]))
heapq.heapify(l)
ls=sum(l)
lsl=[0 for i in range(n+1)]
lsl[0]=ls
for i in range(n,2*n):
    heapq.heappush(l,-a[3*n-i-1])
    ls=ls-a[3*n-i-1]-heapq.heappop(l)
    lsl[2*n-i]=ls
ans=-float('inf')
for i in range(n+1):
    ans=max(ans,fsl[i]-lsl[i])
print(ans)