金声玉亮2.0模型评测报告

问题

You are an expert Python programmer, and here is your task: Write a function to find the similar elements from the given two tuple lists. Your code should pass these tests:

assert similar_elements((3, 4, 5, 6),(5, 7, 4, 10)) == (4, 5)
assert similar_elements((1, 2, 3, 4),(5, 4, 3, 7)) == (3, 4)
assert similar_elements((11, 12, 14, 13),(17, 15, 14, 13)) == (13, 14)

[BEGIN]
'def similar_elements(test_tup1, test_tup2):
res = tuple(set(test_tup1) & set(test_tup2))
return (res)'
[DONE]

You are an expert Python programmer, and here is your task: Write a python function to identify non-prime numbers. Your code should pass these tests:

assert is_not_prime(2) == False
assert is_not_prime(10) == True
assert is_not_prime(35) == True

[BEGIN]
'import math
def is_not_prime(n):
result = False
for i in range(2,int(math.sqrt(n)) + 1):
if n % i == 0:
result = True
return result'
[DONE]

You are an expert Python programmer, and here is your task: Write a function to find the largest integers from a given list of numbers using heap queue algorithm. Your code should pass these tests:

assert heap_queue_largest( [25, 35, 22, 85, 14, 65, 75, 22, 58],3)==[85, 75, 65]
assert heap_queue_largest( [25, 35, 22, 85, 14, 65, 75, 22, 58],2)==[85, 75]
assert heap_queue_largest( [25, 35, 22, 85, 14, 65, 75, 22, 58],5)==[85, 75, 65, 58, 35]

[BEGIN]
'import heapq as hq
def heap_queue_largest(nums,n):
largest_nums = hq.nlargest(n, nums)
return largest_nums'
[DONE]

You are an expert Python programmer, and here is your task: Write a function to find the sum of geometric progression series. Your code should pass these tests:

assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

[BEGIN]

答案

模型	模型答案	测试结果
金声玉亮2.0	```python def sum_gp(a, r, n): if r == 1: return a * n else: return a * (1 - r**n) / (1 - r) ``` [DONE]	wrong_answer 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
deepseekr1	[BEGIN] def sum_gp(a, n, r): if r == 1: return a * n else: return a * (r**n - 1) // (r - 1) [DONE]	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
gpto1	[BEGIN] 'def sum_gp(a, n, r): if r == 1: return n * a else: return a * (r ** n - 1) // (r - 1)' [DONE]	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
GPT-4o	```python def sum_gp(a, n, r): """ Calculate the sum of a geometric progression series. :param a: First term of the geometric progression :param n: Number of terms in the geometric progression :param r: Common ratio of the geometric progression :return: Sum of the geometric progression series """ if r == 1: return a * n else: return a * (1 - r**n) // (1 - r) # Testing the function assert sum_gp(1, 5, 2) == 31 assert sum_gp(1, 5, 4) == 341 assert sum_gp(2, 6, 3) == 728 ```	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
llama3.1-70B-Instruct	def sum_gp(a, n, r): sum = 0 for i in range(n): sum += a * (r ** i) return sum	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
llama3.1-405B-Instruct	def sum_gp(a, n, r): sum = 0 for i in range(n): sum += a * (r ** i) return sum	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
DeepSeek-Coder-V2	```python def sum_gp(a, r, n): if r == 1: return a * n else: return a * (1 - r**n) / (1 - r) ``` [DONE]	wrong_answer 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
Qwen2-72B-Instruct	'def sum_gp(a, r, n): if r == 1: return n * a else: return a * (1 - r**n) / (1 - r)' [DONE]	wrong_answer 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
baichuan4	To find the sum of a geometric progression (GP) series, you can use the formula for the sum of an infinite GP when the absolute value of the ratio (r) is less than 1. For a finite GP with n terms, the sum S_n can be calculated with the formula: S_n = a * (1 - r^n) / (1 - r), where: - a is the first term, - r is the common ratio, - n is the number of terms. Here's the Python function that calculates the sum of a GP series: ```python def sum_gp(a, n, r): if abs(r) >= 1: raise ValueError("The absolute value of the ratio must be less than 1 for the series to converge.") return a * (1 - r**n) / (1 - r) ``` Let's test it with the provided assertions: ```python assert sum_gp(1, 5, 2) == 31 assert sum_gp(1, 5, 4) == 341 assert sum_gp(2, 6, 3) == 728 ``` If you run this code, no exception will be raised, indicating that the function passes the tests.	failed 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
doubao	'def sum_gp(a, n, r): return a * ((r ** n) - 1) / (r - 1)' [DONE]	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
minimax	```python def sum_gp(a, n, r): # a is the first term, n is the number of terms, r is the common ratio sum_gp_series = a * (1 - r**n) / (1 - r) return int(sum_gp_series) # Test cases assert sum_gp(1, 5, 2) == 31 assert sum_gp(1, 5, 4) == 341 assert sum_gp(2, 6, 3) == 728 ``` This function calculates the sum of a geometric progression series using the formula: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. The function returns the sum as an integer.	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
kimi	Sure, let's write the function to find the sum of a geometric progression series. The sum of a geometric progression can be found using the formula: \[ S_n = a \times \frac{1 - r^n}{1 - r} \] where: - \( S_n \) is the sum of the first \( n \) terms, - \( a \) is the first term, - \( r \) is the common ratio, - \( n \) is the number of terms. Here's the Python function: ```python def sum_gp(a, n, r): if r == 1: return a * n else: return a * (1 - r**n) / (1 - r) # Test cases assert sum_gp(1, 5, 2) == 31 assert sum_gp(1, 5, 4) == 341 assert sum_gp(2, 6, 3) == 728 ``` This function calculates the sum of the geometric progression series for the given first term \( a \), number of terms \( n \), and common ratio \( r \). The test cases at the bottom are to ensure the function works correctly for the given examples.	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
wenxinyiyan	'def sum_gp(a, n, r): if r == 1: return n * a return a * (r*n - 1) // (r - 1)' [DONE] 解释：几何级数（Geometric Progression, GP）是一个序列，其中每个项（从第二项开始）是其前一项与一个常数（称为公比）的乘积。几何级数的求和公式是 S = a (r^n - 1) / (r - 1)，其中 S 是和，a 是首项，r 是公比，n 是项数。这个公式在公比 r 不等于 1 时有效。当 r 等于 1 时，几何级数变成算术级数（所有项都相等），所以和就是 n * a。在提供的 `sum_gp` 函数中，我们首先检查公比 `r` 是否等于 1。如果是，我们直接返回 `n * a`。如果不是，我们使用几何级数的求和公式来计算和，并返回结果。注意，在 Python 中，整数除法使用 `//` 运算符，以确保结果也是整数（在 Python 3 中，`/` 运算符执行浮点数除法）。另外，请注意，我在您的代码块中添加了一些缩进和换行符以提高可读性。在 Python 中，缩进是非常重要的，因为它用于定义代码块（例如函数体、循环体等）。现在，您可以将这些函数添加到您的 Python 程序中，并使用提供的测试用例来验证它们的正确性。	failed 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
zhipu	To solve this task, we need to write a function that calculates the sum of a geometric progression series. The formula for the sum of the first \( n \) terms of a geometric progression is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \] where: - \( a \) is the first term, - \( r \) is the common ratio, - \( n \) is the number of terms. Let's implement the function `sum_gp` which takes three arguments: the first term \( a \), the number of terms \( n \), and the common ratio \( r \). ```python def sum_gp(a, n, r): if r == 1: # If the common ratio is 1, the series is just `a` repeated `n` times return a * n else: # Use the geometric progression sum formula return a * (r**n - 1) // (r - 1) # Test cases assert sum_gp(1, 5, 2) == 31 assert sum_gp(1, 5, 4) == 341 assert sum_gp(2, 6, 3) == 728 print("All tests passed!") ``` This function first checks if the common ratio \( r \) is 1, in which case the sum is simply \( a \times n \). Otherwise, it uses the geometric progression sum formula to calculate the sum. The use of integer division `//` ensures that the result is an integer, which is appropriate given the test cases. You can run this code, and it should pass all the provided tests.	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
gpto1	[BEGIN] 'def sum_gp(a, n, r): if r == 1: return n * a else: return a * (r ** n - 1) // (r - 1)' [DONE]	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728
deepseekv3	To find the sum of a geometric progression (GP) series, we can use the formula for the sum of the first \( n \) terms of a GP: \[ S_n = a \frac{r^n - 1}{r - 1} \] where: - \( a \) is the first term, - \( r \) is the common ratio, - \( n \) is the number of terms. Here is the Python function to compute the sum of a geometric progression series: ```python def sum_gp(a, n, r): if r == 1: return a * n else: return a * (r n - 1) / (r - 1) # Test cases assert sum_gp(1, 5, 2) == 31 assert sum_gp(1, 5, 4) == 341 assert sum_gp(2, 6, 3) == 728 ``` ### Explanation: - Case 1: `sum_gp(1, 5, 2)` \( a = 1 \), \( n = 5 \), \( r = 2 \) \( S_5 = 1 \times \frac{2^5 - 1}{2 - 1} = 1 \times \frac{32 - 1}{1} = 31 \) - Case 2: `sum_gp(1, 5, 4)` \( a = 1 \), \( n = 5 \), \( r = 4 \) \( S_5 = 1 \times \frac{4^5 - 1}{4 - 1} = 1 \times \frac{1024 - 1}{3} = \frac{1023}{3} = 341 \) - Case 3**: `sum_gp(2, 6, 3)` \( a = 2 \), \( n = 6 \), \( r = 3 \) \( S_6 = 2 \times \frac{3^6 - 1}{3 - 1} = 2 \times \frac{729 - 1}{2} = 2 \times \frac{728}{2} = 728 \) This function handles the special case where the common ratio \( r \) is 1, in which case the sum is simply \( a \times n \). Otherwise, it uses the standard GP sum formula.	pass 测试用例：assert sum_gp(1,5,2)==31 assert sum_gp(1,5,4)==341 assert sum_gp(2,6,3)==728

模型

模型答案

测试结果

金声玉亮2.0

```python
def sum_gp(a, r, n):
if r == 1:
return a * n
else:
return a * (1 - r**n) / (1 - r)
```
[DONE]

wrong_answer
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

deepseekr1

[BEGIN]
def sum_gp(a, n, r):
if r == 1:
return a * n
else:
return a * (r**n - 1) // (r - 1)
[DONE]

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

gpto1

[BEGIN]
'def sum_gp(a, n, r):
if r == 1:
return n * a
else:
return a * (r ** n - 1) // (r - 1)'
[DONE]

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

GPT-4o

```python
def sum_gp(a, n, r):
"""
Calculate the sum of a geometric progression series.

:param a: First term of the geometric progression
:param n: Number of terms in the geometric progression
:param r: Common ratio of the geometric progression
:return: Sum of the geometric progression series
"""
if r == 1:
return a * n
else:
return a * (1 - r**n) // (1 - r)

# Testing the function
assert sum_gp(1, 5, 2) == 31
assert sum_gp(1, 5, 4) == 341
assert sum_gp(2, 6, 3) == 728
```

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

llama3.1-70B-Instruct

def sum_gp(a, n, r):
sum = 0
for i in range(n):
sum += a * (r ** i)
return sum

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

llama3.1-405B-Instruct

def sum_gp(a, n, r):
sum = 0
for i in range(n):
sum += a * (r ** i)
return sum

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

DeepSeek-Coder-V2

```python
def sum_gp(a, r, n):
if r == 1:
return a * n
else:
return a * (1 - r**n) / (1 - r)
```
[DONE]

wrong_answer
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

Qwen2-72B-Instruct

'def sum_gp(a, r, n):
if r == 1:
return n * a
else:
return a * (1 - r**n) / (1 - r)'
[DONE]

wrong_answer
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

baichuan4

To find the sum of a geometric progression (GP) series, you can use the formula for the sum of an infinite GP when the absolute value of the ratio (r) is less than 1. For a finite GP with n terms, the sum S_n can be calculated with the formula:

S_n = a * (1 - r^n) / (1 - r), where:
- a is the first term,
- r is the common ratio,
- n is the number of terms.

Here's the Python function that calculates the sum of a GP series:

```python
def sum_gp(a, n, r):
if abs(r) >= 1:
raise ValueError("The absolute value of the ratio must be less than 1 for the series to converge.")
return a * (1 - r**n) / (1 - r)
```

Let's test it with the provided assertions:

```python
assert sum_gp(1, 5, 2) == 31
assert sum_gp(1, 5, 4) == 341
assert sum_gp(2, 6, 3) == 728
```

If you run this code, no exception will be raised, indicating that the function passes the tests.

failed
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

doubao

'def sum_gp(a, n, r):
return a * ((r ** n) - 1) / (r - 1)'
[DONE]

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

minimax

```python
def sum_gp(a, n, r):
# a is the first term, n is the number of terms, r is the common ratio
sum_gp_series = a * (1 - r**n) / (1 - r)
return int(sum_gp_series)

# Test cases
assert sum_gp(1, 5, 2) == 31
assert sum_gp(1, 5, 4) == 341
assert sum_gp(2, 6, 3) == 728
```

This function calculates the sum of a geometric progression series using the formula:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. The function returns the sum as an integer.

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

kimi

Sure, let's write the function to find the sum of a geometric progression series. The sum of a geometric progression can be found using the formula:

\[ S_n = a \times \frac{1 - r^n}{1 - r} \]

where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( a \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the number of terms.

Here's the Python function:

```python
def sum_gp(a, n, r):
if r == 1:
return a * n
else:
return a * (1 - r**n) / (1 - r)

# Test cases
assert sum_gp(1, 5, 2) == 31
assert sum_gp(1, 5, 4) == 341
assert sum_gp(2, 6, 3) == 728
```

This function calculates the sum of the geometric progression series for the given first term \( a \), number of terms \( n \), and common ratio \( r \). The test cases at the bottom are to ensure the function works correctly for the given examples.

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

wenxinyiyan

'def sum_gp(a, n, r):
if r == 1:
return n * a
return a * (r**n - 1) // (r - 1)'
[DONE]

解释：
几何级数（Geometric Progression, GP）是一个序列，其中每个项（从第二项开始）是其前一项与一个常数（称为公比）的乘积。几何级数的求和公式是 S = a * (r^n - 1) / (r - 1)，其中 S 是和，a 是首项，r 是公比，n 是项数。这个公式在公比 r 不等于 1 时有效。当 r 等于 1 时，几何级数变成算术级数（所有项都相等），所以和就是 n * a。

在提供的 `sum_gp` 函数中，我们首先检查公比 `r` 是否等于 1。如果是，我们直接返回 `n * a`。如果不是，我们使用几何级数的求和公式来计算和，并返回结果。注意，在 Python 中，整数除法使用 `//` 运算符，以确保结果也是整数（在 Python 3 中，`/` 运算符执行浮点数除法）。

另外，请注意，我在您的代码块中添加了一些缩进和换行符以提高可读性。在 Python 中，缩进是非常重要的，因为它用于定义代码块（例如函数体、循环体等）。

现在，您可以将这些函数添加到您的 Python 程序中，并使用提供的测试用例来验证它们的正确性。

failed
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

zhipu

To solve this task, we need to write a function that calculates the sum of a geometric progression series. The formula for the sum of the first \( n \) terms of a geometric progression is given by:

\[ S_n = a \frac{r^n - 1}{r - 1} \]

where:
- \( a \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the number of terms.

Let's implement the function `sum_gp` which takes three arguments: the first term \( a \), the number of terms \( n \), and the common ratio \( r \).

```python
def sum_gp(a, n, r):
if r == 1:
# If the common ratio is 1, the series is just `a` repeated `n` times
return a * n
else:
# Use the geometric progression sum formula
return a * (r**n - 1) // (r - 1)

# Test cases
assert sum_gp(1, 5, 2) == 31
assert sum_gp(1, 5, 4) == 341
assert sum_gp(2, 6, 3) == 728

print("All tests passed!")
```

This function first checks if the common ratio \( r \) is 1, in which case the sum is simply \( a \times n \). Otherwise, it uses the geometric progression sum formula to calculate the sum. The use of integer division `//` ensures that the result is an integer, which is appropriate given the test cases.

You can run this code, and it should pass all the provided tests.

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

gpto1

[BEGIN]
'def sum_gp(a, n, r):
if r == 1:
return n * a
else:
return a * (r ** n - 1) // (r - 1)'
[DONE]

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728

deepseekv3

To find the sum of a geometric progression (GP) series, we can use the formula for the sum of the first \( n \) terms of a GP:

\[
S_n = a \frac{r^n - 1}{r - 1}
\]

where:
- \( a \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the number of terms.

Here is the Python function to compute the sum of a geometric progression series:

```python
def sum_gp(a, n, r):
if r == 1:
return a * n
else:
return a * (r ** n - 1) / (r - 1)

# Test cases
assert sum_gp(1, 5, 2) == 31
assert sum_gp(1, 5, 4) == 341
assert sum_gp(2, 6, 3) == 728
```

### Explanation:
- **Case 1**: `sum_gp(1, 5, 2)`
\( a = 1 \), \( n = 5 \), \( r = 2 \)
\( S_5 = 1 \times \frac{2^5 - 1}{2 - 1} = 1 \times \frac{32 - 1}{1} = 31 \)

- **Case 2**: `sum_gp(1, 5, 4)`
\( a = 1 \), \( n = 5 \), \( r = 4 \)
\( S_5 = 1 \times \frac{4^5 - 1}{4 - 1} = 1 \times \frac{1024 - 1}{3} = \frac{1023}{3} = 341 \)

- **Case 3**: `sum_gp(2, 6, 3)`
\( a = 2 \), \( n = 6 \), \( r = 3 \)
\( S_6 = 2 \times \frac{3^6 - 1}{3 - 1} = 2 \times \frac{729 - 1}{2} = 2 \times \frac{728}{2} = 728 \)

This function handles the special case where the common ratio \( r \) is 1, in which case the sum is simply \( a \times n \). Otherwise, it uses the standard GP sum formula.

pass
测试用例：assert sum_gp(1,5,2)==31
assert sum_gp(1,5,4)==341
assert sum_gp(2,6,3)==728