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(reverse = True)
V.sort()
# print(V)
result = V[0]

for i in range(1, N):
	result = (V[i] + result) / 2.0
	print(result)

print(result)


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

# V.sort(reverse = True)
V.sort()
# print(V)
result = V[0]

for i in range(1, N):
	result = (V[i] + result) / 2.0
	# print(result)

print(result)


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

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

N = int(input())
v = sorted(list(map(int, input().split())))
ans = v[0]
for i in range(N):
    ans = (ans + v[i])/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 = input()
arr = list(map(int, input().split()))
while len(arr) > 1:
  arr = sorted(arr)
  v1 = arr.pop(0)
  v2 = arr.pop(0)
  arr.appned((v1 + v2) / 2)
print(arr[0])

n = int(input())
arr = list(map(int, input().split()))
while len(arr) > 1:
  arr = sorted(arr)
  v1 = arr.pop(0)
  v2 = arr.pop(0)
  arr.appned((v1 + v2) / 2)
print(arr[0])



n=int(input())
arr=list(map(int, input().split()))
while len(arr)>1:
  arr = sorted(arr)
  v1 = arr.pop(0)
  v2 = arr.pop(0)
  arr.append((v1 + v2) / 2)
print(arr[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):
import heapq
import math


def solve(n, v):
    heapq.heapify(v)
    nn = math.ceil(n / 2)

    while len(nn) != 1:
        x = heapq.heappop(v)
        y = heapq.heappop(v)
        z = (x + y) / 2
        heapq.heappush(v, z)

    return v[0]


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

print(solve(n, v))


import heapq
import math


def solve(n, v):
    heapq.heapify(v)

    while len(v) != 1:
        x = heapq.heappop(v)
        y = heapq.heappop(v)
        z = (x + y) / 2
        heapq.heappush(v, z)

    return v[0]


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

print(solve(n, v))


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()

b = a[0]
for i in range(n - 1):
  b = (b + a[i + 1]) / 2

print(b)

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

b = a[0]
for i in range(n - 1):
  b = (b + a[i + 1]) / 2

print(b)


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 numpy as np
n = int(input())
v = np.array(list(map(float, input().split())))
v.sort()

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

import numpy as np
n = int(input())
v = np.array(list(map(float, input().split())))
v.sort()

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

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

a.sort()

ans = int((a[0]+a[1])/2)

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

print(ans)

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

a.sort()

ans = (a[0]+a[1])/2

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

N = int(input())
List = list(map(int,input().split()))
A = 0
List = sorted(List)
for i in range(1,N):
    A += List[i]/2**(N-i)
print(A+List[0]/2**(N-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 = input()
a = input().split()

a = [float(i) for i in a]#float cast
p = 1
list.sort(a)
print("初期値"+ str(a))
N = int(N)
for j in range(N-1):
    sum = a[j]
    for i in range(1):
        sum = sum + a[p]
        ans = sum / 2
        a[p] = ans
        p += 1
    list.sort(a)
    print(str(j) + "回目:" + str(a))

ans = sum / 2
print(ans)
    

N = input()
a = input().split()

a = [float(i) for i in a]#float cast
p = 1
list.sort(a)

N = int(N)
for j in range(N-1):
    sum = a[j]
    for i in range(1):
        sum = sum + a[p]
        ans = sum / 2
        a[p] = ans
        p += 1
    list.sort(a)
    

ans = sum / 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()))
while len(v) > 2:
  temp = []
  for i in range(len(v)//2):
    temp.append((v[2*i] + v[2*i + 1])/2)
  if n % 2 == 0:
    v = temp
  else:
    temp.append(v[-1])
    v = temp
      
print(v[0]) 

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

while len(v) > 1:
  x = v.pop()
  v[-1] = (x + v[-1]) /2
      
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):
x = int(input())
a = list(map(int,input().split()))
a.sort()
y = 0
for i in range(x):
  y += a[i-0]*(2**i)
print(y/(2**(x+1)))

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

x = int(input())
a = [list(map(int,input().split()))]
b = sorted(a)
y = 0
for i in range(x):
  y += b[i-0]*2**i
print(y/(2**(x+1)))

x = int(input())
a = [list(map(int,input().split()))]
b = sorted(a)
y = 0
for i in range(x):
  y += b[i-0]*(2**i)
print(y//(2**(x+1)))

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

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

Verdicts: Wrong Answer

Wrong Answer

Runtime Error

Runtime Error

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()
a= List[1]
for i in range(len(List)):
  a= (a+List[i])/2
print(a)

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

N=int(input())
List=list(map(int,input().split()))
List.sort()
a= List[0]
for i in range(len(List)):
  a= (a+List[i])/2
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):
n=int(input())
c=list(map(int,input().split()))
c.sort(reverse=True)
value=0
for x in c:
  value=(value+x)/2
print(value)

n=int(input())
c=list(map(float,input().split()))
c.sort()
k=n
for i in range(n-2):
  v=(c[0]+c[1])/2
  c=c[2:]
  if c[k-3]<v:
    c.append(v)
    break
  for j in range(k-2):
    if c[j]>=v:
      c.insert(j,v)
      break
  k=k-1
print((c[0]+c[1])/2)

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()))

from bisect import insort
c=1
while c<N:
    #print(c,v)
    insort(v,(v.pop(0) + v.pop(0)) / 2)
    #print(c,v)
    c+=1
print(v)

N=int(input())
*v,=sorted(map(int,input().split()))

from bisect import insort
c=1
while c<N:
    insort(v,(v.pop(0) + v.pop(0)) / 2)
    c+=1
print(*v)

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()))

v = sorted(v)

print(v)

tmp = 0.0
for i in range(N-1):
    tmp = (v[i]+v[i+1])/2
    v[i+1] = tmp

print(v[-1])

N = int(input())

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

v = sorted(v)

tmp = 0.0
for i in range(N-1):
    tmp = (v[i]+v[i+1])/2
    v[i+1] = tmp

print(v[-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())
nums = list(map(int, input().split()))
nums.sort()
ans = num
for num  in nums[1:]:
  ans = (ans + num)/2
print(ans)
  

N = int(input())
nums = list(map(int, input().split()))
nums.sort()
ans = nums[0]
for num  in nums[1:]:
  ans = (ans + num)/2
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())
vlis=list(map(int,input().split()))
while len(vlis)>=2:
	tlis=[]
	for i in range(len(vlis)-1):
		t=(vlis[i]+vlis[i+1])/2
		tlis.append(t)
	vlis.insert(tlis.index(min(tlis)),min(tlis))
	for i in range(2):
		vlis.pop(tlis.index(min(tlis))+1)
		
print(int(vlis[0]))

n=int(input())
vlis=sorted(list(map(int,input().split())))
while len(vlis)>=2:
	a,b=vlis[0],vlis[1]
	vlis.pop(0)
	vlis.pop(0)
	vlis.insert(0,(a+b)/2)
print(vlis[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=[]
v+=list(map(int,input().split()))
v.sort()
l=len(v)
result=v[0]/(2**(l-1))
for i in range(1,n):
  result=result+v[i]/(2**(l-i))
print(int(result))

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

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())
b = list(map(int, input().split()))
result = 0

b.sort()
print(b)

for i in range(a):
    if i == 0:
        result = b[i]
    else:
        result = result / 2
        result += b[i] / 2

print(result)

a = int(input())
b = list(map(int, input().split()))
result = 0

b.sort()

for i in range(a):
    if i == 0:
        result = b[i]
    else:
        result = result / 2
        result += b[i] / 2

print(result)
    

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 = sorted([int(a) for a in input().split()],reverse=True)
if N  == 2:
  print(round(sum(A)/2))
else :
  print(((A[2] + A[1])/2 + A[0]) / 2)

N =int(input())
A = sorted([int(a) for a in input().split()],reverse=True)
if N  == 2:
  print(round(sum(A)/2))
else :
  print(round(((A[2] + A[1])/2 + A[0]) / 2))

N =int(input())
A = sorted([int(a) for a in input().split()])
if N  == 2:
  print(sum(A)/2)
else :
  ans=A[0]
  for i in (A):
    ans+= i
    anss = ans/2
    ans=anss
  print(anss)
    

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

N=int(input())
v=list(map(int,input().split()))
for i in range(N-1):
  v.sort()
  V=(v[0]+v[1])/2
  del v[0:2]
  v.append(V)
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=list(map(int, input().split()))
v.sort()
for i in range(N-2):
  v1=[(v[0]+v[1])/2]
  v1.extend(v[2:])
  v=v1
print(sum(v1)/2)

N=int(input())
v=list(map(int, input().split()))
v.sort()
for i in range(N-2):
  v1=[(v[0]+v[1])/2]
  v1.extend(v[2:])
  v=v1
print(sum(v1)/2)

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

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()))

for _ in range(n - 1):
    v = sorted(v)
    value = (v.pop(1) + v.pop(0)) // 2
    v.append(value)

print(v[0])


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

for _ in range(n - 1):
    v = sorted(v)
    value = (v.pop(1) + v.pop(0)) / 2
    v.append(value)

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())
vv = list(map(int, input().split(' ')))

vv.sort()

ans += vv[0]
for v in vv:
  ans += v
  ans /= 2
  
print(ans)

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

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

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

Vs.sort()

print(Vs)

initialGousei = (Vs[0] + Vs[1]) / 2

for i in range(2, N):
    initialGousei = (initialGousei + Vs[i]) / 2

print(initialGousei)


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

Vs.sort()

#print(Vs)

initialGousei = (Vs[0] + Vs[1]) / 2

for i in range(2, N):
    initialGousei = (initialGousei + Vs[i]) / 2

print(initialGousei)


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

m = min(lis)
lis.remove(min(lis))
m_2 = min(lis)
lis.remove(min(lis))
v = (m + m_2) / 2   

for i in range(0, N - 1):
    v = (v + min(lis)) / 2
    lis.remove(min(lis))

print(v)
    
    

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

m = min(lis)
lis.remove(min(lis))
m_2 = min(lis)
lis.remove(min(lis))
v = (m + m_2) / 2   

for i in range(0, N - 2):
    v = (v + min(lis)) / 2
    lis.remove(min(lis))

print(v)
    

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_n = list(map(int, input().split()))
v_n_ord = sorted(v_n, reverse=True)
v_n_mod = []
for i in range(len(v_n)):
    if i == (len(v_n) - 1):
        print('b')
        v_n_mod.append(v_n_ord[i] / (2**i))
    else:
        print('a')
        v_n_mod.append(v_n_ord[i] / (2**(i+1)))
print(sum(v_n_mod))

N = int(input())
arr = list(map(int, input().split()))
while len(arr)>1:
    arr=sorted(arr)
    v1 = arr.pop(0)
    v2 = arr.pop(0)
    arr.append((v1+v2)/2)
print(arr[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 = list(map(int, input().split()))

w = v[0]
for i in v[1:]:
    w += (w+i)/2

print(w)


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

w = v[0]
for i in v[1:]:
    w = (w+i)/2

print(w)

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()))
while  n>1:
       v.sort()
       v.append((v[0]+v[1])/2)
       v.remove(v[0])
       v.remove(v[1])
       n-=1
print(int((v[0])))

n=int(input())
v=list(map(int,input().split()))
while  n>1:
       v.sort()
       v.append((v[0]+v[1])/2)
       v.remove(v[0])
       v.remove(v[0])
       n-=1
print(int((v[0])))

n=int(input())
v=list(map(int,input().split()))
while  n>1:
       v.sort()
       v.append((v[0]+v[1])/2)
       v.remove(v[0])
       v.remove(v[0])
       n-=1
print(int((v[0])))
  

n=int(input())
v=list(map(int,input().split()))
while  n>1:
       v.sort()
       v.append((v[0]+v[1])/2)
       v.remove(v[0])
       v.remove(v[0])
       n-=1
print((v[0]))

Verdicts: 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,arr = int(input()),list(map(int,input().split(' ')))
stack = []
arr = sorted(arr)
for i in range(0,len(arr),2):
    if i == len(arr)-1:
        print(int((stack[-1] + arr[i])/2))
    else:
        stack.append((arr[i] + arr[i+1])/2)

n,arr = int(input()),list(map(int,input().split(' ')))
stack = []
arr = sorted(arr)
for i in range(0,len(arr),2):
    if i == len(arr)-1:
        print((stack[-1] + arr[i])/2)
    else:
        stack.append((arr[i] + arr[i+1])/2)

n,arr = int(input()),list(map(int,input().split(' ')))
stack = []
arr = sorted(arr)
stack.append(arr[0])
arr.pop(0)
while arr:
    stack.append((stack[-1] + arr[0])/2)
    arr.pop(0)
print(stack[-1])

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):
n = int(input())
v = sorted(map(int, input().split()))

ans = v[0]

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

print(ans)

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

ans = v[0]

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

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):
#coding utf-8
N=int(input())
v=list(map(int,input().split()))
v.sort()
print(v)

ans=v[0]

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

print(ans)

#coding utf-8
N=int(input())
v=list(map(int,input().split()))
v.sort()

ans=v[0]

for i in range(1,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):
# 

n = int(input())
l = [int(x) for x in input().split()]
i =0
while len(l) >= 2:
    l.sort()
    a = l.pop(i)
    b = l.pop(i)
    c = (a + b) / 2
    l.append (c)

print(l)


# easy for C!!!

n = int(input())
l = [int(x) for x in input().split()]
i =0
while len(l) >= 2:
    l.sort()
    a = l.pop(i)
    b = l.pop(i)
    c = (a + b) / 2
    l.append (c)

print(*l)


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 = input()
V = list(map(int, input().split()))
V = sorted(V)
value = V.pop(-1)
while value:
    value += V.pop(-1)
print(value)

N = input()
V = list(map(int, input().split()))
V = sorted(V)
value = V.pop(0)
while V:
    value += V.pop(0)
print(value)


N = input()
V = list(map(int, input().split()))
V = sorted(V)
value = V.pop(0)
while V:
    value =(value + V.pop(0))/2
print(value)
    

Verdicts: Runtime Error

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

b = a.sort()

for i in range(n-1):
    c=(a[0]+a[1])/2
    a.pop(1)
    a.pop(0)
    a.insert(0,c)
    
print(a)

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

b = a.sort()

for i in range(n-1):
    c=(a[0]+a[1])/2
    a.pop(1)
    a.pop(0)
    a.insert(0,c)
    
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())
L1=sorted(list(map(int,input().split())))
ans=L1[0]
L1.pop(0)
for i in range(len(L1)):
    ans=(ans+L1[i])/2
    print(ans)
print(ans)


N=int(input())
L1=sorted(list(map(int,input().split())))
ans=L1[0]
L1.pop(0)
for i in range(len(L1)):
    ans=(ans+L1[i])//2
print(ans)

N=int(input())
L1=sorted(list(map(int,input().split())))
ans=L1[0]
L1.pop(0)
for i in range(len(L1)):
    ans=(ans+L1[i])/2
print(ans)

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):
N = int(input())
e = list(map(int,input().split()))
for i in range(N):
    x = int(input())
    e.append(x)
e.sort()
for i in range(N - 1):
    a = e.pop(0)
    b = e.pop(0)
    e.insert(0, (a+b)/2)
ans = e[0]
print(ans)

N = int(input())
e = list()
for i in range(N):
    x = int(input())
    e.append(x)
e.sort()
for i in range(N - 1):
    a = e.pop(0)
    b = e.pop(0)
    e.insert(0, (a+b)/2)
ans = e[0]
print(ans)

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

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()))
#h, w = list(map(int, input().split()))
v.sort()
ans = 0
point = (v[0]+v[1])/2
if n == 2:
    print(point)
else:
    for i in range(2, n):
        point = (point+v[i])/2

print(point)


n = int(input())
v  = list(map(int, input().split()))
#h, w = list(map(int, input().split()))
v.sort()
ans = 0
point = (v[0]+v[1])/2
for i in range(2, n):
    point = (point+v[i])/2

print(point)


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):
def  makabe((float)num, (float)ko):
  ky=(num+ko)/2
  return ky



n=int(input())
g=list(map(float,input().split()))
g.sort()
ty=(g[0]+g[1])/2
for h in range(2,n):
  ty= makabe(ty,g[h])
  
print(ty)

def  makabe(nam, co):
  ky=(nam+co)/2.0
  return ky



n=int(input())
g=list(map(int,input().split()))
g.sort()
ty=(g[0]+g[1])/2.0
for h in range(2,n):
  ty= makabe(ty,g[h])
  
print(ty)

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

N = int(input())
v = [int(vi) for vi in input().split()]
vv = [float(vf) for vf in v]
vv.sort()
ans = vv[0]
for i in range(1, N):
  ans = (ans + vv[i]) / 2
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())
h = list(map(int,input().split(" ")))

plsl = []
plsn = 0
for i in range(n-1):
    if h[i]>= h[i+1] and i == n-1:
        plsn += 1
        # print(plsn)
        plsl.append(plsn)
        # print(plsl)
    elif h[i]>= h[i+1] and i < n-1:
        plsn += 1
        # print(plsn)
    else:
        plsl.append(plsn)
        # print(plsl)
        plsn = 0
        # print(plsn)
plsl.sort(reverse = True)
print(plsl[0])

n = int(input())
h = list(map(int,input().split(" ")))

plsl = []
plsn = 0
for i in range(n-1):
    if h[i]>= h[i+1] and i == n-2:
        plsn += 1
        # print(plsn)
        plsl.append(plsn)
        # print(plsl)
    elif h[i]>= h[i+1] and i =< n-1:
        plsn += 1
        # print(plsn)
    else:
        plsl.append(plsn)
        # print(plsl)
        plsn = 0
        # print(plsn)
plsl.sort(reverse = True)
print(plsl[0])

n = int(input())
h = list(map(int,input().split(" ")))

plsl = []
plsn = 0
for i in range(n-1):
    if h[i]>= h[i+1] and i == n-2:
        plsn += 1
        # print(plsn)
        plsl.append(plsn)
        # print(plsl)
    elif h[i]>= h[i+1] and i < n-2:
        plsn += 1
        # print(plsn)
    else:
        plsl.append(plsn)
        # print(plsl)
        plsn = 0
        # print(plsn)
plsl.sort(reverse = True)
print(plsl[0])

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

v.sort()
# print(v)
w = []

w0 = (v[0] + v[1])/2
w.append(w0)

for i in range(1,n-1):
    y = (v[i+1] + w[i-1])/2
    w.append(y)
    # print(w)
# print(w)
print(w[n-2])

Verdicts: Runtime Error

Runtime Error

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 = map(int, input().split())
V = sorted(V)
P = V[0]
for i in range(N-1):
  P += V[i+1]
  P = P/2

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

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()))

v.sort()
print(v)
#subtotal=(v[0]+v[1])/2

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

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

v.sort()
#print(v)
subtotal=(v[0]+v[1])/2

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

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 = [int(i) for i in input().split()]

V.sort()
print(V)

ans = 0

for i in range(N):
	if i == 0 or i == 1:
		ans += V[i] / (2**(N-1))
	else:
		ans += V[i] / (2**(N-i))

print(ans)

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

V.sort()
#print(V)

ans = 0

for i in range(N):
	if i == 0 or i == 1:
		ans += V[i] / (2**(N-1))
	else:
		ans += V[i] / (2**(N-i))

print(ans)

Verdicts: Wrong Answer

Accepted