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):
import numpy as np

def set_list(es):
    return len(list(set(es))) == 1

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

if(len(a)==2):
    print((a[0]+a[1])/2)  
elif(len(a)!=2 and set_list(a)):
    print(a[0])
else:
  for i in range(1,n):
    idx = np.argmin(a)
    num1 = a.pop(idx)
    idx = np.argmin(a)
    num2 = a.pop(idx)
    a.append((num1+num2)/2)
	print(a[0])

import numpy as np

def set_list(es):
    return len(list(set(es))) == 1

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

if(len(a)==2):
    print((a[0]+a[1])/2)
    
if(len(a)!=2 and set_list(a)):
    print(a[0])

for i in range(1,n):
    idx = np.argmin(a)
    num1 = a.pop(idx)
    idx = np.argmin(a)
    num2 = a.pop(idx)
    a.append((num1+num2)/2)
print(a[0])

import numpy as np

def set_list(es):
    return len(list(set(es))) == 1

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

if(len(a)==2):
    print((a[0]+a[1])/2)    
elif(len(a)!=2 and set_list(a)):
    print(a[0])
else:
    for i in range(1,n):
        idx = np.argmin(a)
        num1 = a.pop(idx)
        idx = np.argmin(a)
        num2 = a.pop(idx)
        a.append((num1+num2)/2)
    print(a[0])

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

vs = sorted(v)
ans = vs[0]
for i in range(1,n):
    ans = (ans + vs[i]) // 2

print(ans)

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

vs = sorted(v)
ans = vs[0]
for i in range(1,n):
    ans = (ans + vs[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())
v=list(map(int,input().split()))
list.sort(v)

result=0
for i in range(1,n):
    if i==1:
        result+=(v[i-1]+v[i]/2)
    else:
        result=(result+v[i])/2
print(result)

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

result=0
for i in range(1,n):
    if i==1:
        result+=(v[i-1]+v[i])/2
    else:
        result=(result+v[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 = list(map(int, input().split()))
bo = sum(map(lambda x: 1/x, a))
print(1/bo)



from bisect import insort

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

v.sort()
while len(v) > 1:
    x = v.pop(0)
    y = v.pop(0)
    insort(v, (x + y) / 2)

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()
u = [0]*(N-1)
for i in range(N):
    if i == 0:
        u[0] = (v[0] + v[1]) / 2
    else:
        u[i] = (u[i-1]+v[i+1])/2
print(u[-1])

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

for i in range(N):
  l.append(2**(i))

l = sorted(l, reverse = True)
  
array_multiple = list(map(lambda x,n: x/n, A, l))

print(sum(array_multiple))

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

print((B+A[N-1])/2)

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

for i in range(N-1):
  B = (A[i] + A[i+1])/2

print((B+A[N-1])/2)

N = int(input())
A = sorted(list(map(int, input().split())))
l = []

for i in range(N):
  l.append(2**(i+1))

l = sorted(l, reverse = True)
  
array_multiple = list(map(lambda x,n: x/n, A, l))

print(sum(array_multiple)+A[0]/(2**N))

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


N = int(input())
V = sorted(list(map(int, input().split())))
ans = (V[0] + V[1]) / 2
if N == 2:
  print(ans)
  exit()
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 math
import heapq

N=int(input[0])
V=list(map(int, input[1].split()))  

heapq.heapify(V)
a=heapq.heappop(V)
b=0
for i in range(len(V)):
    b=heapq.heappop(V)
    a=(a+b)/2

print(a)

import math
import heapq

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

heapq.heapify(V)
a=heapq.heappop(V)
b=0
for i in range(len(V)):
    b=heapq.heappop(V)
    a=(a+b)/2

print(a)

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())
vl = list(map(int, input().split()))
vl_s = sorted(vl)
print(vl_s)

n1 = vl_s[0]
for i in range(1, n):
    n2 = vl_s[i]
    ave = (n1 + n2) / 2
    n1 = ave

print(ave)

n = int(input())
vl = list(map(int, input().split()))
vl_s = sorted(vl)

n1 = vl_s[0]
for i in range(1, n):
    n2 = vl_s[i]
    ave = (n1 + n2) / 2
    n1 = ave

print(ave)

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())
data = sorted([int(i) for i in input().split()])
print(data)
ans = data[0]
for i in data[1:]:
  ans = (ans+i)/2
print(ans)


n = int(input())
data = sorted([int(i) for i in input().split()])
ans = data[0]
for i in data[1:]:
  ans = (ans+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):
def main():
  N = int(input())
  v = [int(i) for i in input().split()]
  v.sort()
  ans = 0
  for i in range(len(v)-1):
    ans = (ans+v[i+1])/2 if i > 0 else (v[i]+v[i+1])/2


if __name__ == '__main__':
  main()


def main():
  N = int(input())
  v = [int(i) for i in input().split()]
  v.sort()
  ans = 0
  for i in range(len(v)-1):
    ans = (ans+v[i+1])/2 if i > 0 else (v[i]+v[i+1])/2
  print(ans)

if __name__ == '__main__':
  main()


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())
Vs = input().split()

if N == 2:
    print(sum([int(x) for x in Vs])/N)
else:
    lists = [int(x) for x in Vs]
    while N > 1:
        lists.sort()
        x = lists.pop(0) + lists.pop(0)) / 2
        lists.insert(0, x)
        N -= 1

    print(lists[0])

N = int(input())
Vs = input().split()

if N == 2:
    print(sum([int(x) for x in Vs])/N)
else:
    lists = [int(x) for x in Vs]
    while N > 1:
        lists.sort()
        x = (lists.pop(0) + lists.pop(0)) / 2
        lists.insert(0, x)
        N -= 1

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

from sys import stdin
nii=lambda:map(int,stdin.readline().split())
lnii=lambda:list(map(int,stdin.readline().split()))

n=int(input())
v=lnii()
v.sort()

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

print(ans

from sys import stdin
nii=lambda:map(int,stdin.readline().split())
lnii=lambda:list(map(int,stdin.readline().split()))

n=int(input())
v=lnii()
v.sort()

ans=v[0]
for i in range(1,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,*v=3,10,10,10
v.sort()
print(sum(v[i]<<i+(i==0)for i in range(n))/2**n)

n,*v=map(int,open(0).read().split())
v.sort()
print(sum(v[i]<<i+(i==0)for i in range(n))/2**n)

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())
M = list(map(int,input().split()))
X = sorted(M)
print(X)
c = 0
for i, v in enumerate(X):
    if i >= N-1:
        break
    else:
        a = X[i]+X[i+1]
        b = a/2
        X[i+1]= a/2

print(b)

N = int(input())
M = list(map(int,input().split()))
X = sorted(M)
c = 0
for i, v in enumerate(X):
    if i >= N-1:
        break
    else:
        a = X[i]+X[i+1]
        b = a/2
        X[i+1]= a/2

print(b)

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 numpy as np
n = int(input())
v = [int(i) for i in input().split()]
v.sort()
for i in range(n): 
  v1 = v.pop(-1)
  v2 = v.pop(-1)
  v3 = (v1 + v2) / 2
  v.append(v3)
  v.sort()
print(v[0])
  
  



n = int(input())
v = [int(i) for i in input().split()]
v.sort()
v1 = v.pop(0)
for i in range(n-1): 
  v2 = v.pop(0)
  v1 = (v1 + v2) / 2
print(v1)
  
  


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):
from heapq import heappop,heappush
N = int(input())
V = list(map(int,input().split()))
V.sort()
while True:
    print(V)
    if len(V) == 1:
        break
    a = heappop(V)
    print(V)
    b = heappop(V)
    heappush(V,(a + b) / 2)
print(heappop(V))

from heapq import heappop,heappush
N = int(input())
V = list(map(int,input().split()))
V.sort()
while True:
    if len(V) == 1:
        break
    a = heappop(V)
    b = heappop(V)
    heappush(V,(a + b) / 2)
print(heappop(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())
list_a = list(map(int,input().split()))

list_a.sort()

for m in range(n - 1):
    new = 0
    new += (list_a[0] + list_a[1]) / 2
    print(new)
    del list_a[0:2]
    print(list_a)
    list_a.append(new)
    print(list_a)
    list_a.sort()

print(list_a[0])


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

list_a.sort()

for m in range(n - 1):
    new = 0
    new += (list_a[0] + list_a[1]) / 2
    del list_a[0:2]
    list_a.append(new)
    list_a.sort()

print(list_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())
v = input().split()

v = map(lambda x: int(x), v)
for i in range(n-1):
  v.sort()
  v.append((v.pop()+v.pop())/2)
print(v[0])

n = int(input())
v = input().split()

v = list(map(lambda x: int(x), v))
for i in range(n-1):
  v.sort()
  v.append((v.pop(0)+v.pop(0))/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):
n = int(input())
a = (input()).split(" ")
sumv = 0
temp = []
for i in range(n):
    temp.append(int(a[i]))
temp.sort()
for i in range(n-1):
    if(i==0):
        sumv =  (temp[i] + temp[i+1])/2
    else:
        sumv = (sumv + temp[i+1])/2
    print(sumv)
print(sumv)

n = int(input())
a = (input()).split(" ")
sumv = 0
temp = []
for i in range(n):
    temp.append(int(a[i]))
temp.sort()
for i in range(n-1):
    if(i==0):
        sumv =  (temp[i] + temp[i+1])/2
    else:
        sumv = (sumv + temp[i+1])/2
print(sumv)


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(float,input().split())) 
v.sort()
for i in range(N):
    v/=2**(N-i)

print(sum(v)+v[0])

N=int(input())
v=list(map(float,input().split())) 
v.sort()
for i in range(N):
    v[i]/=2**(N-i)

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

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

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

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):
a = int(input())
l=list(map(int ,input().split()))
ll=[]
for i in range(a-2):
        k=(l[i]+l[i+1])/2+(l[i+2])
        ll.append(k/2)

print(max(ll))

a = int(input())
l=(list(map(int ,input().split())))
l.sort()
ll=[]
j=0
r=(l[0]+l[1])/2
ll.append(r)
for i in range(a-2):
        j=(ll[0]+l[i+2])/2
        ll.append(j)
        ll.remove(ll[0])
print((ll[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
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def S(): return sys.stdin.readline().rstrip()

N = I()
v = LI()

v = sorted(v,reverse=True)

ans = 0

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

print(ans + v[N-1]//(2**(N-1)))

import sys
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def S(): return sys.stdin.readline().rstrip()

N = I()
v = LI()

v = sorted(v,reverse=True)

ans = 0

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

print(ans + v[N-1]/(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):
a = int(input())
b = list(map(int, input().split()))

while (len(b) > 1):
	b = sorted(b)
    x = b.pop(0)
    y = b.pop(0)
    b.append((x+y)/2)
print(b[0])
pass

a = int(input())
b = list(map(int, input().split()))
while (len(b) > 1):
    b = sorted(b)
    x = b.pop(0)
    y = b.pop(0)
    b.append((x+y)/2)
print(b[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()

for i in range(n-1):
    num1 = v.pop(0)
    num2 = v.pop(0)
    print(v)
    v.append( (num1+num2)/2)
    print(v)
print(v[0])


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

for i in range(n-1):
    v.sort()
    num1 = v.pop(0)
    num2 = v.pop(0)
    v.append( (num1+num2)/2)
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):
import decimal
n = int(input())
v = list(map(int, input().split()))
v = sorted(v)

temp = 0

for i in range(1,n):
    if i == 1:
        temp += decimal.Decimal((v[0] + v[1])/2)
    else:
        print(temp, v[i])
        sum_ = decimal.Decimal((temp + v[i])/2)

print(sum_)

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

temp = 0

for i in range(1,n):
    if i == 1:
        temp += decimal.Decimal((v[0] + v[1])/2)
    else:
        sum_ = decimal.Decimal((temp + v[i])/2)

print(sum_)


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

temp = 0
sum_ = 0

for i in range(1,n):
    if i == 1:
        temp += decimal.Decimal((v[0] + v[1])/2)
    else:
        temp = decimal.Decimal((temp + v[i])/2)

print(temp)

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(reverce=True)
for i in range(n-1):
    a=v.pop()
    b=v.pop()
    v.append((a+b)/2)
    v.sort(reverse=True)
print(v[0])

n=int(input())
v=list(map(int,input().split()))
v.sort(reverse=True)
for i in range(n-1):
    a=v.pop()
    b=v.pop()
    v.append((a+b)/2)
    v.sort(reverse=True)
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()) 
lst=list(map(int,input().split())) 
#print(lst) 
a=[] 

for i in range(n): 
	if i==0: 
		#print(lst[i]) 
		a.append(lst[i]) 
	else: 
		#print(lst[i]) 
		a.append((a[-1]+lst[i])/2)   
print(a[-1])

n=int(input()) 
lst=list(map(int,input().split()))
lst.sort()
#print(lst) 
a=[] 

for i in range(n): 
	if i==0: 
		#print(lst[i]) 
		a.append(lst[i]) 
	else: 
		#print(lst[i]) 
		a.append((a[-1]+lst[i])/2)   
print(a[-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())
V = list(map(int,input().split()))
ans=0
V.sort()
S=V[0]
for i in N-1:
  ans += (S + V[i])/2
  S=V[i]
print(ans)  


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

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


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


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

Verdicts: Runtime Error

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())
v = sorted(list(map(int, input().split())))
while len(v) > 1:
  v.insert(0, v.pop(0) + v.pop(0) / 2)
  print(v[0])


n = int(input())
v = sorted(list(map(int, input().split())))
while len(v) > 1:
  v.insert(0, (v.pop(0) + v.pop(0)) / 2)
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 = sorted(list(map(int, input().split())))
ans = V[0]
print(ans)
for i in range(1, n):
  ans = (V[i] + ans)/2

print(ans)

n = int(input())
V = sorted(list(map(int, input().split())))
print(ans)
for i in range(1, n):
  ans = (V[i] + ans)/2

print(ans)

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

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()))
while len(v)>1:
  v=v.sort()
  v1=v[0]
  del v[0]
  v2=v[0]
  del v[0]
  v.append((v1+v2)/2)
print(v)

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

N=int(input())
v=list(map(int,input().split()))
while len(v)>1:
  v=sorted(v)
  v_1=v.pop(0)
  v_2=v.pop(0)
  v.append((v_1+v_2)/2)
print(v)

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


N=int(input())
v=list(map(int,input().split()))
while len(v)>1:
  v=sorted(v)
  v_1=v.pop(0)
  v_2=v.pop(0)
  v.append((v_1+v_2)/2)
print(v[0])

Verdicts: Runtime Error

Runtime Error

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):
def Input():
    return mapt(int, input().split(" "))


def main():
    n = int(input())
    a = sorted(Input())
    while len(a) > 1:
        x, y = a[0], a[1]
        a[0:2] = [(x+y)/2]

    return a[0]


print(main())

def mapt(fn, *args):
    return tuple(map(fn, *args))


def Input():
    return mapt(int, input().split(" "))

  
def main():
    n = int(input())
    a = sorted(Input())
    while len(a) > 1:
        x, y = a[0], a[1]
        a[0:2] = [(x+y)/2]
    return a[0]


print(main())

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

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

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

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

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

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

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

Verdicts: Runtime Error

Runtime Error

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 = iin()
a = [int(x) for x in input().split()]
for i in range(n-1):
    a.sort()
    a[i + 1] = (a[i]+a[i + 1])/2
print(a[n-1])

n = int(input())
v = list(map(int,input().split()))
v = sorted(v)
for i in range(n-1):
    v[i+1] = (v[i] + v[i+1])/2
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):
import numpy as np

N = int(input())
vn_list = input().split()

vn = np.sort(np.array(vn_list, dtype="float"))[::-1]
print(vn)
w = np.logspace(1, N, base=1/2, num = N)
w[N-1] = w[N-2]
result = np.dot(vn, w)
print(result)

import numpy as np

N = int(input())
vn_list = input().split()

vn = np.sort(np.array(vn_list, dtype="float"))[::-1]
w = np.logspace(1, N, base=1/2, num = N)
w[N-1] = w[N-2]
result = np.dot(vn, w)
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())
value = [int(i) for i in input().split()]

now = 0
a = sorted(value) 

while len(a) >= 2:
  a = sorted(a)
  print(a)
  now = (float(a[0]) + float(a[1])) / float(2)
  print(now)
  if len(a) >= 3:
    del a[0],a[0] 
    a.append(now)
  elif len(a) == 2:
    print(now)
    break
    

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

now = 0
a = sorted(value) 

while len(a) >= 2:
  a = sorted(a)
  now = (float(a[0]) + float(a[1])) / float(2)
  if len(a) >= 3:
    del a[0],a[0] 
    a.append(now)
  elif len(a) == 2:
    print(now)
    break
    

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)

while len(V) > 1:
    V.append((V.pop(0) + V.pop(0)) // 2)
    V = sorted(V)

print(*V)

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

while len(V) > 1:
    V.append((V.pop(0) + V.pop(0)) / 2)
    V = sorted(V)

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):
input()
A = list(map(int, input().split()))
A.sort()
while len(A)>1:
  b = A[0:2]
  c = sum(i for i in b)
  A.append(c/2)
  del(A[0:2])
  A.sort()

print(A)

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

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

input()
A = list(map(int, input().split()))
A.sort()
while len(A)>1:
  b = A[0:2]
  c = sum(i for i in b)
  A.append(c/2)
  del(A[0:2])
  A.sort()
z = [str(a) for a in A]
z = ''.join(z)
print(z)

Verdicts: Wrong Answer

Runtime Error

Time Limit Exceeded

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)
a = v[0]
for i in range(1, N):
    a = (a+v[i])/2

print(a)

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

v.sort()
a = v[0]
for i in range(1, N):
    a = (a+v[i])/2

print(a)

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()
ans = a[0]
for i in range(1,n):
    ans = (ans+a[i])/2
pritn(ans)
    

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: 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_sorted = sorted(V, reverse = True)

ans = 0
for i in range(N):
    ans += V_sorted[i] /(2 ** (i+1))
    
print(ans)

N = int(input())
V = list(map(int, input().split()))
V_sorted = sorted(V, reverse = True)

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

Verdicts: Wrong Answer

Accepted