问题

### Instructions: You are an expert Python programmer. You will be given a question (problem specification) and will generate a correct Python program that matches the specification and passes all tests. You will NOT return anything except for the program.
### Question:
This is an interactive problem (a type of problem where your program interacts with the judge program through Standard Input and Output).
There are N bottles of juice, numbered 1 to N. It has been discovered that exactly one of these bottles has gone bad. Even a small sip of the spoiled juice will cause stomach upset the next day.
Takahashi must identify the spoiled juice by the next day. To do this, he decides to call the minimum necessary number of friends and serve them some of the N bottles of juice. He can give any number of bottles to each friend, and each bottle of juice can be given to any number of friends.
Print the number of friends to call and how to distribute the juice, then receive information on whether each friend has an upset stomach the next day, and print the spoiled bottle's number.

Input/Output

This is an interactive problem (a type of problem where your program interacts with the judge program through Standard Input and Output).
Before the interaction, the judge secretly selects an integer X between 1 and N as the spoiled bottle's number. The value of X is not given to you. Also, the value of X may change during the interaction as long as it is consistent with the constraints and previous outputs.
First, the judge will give you N as input.
N

You should print the number of friends to call, M, followed by a newline.
M

Next, you should perform the following procedure to print M outputs.
For i = 1, 2, \ldots, M, the i-th output should contain the number K_i of bottles of juice you will serve to the i-th friend, and the K_i bottles' numbers in ascending order, A_{i, 1}, A_{i, 2}, \ldots, A_{i, K_i}, separated by spaces, followed by a newline.
K_i A_{i, 1} A_{i, 2} \ldots A_{i, K_i}

Then, the judge will inform you whether each friend has a stomach upset the next day by giving you a string S of length M consisting of 0 and 1.
S

For i = 1, 2, \ldots, M, the i-th friend has a stomach upset if and only if the i-th character of S is 1.
You should respond by printing the number of the spoiled juice bottle X', followed by a newline.
X'

Then, terminate the program immediately.
If the M you printed is the minimum necessary number of friends to identify the spoiled juice out of the N bottles, and the X' you printed matches the spoiled bottle's number X, then your program is considered correct.

Input/Output

This is an interactive problem (a type of problem where your program interacts with the judge program through Standard Input and Output).
Before the interaction, the judge secretly selects an integer X between 1 and N as the spoiled bottle's number. The value of X is not given to you. Also, the value of X may change during the interaction as long as it is consistent with the constraints and previous outputs.
First, the judge will give you N as input.
N

You should print the number of friends to call, M, followed by a newline.
M

Next, you should perform the following procedure to print M outputs.
For i = 1, 2, \ldots, M, the i-th output should contain the number K_i of bottles of juice you will serve to the i-th friend, and the K_i bottles' numbers in ascending order, A_{i, 1}, A_{i, 2}, \ldots, A_{i, K_i}, separated by spaces, followed by a newline.
K_i A_{i, 1} A_{i, 2} \ldots A_{i, K_i}

Then, the judge will inform you whether each friend has a stomach upset the next day by giving you a string S of length M consisting of 0 and 1.
S

For i = 1, 2, \ldots, M, the i-th friend has a stomach upset if and only if the i-th character of S is 1.
You should respond by printing the number of the spoiled juice bottle X', followed by a newline.
X'

Then, terminate the program immediately.
If the M you printed is the minimum necessary number of friends to identify the spoiled juice out of the N bottles, and the X' you printed matches the spoiled bottle's number X, then your program is considered correct.

Constraints


- N is an integer.
- 2 \leq N \leq 100



### Format: Read the inputs from stdin solve the problem and write the answer to stdout (do not directly test on the sample inputs). Enclose your code within delimiters as follows.
```python
# YOUR CODE HERE
```

### Answer: (use the provided format with backticks)

答案

模型 模型答案 正确答案
金声玉亮2.0 ```python
import math

def find_spoiled_juice():
N = int(input())
M = math.ceil(math.log2(N))
print(M)

for i in range(M):
bottles = [j + 1 for j in range(N) if (j >> i) & 1]
print(len(bottles), *bottles)

S = input().strip()
spoiled_bottle = 0
for i in range(M):
if S[i] == '1':
spoiled_bottle |= (1 << i)

print(spoiled_bottle + 1)

find_spoiled_juice()
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
deepseek-r1

```python
# YOUR CODE HERE
n = int(input())
m = (n - 1).bit_length()
print(m)
for j in range(m):
group = []
for i in range(1, n + 1):
if ((i - 1) >> j) & 1:
group.append(i)
print(len(group), *group)
s = input().strip()
x = 0
for j in range(len(s)):
if s[j] == '1':
x += (1 << j)
print(x + 1)
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
gpto1 ```python
# YOUR CODE HERE
import math

import sys

def main():
import sys
input = sys.stdin.readline
N = int(sys.stdin.readline())
M = math.ceil(math.log2(N))
friends_bottles = [[] for _ in range(M)]
for bottle in range(1, N+1):
adjusted_bottle = bottle - 1
bits = format(adjusted_bottle, '0{}b'.format(M))
for i in range(M):
if bits[i] == '1':
friends_bottles[i].append(bottle)
print(M)
for i in range(M):
Ki = len(friends_bottles[i])
bottles = friends_bottles[i]
print(Ki, *bottles)
sys.stdout.flush()
S = sys.stdin.readline().strip()
bottle_number = int(S, 2) + 1
print(bottle_number)
sys.stdout.flush()

if __name__ == "__main__":
main()
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
gpt-4o-2024-08-06 ```python
# YOUR CODE HERE
import sys
input = sys.stdin.read

def main():
# Read the number of bottles
N = int(input().strip())

# Calculate the minimum number of friends needed
# We need to find the smallest M such that 2^M >= N
M = 0
while (1 << M) < N:
M += 1

# Print the number of friends
print(M)

# Distribute the bottles to friends
for i in range(M):
# For each friend, we will give them bottles where the i-th bit is set
bottles = []
for j in range(1, N + 1):
if (j >> i) & 1:
bottles.append(j)

# Print the number of bottles and the bottles themselves
print(len(bottles), ' '.join(map(str, bottles)))

# Read the response from the judge
S = input().strip()

# Determine the spoiled bottle
spoiled_bottle = 0
for i in range(M):
if S[i] == '1':
spoiled_bottle |= (1 << i)

# The spoiled_bottle is 0-indexed, we need to add 1 to make it 1-indexed
print(spoiled_bottle + 1)

if __name__ == "__main__":
main()
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
DeepSeek-Coder-V2-Lite-Instruct ```python
# YOUR CODE HERE
import sys

# Read the number of bottles N
N = int(sys.stdin.readline().strip())

# Binary search to find the spoiled bottle
low, high = 1, N
while low < high:
mid = (low + high) // 2
print(1) # Call 1 friend
sys.stdout.flush()
# Serve mid bottle to the friend
print(mid)
sys.stdout.flush()
# Read the result from the judge
result = sys.stdin.readline().strip()
if result == '1':
high = mid
else:
low = mid + 1

# The spoiled bottle is found
print(low)
sys.stdout.flush()
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
Meta-Llama-3.1-405B-Instruct-Turbo ```python
import math

def get_sick_friend(friends):
for i, friend in enumerate(friends):
if friend == '1':
return i
return -1

def get_sick_bottle(bottles, friends):
sick_friend = get_sick_friend(friends)
if sick_friend == -1:
return -1
return bottles[sick_friend][0]

def get_bottles(n):
num_friends = math.ceil(math.log2(n))
bottles = []
for i in range(num_friends):
friend_bottles = []
for j in range(n):
if (j >> i) & 1:
friend_bottles.append(j + 1)
bottles.append(friend_bottles)
return bottles

def main():
n = int(input())
bottles = get_bottles(n)
print(len(bottles))
for bottle in bottles:
print(len(bottle), *bottle)
friends = input()
sick_bottle = get_sick_bottle(bottles, friends)
print(sick_bottle)

if __name__ == "__main__":
main()
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
Meta-Llama-3.1-70B-Instruct-Turbo ```python
import sys
import math

n = int(input())

m = math.ceil(math.log2(n))

print(m)

friends = [[] for _ in range(m)]

for i in range(n):
for j in range(m):
if (i >> j) & 1:
friends[j].append(i + 1)

for friend in friends:
print(len(friend), *friend)

s = input()

res = 0
for i in range(m):
if s[i] == '1':
res += 2 ** i

print(res + 1)
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
Qwen2-72B-Instruct ```python
import math

def main():
n = int(input())
m = math.ceil(math.log2(n))

print(m)
print flush()

for i in range(m):
start = 1 << i
end = min(1 << (i + 1), n + 1)
print(end - start, *range(start, end))
print flush()

s = raw_input()

x = 0
for i in range(m):
if s[i] == '1':
x += 1 << i

print(x + 1)
print flush()

if __name__ == '__main__':
main()
```
Note: The `flush` function is used to ensure that the output is flushed immediately after printing, which is necessary for interactive problems. However, `flush` is not a built-in function in Python 3, so you might need to use `sys.stdout.flush()` instead, depending on the environment. Also, `raw_input` is not available in Python 3, so you would need to use `input()` instead.		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
baichuan4 To solve this problem, we need to find the minimum number of friends (M) required to identify the spoiled juice bottle among N bottles. The key insight is to use a binary search approach to divide the set of bottles into two halves and observe which half causes stomach upset. By recursively applying this process, we can narrow down the spoiled bottle efficiently.

Here's the Python program that implements this strategy:

```python
def find_spoiled_bottle(n):
if n == 1:
return 1
mid = n // 2
print(1)
print(f"{mid} 1 2 3 ... {mid}")
stomach_upset = input()
if stomach_upset == '1':
return find_spoiled_bottle(mid)
else:
return find_spoiled_bottle(n - mid) + mid

N = int(input())
M = find_spoiled_bottle(N)
print(M)
```

This program defines a recursive function `find_spoiled_bottle` that takes the number of bottles `n` and returns the number of the spoiled bottle. If `n` is 1, it means there's only one bottle left, so it must be the spoiled one. Otherwise, it divides the bottles into two halves, serves the first half to a friend, and checks if the friend has a stomach upset. Depending on the result, it recursively searches either the first or second half.

The main part of the program reads the initial number of bottles `N`, calls the `find_spoiled_bottle` function to find the spoiled bottle, and prints the result.

Please note that this is a simplified version of the solution and assumes that the input/output operations work as described. In a real-world scenario, you might need to handle exceptions and validate inputs more thoroughly.		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
doubao ```python
import math

N = int(input())
M = math.ceil(math.log2(N))

print(M)

for i in range(M):
print(1, i + 1)

S = input()

for i in range(M):
if S[i] == '1':
print(i + 1)
break
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
gpto1 ```python
# YOUR CODE HERE
import math

import sys

def main():
import sys
input = sys.stdin.readline
N = int(sys.stdin.readline())
M = math.ceil(math.log2(N))
friends_bottles = [[] for _ in range(M)]
for bottle in range(1, N+1):
adjusted_bottle = bottle - 1
bits = format(adjusted_bottle, '0{}b'.format(M))
for i in range(M):
if bits[i] == '1':
friends_bottles[i].append(bottle)
print(M)
for i in range(M):
Ki = len(friends_bottles[i])
bottles = friends_bottles[i]
print(Ki, *bottles)
sys.stdout.flush()
S = sys.stdin.readline().strip()
bottle_number = int(S, 2) + 1
print(bottle_number)
sys.stdout.flush()

if __name__ == "__main__":
main()
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
kimi ```python
import math

# YOUR CODE HERE
n = int(input())
m = int(math.log2(n))
print(m + 1)
for i in range(m):
print(1, 2**i + 1)
print(1, n - 2**m + 1)
s = input()
spoiled = 0
for i in range(m):
if s[i] == '1':
spoiled |= 2**i
if s[-1] == '1':
spoiled |= 1 << m
print(spoiled + 1)
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
minimax ```python
# YOUR CODE HERE
import math

def find_min_friends(N):
# The minimum number of friends required is the base-2 logarithm of N, rounded up.
return math.ceil(math.log2(N))

def find_spoiled_bottle(N, friends_info):
spoiled_bottle = 0
for friend_info in friends_info:
for bottle in friend_info[1]:
spoiled_bottle ^= bottle
return spoiled_bottle

# Read the number of bottles N from stdin
N = int(input())

# Print the number of friends to call
M = find_min_friends(N)
print(M)

# Prepare to receive information about which friends have stomach upsets
friends_info = []

# For each friend, receive the number of bottles they were given and the bottles' numbers
for i in range(M):
K_i, *A_i = map(int, input().split())
friends_info.append((K_i, A_i))

# Receive information about which friends have stomach upsets
S = input()

# Print the number of the spoiled juice bottle
X_prime = find_spoiled_bottle(N, friends_info)
print(X_prime)
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
wenxinyiyan ```python
# Read the number of bottles N from stdin
N = int(input())

# Calculate the minimum number of friends M required to identify the spoiled juice
# Using binary encoding, we need enough bits to represent all N bottles
M = len(bin(N-1)) - 2 # Subtract 2 to exclude the '0b' prefix

# Print the number of friends M to call
print(M)

# Prepare a list to store the juice distribution for each friend
juice_distribution = []

# Distribute the juice among the friends using binary encoding
for i in range(M):
distribution = []
for j in range(N):
# Check if the j-th bit is set in the i-th position of the binary representation
if (j >> i) & 1:
distribution.append(j + 1) # Add the bottle number (offset by 1 since bottles are numbered from 1 to N)
juice_distribution.append(distribution)
# Print the distribution for the i-th friend
print(len(distribution), *distribution)

# Read the stomach upset information for each friend as a string of 0s and 1s
stomach_upset = input()

# Identify the spoiled juice bottle based on the stomach upset information
spoiled_bottle = 0
for i in range(M):
if stomach_upset[i] == '1':
# If the i-th friend has a stomach upset, set the i-th bit in the spoiled_bottle variable
spoiled_bottle |= (1 << i)

# Print the number of the spoiled juice bottle (offset by 1 since bottles are numbered from 1 to N)
print(spoiled_bottle + 1)
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
zhipu ```python
import sys
import math

def main():
input = sys.stdin.read
data = input().split()

N = int(data[0])

# Calculate the minimum number of friends needed using binary representation
M = math.ceil(math.log2(N))
print(M)

# Prepare the distribution of bottles to friends
# Each friend will be responsible for bottles where the bit at their position is 1
for i in range(M):
bottles = []
for j in range(1, N + 1):
if (j >> i) & 1:
bottles.append(j)
print(len(bottles), *bottles)

# Read the results from the judge
S = input().strip()

# Determine the spoiled bottle using the results
spoiled_bottle = 0
for i in range(M):
if S[i] == '1':
spoiled_bottle |= (1 << i)

print(spoiled_bottle + 1)

if __name__ == "__main__":
main()
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]
deepseek-v3 ```python
# YOUR CODE HERE
import sys
import math

def main():
# Read N
N = int(sys.stdin.readline())

# Calculate the minimum number of friends needed
M = math.ceil(math.log2(N))

# Print M
print(M)
sys.stdout.flush()

# Prepare the distribution of bottles to friends
# Each friend corresponds to a bit in the binary representation of the bottle number
distribution = []
for i in range(M):
bottles = []
for j in range(1, N+1):
if (j >> i) & 1:
bottles.append(j)
distribution.append(bottles)

# Print the distribution
for bottles in distribution:
print(len(bottles), end=' ')
print(' '.join(map(str, sorted(bottles))))
sys.stdout.flush()

# Read the response S
S = sys.stdin.readline().strip()

# Determine the spoiled bottle
X = 0
for i in range(M):
if S[i] == '1':
X |= (1 << i)

# Print the spoiled bottle number
print(X)
sys.stdout.flush()

if __name__ == "__main__":
main()
```		 [{"input": "3 1\n", "output": "2\n1 2 \n1 3 \n2\n", "testtype": "stdin"}]