TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
v = list(map(int, input().split()))
v.sort()

while len(v) != 1:
    new_v = (v[0] + v[1]) / 2
    del v[0]
    del v[0]
    v.append(new_v)
    v.sort()
print(v)

N = int(input())
v = list(map(int, input().split()))
v.sort()

while len(v) != 1:
    new_v = (v[0] + v[1]) / 2
    del v[0]
    del v[0]
    v.append(new_v)
    v.sort()
print(v[0])

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
from heapq import heappush, heappop N = int(input()) vLst = list(map(int, input().split())) v = [] for i in range(N): heappush(v, vLst[i]) ans = 0 for i in range(N - 1): tmp1 = heappop(v) tmp2 = heappop(v) ans = float((tmp1 + tmp2) / 2) heappush(v, ans) print(v[0])


from heapq import heappush, heappop

N = int(input())
vLst = list(map(int, input().split()))
v = []
for i in range(N):
    heappush(v, vLst[i])

ans = 0
for i in range(N - 1):
    tmp1 = heappop(v)
    tmp2 = heappop(v)
    ans = float((tmp1 + tmp2) / 2)
    heappush(v, ans)

print(v[0])


Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
v = list(map(int,input().split()))
v.sort()
ans = (v[0] + v[1])/2
for i in range(2,N):
  ans = (ans + v[i])/2
print(ans)
print(v)

N = int(input())
v = list(map(int,input().split()))
v.sort()
ans = (v[0] + v[1])/2
for i in range(2,N):
  ans = (ans + v[i])/2
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
import heapq
N = int(input())
V = list(map(int, input().split()))
 
heapq.heapfy(V)
for i in range(N-2):
  heapq.heappush((heapq.heappop(V)+heapq.heappop(V))/2)
print(sum(V)/2)
  

import heapq
N = input()
V = input().split()

heapq.heapfy(V)
for i in range(N-2):
  heapq.heappush((heapq.heappop(V)+heapq.heappop(V))/2)
print(sum(V)/2)
  

import heapq
N = int(input())
V = list(map(int, input().split()))
 
heapq.heapify(V)
for i in range(N-2):
  heapq.heappush(V, (heapq.heappop(V)+heapq.heappop(V))/2)
print(sum(V)/2)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
*V, = map(int, input().split())
V.sort()
ans = V[0]
for i in range(1, N):
    ans = mix(ans,V[i])
print(ans)

def mix(x, y):
    return (x + y) / 2
    
N = int(input())
*V, = map(int, input().split())
V.sort()
ans = V[0]
for i in range(1, N):
    ans = mix(ans,V[i])
print(ans)

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())

List = list(map(int,input().split()))
List.sort()
ans = 0

for i in range(N):
  a = List[i]/2**(N-i)
  if i == 0:
    a = a*2
  ans = a + ans
  
print(int(ans))

N = int(input())

List = list(map(int,input().split()))
List.sort()
ans = 0

for i in range(N):
  a = List[i]/2**(N-i)
  if i == 0:
    a = a*2
  ans = a + ans
  
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
a = int(input())
Lists = sorted(list(map(int, input().split())))
print(Lists)

ans = Lists[0]

for i in range(1,a):
    ans += Lists[i]
    ans /= 2

print(ans)


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

ma = []

for j in range(a):
    test = Lists.pop(0)
    for i in Lists:
        ma.append((i+test)/2)

ma2 = []
for j in ma:
    test = ma.pop(0)
    for i in ma:
        ma2.append((i+test)/2)

print (max(ma2))


a = int(input())
Lists = sorted(list(map(int, input().split())))

ans = Lists[0]

for i in range(1,a):
    ans += Lists[i]
    ans /= 2

print(ans)


Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N=int(input())
values=[int(input()) for i in range(N)]
values.sort()
ret=0
for i in values:
    ret=(i+ret)*0.5
print(ret)

N=int(input())
values=[int(thing) for thing in input().split(" ")]
values.sort()
ret=0
for i in values:
    if ret!=0:
        ret=(i+ret)*0.5
    else:
        ret=i
print(ret)

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N,*V=map(int,open(0).read().split())
A=sorted(V)
for n in range(N-1):
    print(A)
    A=sorted([(A[0]+A[1])/2,]+A[2:])
print(A[0])

N,*V=map(int,open(0).read().split())
A=sorted(V)
for n in range(N-1):
    A=sorted([(A[0]+A[1])/2,]+A[2:])
print(A[0])

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n = int(input())

s = list(map(int, input().split()))

s.sort()

v = 0

print(s)

if n == 1:
  print(s[0])
  exit()
elif n == 2:
  print((s[0]+s[1])/2)
  exit()

c = (s[0]+s[1])/2

del s[0]
del s[0]

print(c)
print(s)

for i in range(n-2):
  c = (c + s[0])/2
  del s[0]
  print(c)
  
print(c)

n = int(input())

s = list(map(int, input().split()))

s.sort()

v = 0

if n == 1:
  print(s[0])
  exit()
elif n == 2:
  print((s[0]+s[1])/2)
  exit()

c = (s[0]+s[1])/2

del s[0]
del s[0]

for i in range(n-2):
  c = (c + s[0])/2
  del s[0]
  
  
print(c)

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
a = map(int, input().split())

sorted_list = sorted(a)
print(sorted_list)
num = 0
last_sum = 0
for i in range(N):
    if len(sorted_list) == 1:
        last_sum = (sorted_list[num] + last_sum) / 2
        break
    x = sorted_list[num]
    y = sorted_list[num+1]
    z = (x + y) / 2
    last_sum = z
    sorted_list.pop(num)
    sorted_list.pop(num)

print(last_sum)


N = int(input())
a = map(int, input().split())
# N = 3
# a = [200, 300, 500]

s = sorted(a)
t = 0
result = 0
for i in range(0, N-1):
    if i == N - (N - 1):
        result = (t + s[i+1]) / 2
    z = (s[i] + s[i+1]) / 2
    t = z
print(result)

N = int(input())
a = map(int, input().split())
# N = 3
# a = [200, 300, 500]

s = sorted(a)
t = 0
result = 0
if len(s) == 1:
    result = s[0]
    
for i in range(0, N-1):
    if i == N - (N - 1):
        result = (t + s[i+1]) / 2
    z = (s[i] + s[i+1]) / 2
    t = z
print(result)

N = int(input())
a = map(int, input().split())

sorted_list = sorted(a)
print(sorted_list)
num = 0
last_num = 0

for i in range(N):
    if len(sorted_list) == 1:
        if i == 0:
            last_num = sorted_list[num]
        last_num = (sorted_list[num] + last_num) / 2
        break
    x = sorted_list[num]
    y = sorted_list[num+1]
    z = (x + y) / 2
    last_num = z
    sorted_list.pop(num)
    sorted_list.pop(num)

print(last_num)


N = int(input())
a = map(int, input().split())

sorted_list = sorted(a)
print(sorted_list)
num = 0
last_num = 0

for i in range(N):
    if len(sorted_list) == 1:
        if i == 0:
            last_num = sorted_list[num]
        last_num = (sorted_list[num] + last_num) / 2
        break
    x = sorted_list[num]
    y = sorted_list[num+1]
    z = (x + y) / 2
    last_sum = z
    sorted_list.pop(num)
    sorted_list.pop(num)

print(last_num)


n = int(input())
v = sorted(list(map(int, input().split())))
ans = v[0]

for i in range(n-1):
    ans = (ans + v[i+1]) / 2
print(ans)


Verdicts: Runtime Error

Wrong Answer

Wrong Answer

Runtime Error

Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
A = list(map(float,input().split()))
tmp = 0
A.sorted()
tmp = A[0]
for i in range(1,N):
    tmp = (tmp+A[i])/2
print(tmp)


N = int(input())
A = list(map(float,input().split()))
A.sort()
tmp = A[0]
for i in range(1,N):
    tmp = (tmp+A[i])/2
print(tmp)

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n=int(input())
v=sorted(list(map(int, input().split())))
s=v[0]/(2**(n-1))
for i in  range(1,n):
  s+=v[n]/(2**(n-i))
print(s)

n=int(input())
v=sorted(list(map(int, input().split())))
s=v[0]/(2**(n-1))
for i in  range(1,n):
  s+=v[i]/(2**(n-i))
print(s)


Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
v = list(map(int, input().split()))

v.sort()
ans = 0

for i in range(N - 1):
    v1 = v.pop(0)
    v2 = v.pop(0)
    vv = (v1 + v2) / 2
    v.insert(0, vv)
print(int(v[0]))

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

v.sort()
ans = 0

for i in range(N - 1):
    v1 = v.pop(0)
    v2 = v.pop(0)
    vv = (v1 + v2) / 2
    v.insert(0, vv)
print(v[0])

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
v = input().split()

v.sort()

i = 0
while i <= N-2:
  v[i+1]=(int(v[i])+int(v[i+1]))/2
  i += 1
  

print(int(v[-1]))

N = int(input())
v = input().split()
v.sort(key=float)

i = 0
while i <= N-2:
  v[i+1]=(float(v[i])+float(v[i+1]))/2
  i += 1
  

print(v[i])


Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
Vn = list(map(int, input().split()))
print(Vn)
sorted_Vn = sorted(Vn)
average = sorted_Vn[0]
for i in range(1,N):
    average = (average + sorted_Vn[i])/2
    
print(average)

N = int(input())
Vn = list(map(int, input().split()))
sorted_Vn = sorted(Vn)
average = sorted_Vn[0]
for i in range(1,N):
    average = (average + sorted_Vn[i])/2
    
print(average)

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
V_list = list(map(int, input().split()))


import heapq

heapq.heapify(V_list)


for i in range(N-1):
        v1=heapq.heappop(V_list)
        v2=heapq.heappop(V_list)
        new_v = (v1+v2)/2
        heapq.heappush(V_list,new_v)





print(V_list[0])
~

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


import heapq

heapq.heapify(V_list)


for i in range(N-1):
        v1=heapq.heappop(V_list)
        v2=heapq.heappop(V_list)
        new_v = (v1+v2)/2
        heapq.heappush(V_list,new_v)





print(V_list[0])

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
import sys
import itertools

# import numpy as np

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

N, *H = map(int, read().split())
cnt = 1

H.sort()

ans = H[0]
print(H)
for i in H[1:]:
    print(ans, cnt)

    ans += cnt * i
    cnt *= 2
    print(ans, cnt)
print(ans / cnt)


import sys
import itertools

# import numpy as np

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

N, *H = map(int, read().split())
H.sort(reverse=True)
while len(H) > 1:
    x = H.pop()
    y = H.pop()
    H.append((x+y)/2)
print(H[0])


Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
v = sorted(map(int,input().split()))
value = v[0]
for i in range(1,N):
  value=(value+v[i])/2
  print(value)
print(value)

N=int(input())
v=sorted(map(int,input().split()))
value=v[0]
for i in range(1,N):
    value=(value+v[i])/2
print(value)

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n=int(input())
v1=map(int,input().split())
v=list(v1)
for i in range(n):
    for j in range(i+1,n):
        if v[i]>v[j]:
            a=v[i]
            v[i]=v[j]
            v[j]=a
for l in range(n-1):
    a=(v[0]+v[1])/2
    v=a+v[2,n-l]
print(v[0])


n=int(input())
v1=map(int,input().split())
v=list(v1)

for i in range(n):
    for j in range(i+1,n):
        if v[i]>v[j]:
            a=v[i]
            v[i]=v[j]
            v[j]=a
for l in range(n-1):
    a=(v[0]+v[1])/2
    v=a+v[2,n-l]
print(v[0])


n=int(input())
v1=map(int,input().split())
v=list(v1)

for i in range(n):
    for j in range(i+1,n):
        if v[i]>v[j]:
            a=v[i]
            v[i]=v[j]
            v[j]=a
for l in range(n-1):
    v2=[]
    a=(v[0]+v[1])/2
    v2.append(a)
    v2=v2+v[2:n]
    v=v2
print(v[0])

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n=int(input())
v=list(map(int,input().split()))
v.sort(key=int)
for i in range(n-1):
  z=(v[i]+v[i+1])/2
  v[i]=z
  v[i+1]=z
print(v[n])

n=int(input())
v=list(map(int,input().split()))
v.sort(key=int)
for i in range(n-1):
  z=(v[i]+v[i+1])/2
  v[i]=z
  v[i+1]=z
print(v[n-1])


Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):

N = int(input())

A = list(map(int,input().split()))
A = sorted(A)
ans = 0
for i in range(N):
    if i == 0:
        ans += (A[i]+A[i+1])/2
        print(ans)
    elif i<N-1:
        ans = (ans+A[i+1])/2
print(ans)



N = int(input())

A = list(map(int,input().split()))
A = sorted(A)
ans = 0
for i in range(N):
    if i == 0:
        ans += (A[i]+A[i+1])/2
    elif i<N-1:
        ans = (ans+A[i+1])/2
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n = int(input())
v = list(map(int,input().split())

w = sorted(v)
ans = [] 

for i in range(1,n):
  ans.append(0)

for i in range(1,n):
  ans[i-1] = (w[i]+w[i-1])/2
  w[i] = ans[i-1]

print(ans[-1])

n = int(input())
a = list(map(int,input().split()))
a.sort()
now = a[0]
for i in range(1,n):
  now = ( now + a[i] ) / 2
print(now)

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):

N = int(input())
#s = input()
#S = []

S = input().split(" ")
for i in range(N):
    S[i] = int(S[i])
for i in range(N):
    S[i+1] = (int(S[i])+int(S[i+1])) / 2
print(S[N-1])





N = int(input())
#s = input()
#S = []

S = input().split(" ")
for i in range(N):
    S[i] = float(S[i])
S.sort()
for i in range(N-1):
    S[i+1] = (float(S[i])+float(S[i+1])) / 2
print(S[N-1])




Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
import heapq

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

heapq.heapify(v)

res = 0
for _ in range(int(N/2)+1):
  a = heapq.heappop(v)
  b = heapq.heappop(v)
  res = (a+b)/2
  heapq.heappush(v, res)
    
print(res)

import heapq

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

heapq.heapify(v)

res = 0
for _ in range(N-1):
  a = heapq.heappop(v)
  b = heapq.heappop(v)
  res = (a+b)/2
  heapq.heappush(v, res)
    
print(res)

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n = int(input())
arr = list(map(int,input().split()))

arr.sort()

prev = arr[0]

for i in range(n-1)
  prev = (prev+arr[i+1])/2
  
print(prev)



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

arr.sort()

prev = arr[0]

for i in range(n-1):
  prev = (prev+arr[i+1])/2
  
print(prev)



Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
# coding: utf-8
n = int(input())

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

a.sort()

for k in range(n-1):
    b=(a[0]+a[1])/2
    a.append(b)
    a.pop(0)
    a.pop(0)
    a.sort

A =a[0]
print(int(A))

# coding: utf-8
n = int(input())

a = list(map(float,input().split())) 

a.sort()

for k in range(n-1):
    b=(a[0]+a[1])/2
    a.append(b)
    a.pop(0)
    a.pop(0)
    a.sort()
    print(a)

A =a[0]
print(A)

# coding: utf-8
n = int(input())

a = list(map(float,input().split())) 

a.sort()

for k in range(n-1):
    b=(a[0]+a[1])/2
    a.append(b)
    a.pop(0)
    a.pop(0)
    a.sort()

A =a[0]
print(A)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
from heapq import *

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

h = []
for i in range(n):
    heappush(h, v[i])

for i in range(n-1):
    x, y = heappop(h), heappop(h)
    heappush(h, (x + y) / 2)

print(h[0]//1)

from heapq import *

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

h = []
for i in range(n):
    heappush(h, v[i])

for i in range(n-1):
    x, y = heappop(h), heappop(h)
    heappush(h, (x + y) / 2)

print(h[0])

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n = int(input())
arr = list(map(int,input().split()))

arr.sort()

for i in range(len(arr)-1):
    arr[i+1] = (arr[i]+arr[i+1])/2

print(int(arr[-1]))

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

arr.sort()

for i in range(len(arr)-1):
    arr[i+1] = (arr[i]+arr[i+1])/2

print(arr[-1])

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
a = map(int,input().split())
heap.heapify(a)
for i in range(N-1):
  x = heap.heappop(a)
  y = heap.heappop(a)
  z = (x+y)/2
  heap.heappush(a,z)
answer = heap.heappop(a)
print(answer)

N = int(input())
a = list(map(int,input().split()))
import heapq
heapq.heapify(a)
for i in range(N-1):
  x = heapq.heappop(a)
  y = heapq.heappop(a)
  z = (x+y)/2
  heapq.heappush(a,z)
answer = heapq.heappop(a)
print(answer)

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n = int(input()) 
v = list(map(int,(input().split()))) 
sum_now = 0
sum_old = 0
v.sort()
print(v)

tmp = v[0]
for i in range(1,n):
    tmp = ((tmp + v[i])  /2) 
print(tmp)

n = int(input()) 
v = list(map(int,(input().split()))) 
sum_now = 0
sum_old = 0
v.sort()

tmp = v[0]
for i in range(1,n):
    tmp = ((tmp + v[i])  /2) 
print(tmp)

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
V = [int(zz) for zz in input().split()]

V.sort()
ans = V[0]+V[1]
for i in range(2,N):
    ans += V[i]*(2**(i-1))
print(ans//(2**(N-1)))


N = int(input())
V = [int(zz) for zz in input().split()]

V.sort()
ans = V[0]+V[1]
for i in range(2,N):
    ans += V[i]*(2**(i-1))
ans = ans/(2**(N-1))

if ans == int(ans):
    print(int(ans))
else:
    print(ans)


Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
V = list(map(int, input().split()))

C = sorted(V)
print(C)
ans = 0

a = C[0]
for i in range(1, N):
    a = (a + C[i]) / 2
ans = max(ans, a)
print(ans)

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

C = sorted(V)

ans = 0
a = C[0]
for i in range(1, N):
    a = (a + C[i]) / 2
ans = max(ans, a)
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n = int(input())
lists = list(map(int,input().split()))

lists.sort()

for i in range(n - 1):
    lists.insert(0, (lists.pop(0) + lists.pop(0))/2)

print(int(lists[0]))


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

lists.sort()

for i in range(n - 1):
    lists.insert(0, (lists.pop(0) + lists.pop(0))/2)

print(lists[0])


Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
values = [int(x) for x in input().split()]

combined_value = values[0]
sliced_values = values[1:]
for value in sliced_values:
  combined_value = (combined_value + value) / 2

print(round(combined_value, 5))

N = int(input())
values = [int(x) for x in input().split()]

combined_value = 0
for value in values:
  combined_value = (combined_value + value) / 2

print(round(combined_value, 5))

values = [int(x) for x in input().split()]

combined_value = values[0]
sliced_values = values[1:]
for value in sliced_values:
  combined_value = (combined_value + value) / 2

print(combined_value)

values = [int(x) for x in input().split()]
combined_value = values[0]
sliced_values = values[1:]
for value in sliced_values:
  combined_value = (combined_value + value) / 2
print(round(combined_value, 5))

N = int(input())
values = sorted([int(x) for x in input().split()])

combined_value = values[0]
sliced_values = values[1:]
for value in sliced_values:
  combined_value = (combined_value + value) / 2

print(round(combined_value, 5))

Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
vs = sorted(map(int, input().split(" ")), reverse=True)

r = [v * (0.5 ** i) for i, v in enumerate(vs)]

return sum(r)

N = int(input())
vs = sorted(map(int, input().split(" ")), reverse=True)
 
r = [v * (0.5 ** (i + 1)) for i, v in enumerate(vs)]
r [-1] = r[-1] * 2
print(sum(r))

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
u = list(map(int, input().split()))

for i in range(N-1):
  a = u.pop(u.index(min(u)))
  b = u.pop(u.index(min(u)))
  u.append((a + b) / 2)

print(u)

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

for i in range(N-1):
  a = u.pop(u.index(min(u)))
  b = u.pop(u.index(min(u)))
  u.append((a + b) / 2)

print(u[0])

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
hoge = input().split()
v = []
for i in range(N):
  v.append(int(hoge[i]))

_v = sorted(v, reverse=True)

gu = _v.pop(N-1)
if N == 2:
  print((_v[0]+_v[1])/2)
else:
  for i in range(N-2,-1,-1):
    gu = (gu + _v.pop(i))/2
  print(gu)

N = int(input())
hoge = input().split()
v = []
for i in range(N):
  v.append(int(hoge[i]))

_v = sorted(v, reverse=True)

gu = _v.pop(N-1)
if N == 2:
  print((_v[0]+gu)/2)
else:
  for i in range(N-2,-1,-1):
    gu = (gu + _v.pop(i))/2
  print(gu)

Verdicts: Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
V = list(map(int, input().split()))

for i in range(N-1):
    V.sort()
    tmp = (V.pop(0) + V.pop(0)) / 2
    V.append(tmp)
    print(V)

print(V[0])


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

for i in range(N-1):
    V.sort()
    tmp = (V.pop(0) + V.pop(0)) / 2
    V.append(tmp)

print(V[0])


Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N = int(input())
values = list(map(int, input().strip().split()))
total = 0
values.sort()
step = 0
print(values)
while len(values) > 1:
    new = (values[step] + values[step + 1]) / 2
    del values[0]
    del values[0]
    values = [new] + values
    print(values)
print(values[0])

N = int(input())
values = list(map(int, input().strip().split()))
total = 0
values.sort()
step = 0
#print(values)
while len(values) > 1:
    new = (values[step] + values[step + 1]) / 2
    del values[0]
    del values[0]
    values = [new] + values
    #print(values)
print(values[0])

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
N-int(input())
s=list(map(int, input().split()))
s.sort()
ans=0
for i in range(N-1):
    if i==0:
        ans=s[0]
    else:
        ans=((ans+s[i])/2)
print(ans)

N-int(input())
s=list(map(int, input().split()))
ans=0
for i in range(N-1):
    if i==0:
        ans=s[0]
    else:
        ans=((ans+s[i])/2)
print(ans)

N-int(input())
s=list(map(int, input().split()))
ans=0
for i in range(N-1):
    if i==0:
        ans=s[0]
    else:
        ans=((ans+s[i])/2)
print(ans)

N-int(input())
s=list(map(int, input().split()))
s.sort()
ans=0
for i in range(N-1):
    if i==0:
        ans=+s[0]
    else:
        ans=((ans+s[i])/2)
print(ans)

N=int(input())
s=list(map(int, input().split()))
s.sort()
ans=s[0]
for i in range(N-1):
    ans=((ans+s[i+1])/2)
print(ans)

Verdicts: Runtime Error

Runtime Error

Runtime Error

Runtime Error

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
a=int(input())
b=list(map(int,input().split()))
b.sort()
while len(b)>1:
    c = (b[0]+b[1])/2
    del b[:2]
    b.insert(0,c)
    b.sort()
print(int(b[0]))

a=int(input())
b=list(map(int,input().split()))
b.sort()
while len(b)>1:
    c = (b[0]+b[1])/2
    del b[:2]
    b.insert(0,c)
    b.sort()
print(b[0])

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

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

list.sort()
print(list)
num = list[0]
for i in range(0, N):
    num = (num + list[i]) / 2
print(num)

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

list.sort()
num = list[0]
for i in range(0, N):
    num = (num + list[i]) / 2
print(num)

Verdicts: Wrong Answer

Accepted
Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is v_i.  When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.  After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Submission attempts (in order):
n = int(input())
vn = list(map(int, input().split())).sort()

sumV = vn[0]

for v in vn[1:]:
    sumV = (sumV + v)/2
print(sumV)

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

vn.sort()
sumV = vn[0]

for v in vn[1:]:
    sumV = (sumV + v)/2
print(sumV)

Verdicts: Runtime Error

Accepted