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):
a = int(input())
n = list(map(int, input().split()))
b=sorted(n,reverse=True)
ans=0
for i,j in enumerate (b):
    print(i,j)
    if i < a-1:
        ans += j*((1/2)**(i+1))
    if i ==a-1:
        ans += j*((1/2)**(i))

a = int(input())
n = list(map(int, input().split()))
b=sorted(n,reverse=True)
ans=0
for i,j in enumerate (b):
    print(i,j)
    if i < a-1:
        ans += j*((1/2)**(i+1))
    if i ==a-1:
        ans += j*((1/2)**(i))
print(ans)

a = int(input())
n = list(map(int, input().split()))
b=sorted(n,reverse=True)
ans=0
for i,j in enumerate (b):
    if i < a-1:
        ans += j*((1/2)**(i+1))
    if i ==a-1:
        ans += j*((1/2)**(i))
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())
v = list(map(int, input().split()))
vsort = sorted(v)

for i in range(N):
    ans = (vsort[0]+vsort[1])/2
    if N <= 2:
        print(ans)
    else:
        ans = (ans + vsort[i])/2  # first term
        
print(ans)

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

for i in range(N):
    ans = (vsort[0]+vsort[1])/2
    if N <= 2:
        print(ans)
    else:
        ans = (ans + vsort[i])/2  # first term
        
print(int(ans))

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

for i in range(1, N, 1):
    ans = (ans + v[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())
v = list(map(int, input().split()))

v1 = v.sort().reverse()
k = 0

for i in range(n):
  k += v[i] * (2 ** ( - (i + 1)))

print(k + v[n] * (2 ** ( - (n + 1)))

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

v1 = list(sorted(v))
v2 = list(reversed(v1))
k = 0

for i in range(n):
  k += v2[i] * (2 ** ( -i - 1))

k += v2[n - 1] * (2 ** ( -n)) 

print(k)

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())
value= list(map(int,input().split()))
value.sort()
for i in range(N):
  if i<N-1:
  	value[0]=(value[0]+value[1])/2
  	value.pop(1)
  if i==N-1:
    pass
print(int(value[0]))  

N=int(input())
value= list(map(int,input().split()))
value.sort()
for i in range(N):
  if i<N-1:
  	value[0]=(value[0]+value[1])/2
  	value.pop(1)
  if i==N-1:
    pass
print(float(value[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):


#include <algorithm>

#include <set>
#include <map>
#include <string>
#include <cmath>
#include <climits>
#include <iomanip>

using namespace std;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

typedef long long LL;
typedef vector<int> VI;

int main() {
  int N;
  cin >> N;
  priority_queue<double, vector<double>,  greater<double>> q;
  int v;
  for (int i = 0; i < N; i++) {
    cin >> v;
    q.push((double)v);
  }
  while (q.size() > 1) {
    double v1 = q.top();
    q.pop();
    double v2 = q.top();
    q.pop();
    q.push((v1+v2)/2);
  }
  cout << setprecision(10) << q.top() << endl;
  return 0;
}




#include <algorithm>

#include <set>
#include <map>
#include <string>
#include <cmath>
#include <climits>
#include <iomanip>

using namespace std;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

typedef long long LL;
typedef vector<int> VI;

int main() {
  int N;
  cin >> N;
  priority_queue<double, vector<double>,  greater<double>> q;
  int v;
  for (int i = 0; i < N; i++) {
    cin >> v;
    q.push((double)v);
  }
  while (q.size() > 1) {
    double v1 = q.top();
    q.pop();
    double v2 = q.top();
    q.pop();
    q.push((v1+v2)/2);
  }
  cout << fixed << setprecision(10) << q.top() << endl;
  return 0;
}


import heapq

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

heapq.heapify(q)

while (len(q) > 1):
    v1 = heapq.heappop(q)
    v2 = heapq.heappop(q)
    heapq.heappush(q, (v1+v2)/2.0)

print(heapq.heappop(q))



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

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

import heapq

heapq.heapify(V)

while len(V) >= 2:
    x = heapq.heappop(V)
    y = heapq.heappop(V)
    heapq.heappush(V, (x + y) // 2)

print(V[0])






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

import heapq

heapq.heapify(V)

while len(V) >= 2:
    x = heapq.heappop(V)
    y = heapq.heappop(V)
    heapq.heappush(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):
##C problem
N = int(input())
v = list(map(float, input().split()))

a=min(v)

#v.remove(a)
#v.remove(b)
#v.append( (a+b)/2 )

for i in range(N-1):
        a=v[-1]
        remove(a)
        b=min(v)
        v.remove(b)
        v.append((a+b)/2)

##C problem
N = int(input())
v = list(map(float, input().split()))

vs = sorted(v)

for i in range(N-1):

    a=vs[0]

    b=vs[1]
    c=(a+b)/2
    vs.remove(b)
    vs.insert(0,c)
    vs.remove(a)
    #print(i)
    #print(vs)

print(vs[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())
listV =  list(map(int,input().split()))
sortedListV = sorted(listV)
result = (sortedListV[0] + sortedListV[1]) / 2

for i range (0,3):
    result = (sortedListV[i] + sortedListV[i+1]) / 2
print(result)

n = int(input())
listV =  list(map(int,input().split()))
sortedListV = sorted(listV)
result = (sortedListV[0] + sortedListV[1]) / 2

for i range (2, n-1):
    result += (sortedListV[i] + sortedListV[i+1]) / 2
print(result)

n = int(input())
listV =  list(map(int,input().split()))
sortedListV = sorted(listV)
result = (sortedListV[0] + sortedListV[1]) / 2

for i range (2, n-1):
    result = (sortedListV[i] + sortedListV[i+1]) / 2
print(result)

n = int(input())
listV =  list(map(int,input().split()))
sortedListV = sorted(listV)
result = (sortedListV[0] + sortedListV[1]) / 2

for i range (2, n):
    result += (result + sortedListV[i]) / 2
print(result)

n = int(input())
listV =  list(map(int,input().split()))
sortedListV = sorted(listV)
result = (sortedListV[0] + sortedListV[1]) / 2

for i range (2, n-1):
    result += (result + sortedListV[i]) / 2
print(result)

n = int(input())
listV =  list(map(int,input().split()))
sortedListV = sorted(listV)

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

n = int(input())
listV =  list(map(int,input().split()))
sortedListV = sorted(listV)

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

Verdicts: Runtime Error

Runtime Error

Runtime Error

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

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


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

Verdicts: Runtime Error

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

Submission attempts (in order):
import heapq

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

def nabe(v):
    
    if len(v) == 1:
        return v[0]

    smallests = heapq.nsmallest(2, v)

    print(smallests)

    for s in smallests:
        v.remove(s)
    v.append(sum(smallests) / 2)

    return nabe(v)

print(nabe(v))

import heapq

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

def nabe(v):
    
    if len(v) == 1:
        return v[0]

    smallests = heapq.nsmallest(2, v)

    for s in smallests:
        v.remove(s)
    v.append(sum(smallests) / 2)

    return nabe(v)

print(nabe(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,*a=map(int,open(0).read().split())
for i in range(n-1):
  a.sort()
  b=(a[0]+a[1])/2
  a[:2]=[b]
print(int(a[0])) 

n,*a=map(int,open(0).read().split())
a.sort()
tmp=(a[0]+a[1])/2
for i in range(2,n):
  tmp=(tmp+a[i])/2
print(int(tmp))

n,*a=map(int,open(0).read().split())
for i in range(n-1):
  a.sort()
  _min=a[-1]
  jj,kk=0,0
  for j,k in zip(a,a[1:]):
    if _min>min(_min,k-j):
      _min=min(_min,k-j)
      jj,kk=j,k  
  a.remove(jj)
  a.remove(kk)
  a.append((jj+kk)/2)
print(int(a[0]))    
    
    
  

n,*a=map(int,open(0).read().split())
a.sort()
tmp=(a[0]+a[1])/2
for i in range(2,n):
  tmp=(tmp+a[i])/2
print(tmp)

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 = list(map(int, input().split()))
V.sort()
V.reverse()

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

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

if len(V) == 2:
  print((V[0]+V[1])/2)
else:
  ans = 0
  for i in range(n-1):
    ans += V[i]/(2**(i+1))
  ans += V[n-1]/(2**(n-1))
  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.sort()

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

print(v[n-1])

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

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

print(v[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):
def C():
    N = int(input())
    values = list(map(int, input().split()))

    values = values.sort()
    total_value = [values[i]*((1/2)**i) for i in range(N)]    
    
    print(total_value)

    
if __name__ == '__main__':
    C()


def C():
    N = int(input())
    values = list(map(int, input().split()))

    values = sorted(values, reverse=True)
    total_value = sum([values[i]*((1/2)**i) for i in range(N)])

    print(total_value)


if __name__ == '__main__':
    C()


def C():
    N = int(input())
    values = list(map(int, input().split()))

    values = sorted(values, reverse=True)
    weights = [(1/2)**(i+1) for i in range(N-1)] + [(1/2)**(N-1)]

    total_value = sum([values[i]*weights[i] for i in range(N)])

    print(total_value)


if __name__ == '__main__':
    C()


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()))
A.sort()
while len(A)>=2: 
    X = int((A[0]+A[1])/2)
    A.pop(0)
    A.pop(0)
    A.append(X)
    A.sort()
else:
    print(*A)

N = int(input())
A = list(map(int, input().split()))
A.sort()
while len(A)>=2: 
    X = (A[0]+A[1])/2
    A.pop(0)
    A.pop(0)
    A.append(X)
    A.sort()
else:
    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):
if __name__ == '__main__':
    n = int(input())

    Arrnums = sorted([float(x) for x in input().split(' ')])
    j = 1
    print(*Arrnums)
    q = Arrnums[0]
    for i in range(j, n):
        q = (Arrnums[i] + q)/2

    print(q)


if __name__ == '__main__':
    n = int(input())
    Arrnums = sorted([float(x) for x in input().split(' ')])
    j = 1
    q = Arrnums[0]
    for i in range(j, n):
        q = (Arrnums[i] + q)/2

    print(q)

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()))
sum = 0

a.sort()
a_1 = (a[0] + a[1])/2
b = [a_1]

for i in range(2, n):
    b.append((b[-1]+a[i])/2)

print(int(b[-1]))

n = int(input())
a = list(map(int,  input().split()))
sum = 0

a.sort()
a_1 = (a[0] + a[1])/2
b = [a_1]

for i in range(2, n):
    b.append((b[-1]+a[i])/2)

print(b[-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()
n=int(n)
val=list(map(int,input().strip().split(' ')))
val.sort()
res=0
if len(val)>0:
    if len(val)==1:
        print(val[0]/2)
    elif len(val)==2:
        print(sun(val)/2)
    else:
        #res=(val[0]+val[1])/2
        res=(val[0]+val[1])/2
        for i in range(2,len(val)):
            res=(res+val[i])/2
            #print(res)
        print(res)
else:
    #print('error')
    pass

n=input()
n=int(n)
val=list(map(int,input().strip().split(' ')))
val.sort()
res=0
if val:
    if len(val)==1:
        print(val[0]/2)
    elif len(val)==2:
        print(sun(val)/2)
    else:
        #res=(val[0]+val[1])/2
        res=(val[0]+val[1])/2
        for i in range(2,len(val)):
            res=(res+val[i])/2
            #print(res)
        print(res)
else:
    #print('error')
    pass

n=input()
n=int(n)
val=list(map(int,input().strip().split(' ')))
val.sort()
res=0
if val and n>=2:
    if len(val)==1:
        print(val[0]/2)
    elif len(val)==2:
        print((val[0]+val[1])/2)
    else:
        #res=(val[0]+val[1])/2
        res=(val[0]+val[1])/2
        for i in range(2,len(val)):
            res=(res+val[i])/2
           # print(res)
        print(res)
else:
    #print('error')
    pass

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 = input()
v = sorted(map(int, input().split()))
vt = v[0]
vt = list(map(lambda x:(vt+x)/2, v[1:]))
for vn in v[1:]:
	vt = (vt + vn)/2
print(vt)

n = input()
v = sorted(map(int, input().split()))
vt = v[0]
for vn in v[1:]:
	vt = (vt + vn)/2
print(vt)

Verdicts: Runtime Error

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

Submission attempts (in order):
n = int(input())
v = sorted(list(map(int, input().split())), reverse=True)
ans = 0
print(v)
for i, j in enumerate(v):
  if i!=len(v)-1:
    ans += j/(2**(i+1))
  else:
    ans += j/(2**(i))
print(ans)

n = int(input())
v = sorted(list(map(int, input().split())), reverse=True)
ans = 0
for i, j in enumerate(v):
  if i!=len(v)-1:
    ans += j/(2**(i+1))
  else:
    ans += j/(2**(i))
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=sorted(list(map(int,input().split())))
for i in range(N-1):
  v[i+1]=(v[i]+v[i+1])/2

print(int(v[N-1]))
  

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

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

N = int(input())
a = sorted(list(map(int, input().split())))
b = (a[0] + a[1]) / 2
if N > 2:
  for c in a[2:]:
    b = (b + c) / 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):
def calculate(arr):

    if len(arr) == 1:
        return arr[0]

    if len(arr) == 2:
        return (arr[0] + arr[1]) / 2

    result = (calculate(arr[:-2]) + (arr[-1] + arr[-2])/2) / 2

    return result

def rank(arr):
    lastIndex = len(arr) - 1
    while lastIndex > 0:
        for i in range(lastIndex):
            if arr[i + 1] > arr[i]:
                tmp = arr[i + 1]
                arr[i + 1] = arr[i]
                arr[i] = tmp

        lastIndex = lastIndex - 1
    return arr

N = int(input())
S = input().split(" ")
arr = list(map(int, list(S)))

arr2 = rank(arr)

result = calculate(list(arr2))
print(int(result))

def calculate(arr):

    if len(arr) == 1:
        return arr[0]
    if len(arr) == 2:
        result = (arr[0] + arr[1]) / 2
        return result

    if len(arr) > 2:
        res = (arr[-1] + arr[-2]) / 2
        newArr = arr[:-2]
        newArr.append(res)
        return calculate(newArr)

def rank(arr):
    lastIndex = len(arr) - 1
    while lastIndex > 0:
        for i in range(lastIndex):
            if arr[i + 1] > arr[i]:
                tmp = arr[i + 1]
                arr[i + 1] = arr[i]
                arr[i] = tmp

        lastIndex = lastIndex - 1
    return arr

N = int(input())
S = input().split(" ")
arr = list(map(int, list(S)))
newArr = rank(arr)

result = calculate(list(arr))
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):
import numpy as np

N = input()
N = int(N)
stN = str(N)
p = input().split()
p_l = [int(s) for s in p]

p_sort = sorted(p_l,reverse=True)
print(p_sort)
for i in range(0,N-1):
    x = p_sort.pop()
    y = p_sort.pop()
    z = (x+y)/2
    p_sort.append(z)
    #print(p_sort)

print(p_sort[0])

import numpy as np

N = input()
N = int(N)
stN = str(N)
p = input().split()
p_l = [int(s) for s in p]

p_sort = sorted(p_l,reverse=True)
#print(p_sort)
for i in range(0,N-1):
    x = p_sort.pop()
    y = p_sort.pop()
    z = (x+y)/2
    p_sort.append(z)
    #print(p_sort)

print(p_sort[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())
lst = list(map(int,input().split(" ")))
lst.sort()
print(lst)
val = min(lst)
for i in range(1,len(lst)):
  print("i=%d" %i)
  val += (2**(i-1))*lst[i]
  print(val)
print(val/(2**(N-1)))

N = int(input())
lst = list(map(int,input().split(" ")))
lst.sort()
val = min(lst)
for i in range(1,len(lst)):
  val += (2**(i-1))*lst[i]
print(val/(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 = int(input())
v = sotred(list(map(int,input().split())))


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

print(v[n-1])


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

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

v.sort()

while len(v) > 1:
    tmp = v[0] + v[1]
    del v[1]
    v[0] = tmp / 2
print(v)

N = int(input())

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

v.sort()

while len(v) > 1:
    tmp = v[0] + v[1]
    del v[1]
    v[0] = tmp / 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()
r = v[0]
for i in r:
  r = (r + i) / 2
print(r)

n = int(input())
v = [int(i) for i in input().split()]
v.sort()
ans = (v[0] + v[1]) / 2
for i in range(2, len(v)):
    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())
Values=sorted(list(map(int,input().split())))
i=1
while i<=n-1:
    x=Values.pop(0)
    y=Values.pop(0)
    new=(x+y)/2
    Values.append(new)
    Values=sorted(Values)
    i+=1
print(int(*Values))

n=int(input())
Values=list(map(int,input().split()))
Values=sorted(Values)
i=1
while i<=n-1:
    x=Values.pop(0)
    y=Values.pop(0)
    new=(x+y)/2
    Values.append(new)
    Values=sorted(Values)
    i+=1
print(int(*Values))

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

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()))

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

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

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

Verdicts: Wrong Answer

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

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

for i in range(N):
    v[i] = float(v[i])
    
v.sort()
ite = 0
while(ite < N-1):
    v[ite+1] = (v[ite]+v[ite+1])/2
    v[ite] = 0
    v.sort()
    print(v)
    ite += 1

print(v[-1])



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

SUM = 0
for i in range(N):
    v[i] = int(v[i])
v.sort()
ite = 0
while(ite < N-1):
    v[ite+1] = (int(v[ite])+int(v[ite+1]))/2
    v.sort()
    print(v)
    ite += 1

print(v[-1])

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

for i in range(N):
    v[i] = float(v[i])
    
v.sort()
ite = 0
while(ite < N-1):
    v[ite+1] = (v[ite]+v[ite+1])/2
    v[ite] = 0
    v.sort()
    #print(v)
    ite += 1

print(v[-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 = list(map(int, input().split()))
v.sort()

while len(v) > 1:
    new = (v[0] + v[1])/2
    del v[0:2]
    v.insert(0, new)

print(v)

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

while len(v) > 1:
    new = (v[0] + v[1])/2
    del v[0:2]
    v.insert(0, new)

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 sys
import math
import itertools
import collections
import heapq
import re
import numpy as np

rr = lambda: str(sys.stdin.buffer.readline().rstrip())
rs = lambda: map(str, sys.stdin.buffer.readline().split())
ri = lambda: int(sys.stdin.buffer.readline())
rm = lambda: map(int, sys.stdin.buffer.readline().split())
rl = lambda: list(map(int, sys.stdin.buffer.readline().split()))

n = ri()
a = rl()
def heapsort(l):
  li = []
  for v in l:
    heapq.heappush(li, v)
  return [heapq.heappop(li) for _ in range(len(li))]
a = heapsort(a)
print(a)
ans = a[0]
for i in a:
  ans = (ans + i)/2
print(ans)


import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np

rr = lambda: str(sys.stdin.buffer.readline().rstrip())
rs = lambda: map(str, sys.stdin.buffer.readline().split())
ri = lambda: int(sys.stdin.buffer.readline())
rm = lambda: map(int, sys.stdin.buffer.readline().split())
rl = lambda: list(map(int, sys.stdin.buffer.readline().split()))

n = ri()
a = rl()
def heapsort(l):
  li = []
  for v in len(l):
    heappush(li, v)
  return [heappop(li) for _ in range(li)]
a = heapsort(a)
a.reverse()
ans = a[0]
for i in a:
  ans = (ans + i)/2
print(ans)


import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np

rr = lambda: str(sys.stdin.buffer.readline().rstrip())
rs = lambda: map(str, sys.stdin.buffer.readline().split())
ri = lambda: int(sys.stdin.buffer.readline())
rm = lambda: map(int, sys.stdin.buffer.readline().split())
rl = lambda: list(map(int, sys.stdin.buffer.readline().split()))

n = ri()
a = rl()
def heapsort(l):
  li = []
  for v in l:
    heapq.heappush(li, v)
  return [heapq.heappop(li) for _ in range(len(li))]
a = heapsort(a)
ans = a[0]
for i in a:
  ans = (ans + 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=sorted(list(map(float,input().split())))
z = v[0]
for i in range(1,n):
    print(i)
    z=(z+v[i])/2
print(z)

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

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

sum = 0
a = 0

if N == 1:
    sum += v_list[0]

elif N == 2:
    sum += (v_list[0] + v_list[1])/2

else:
    a = v_list[0] + v_list[1]
    for i in range(N-1):
        sum += a *2 ** (-N + i +1)
     
print(sum)

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

sum = 0
a = 0

if N == 1:
    sum += v_list[0]

elif N == 2:
    sum += (v_list[0] + v_list[1])/2

else:
    sum += (v_list[0] + v_list[1]) * 2 ** (-N + 1)
    for i in range(N-2):
        sum += v_list[i+2] * 2 ** (-N + i + 2)
        
print(sum)

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())
D = sorted([int(i) for i in input().split()])
D = [D[i]*2**max(0,i-1) for i in range(N)]
print(D)
print(sum(D)/2**(N-1))

N = int(input())
D = sorted([int(i) for i in input().split()])
D = [D[i]*2**max(0,i-1) for i in range(N)]
print(sum(D)/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 = int(input())
a = list(map(int,input().split()))
a = sorted(a)
res = a[0]
for i in range(1, n):
    res += a[i]
    res /= 2
print(resz)

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

Verdicts: Runtime Error

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

Submission attempts (in order):
n = input()
values = list((input()).split(" "))
v = list(map(lambda x: int(x), values))
result = 0
while len(v) > 0 :
    el = v.pop(0)
    result = (result + el)/2
print(result)

n = input()
values = list((input()).split(" "))

v = sorted(list(map(lambda x: int(x), values)))

result = v.pop(0)
while len(v) > 0 :
    el = v.pop(0)
    result = (result + el)/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())
arr = list(map(int, input().split()))

for _ in range(n - 1):
    arr.sort()
    new_v = (arr[0] + arr[1]) / 2
    del arr[0:2]
    arr.append(new_v)

print(int(arr[0]))


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

for _ in range(n - 1):
    arr.sort()
    new_v = (arr[0] + arr[1]) / 2
    del arr[0:2]
    arr.append(new_v)

print(arr[0])


Verdicts: Wrong Answer

Accepted