| 模型 | 模型答案 | 测试结果 |
|---|---|---|
| 金声玉亮2.0 | ```python def sum_Of_Series(n): return sum(i**3 for i in range(1, n+1)) ``` [DONE] |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| deepseekr1 | To solve this problem, we need to compute the cube sum of the first n natural numbers efficiently using a mathematical formula. This approach avoids the need for iterative summation, thus optimizing both time and space complexity. ### Approach The cube sum of the first n natural numbers (1, 2, 3, ..., n) can be calculated using a well-known mathematical formula. The formula for the sum of cubes of the first n natural numbers is derived as \((\frac{n(n+1)}{2})^2\). This formula allows us to compute the sum in constant time, O(1), making the solution very efficient even for large values of n. ### Solution Code ```python def sum_Of_Series(n): return (n * (n + 1) // 2) ** 2 ``` ### Explanation 1. **Mathematical Formula**: The formula \((\frac{n(n+1)}{2})^2\) is used to compute the sum of cubes of the first n natural numbers. This formula is derived from the observation that the sum of cubes can be represented as the square of the sum of the first n natural numbers. 2. **Efficiency**: By using this formula, we avoid the need for loops or iterative summation, which would have a time complexity of O(n). Instead, the solution is computed in constant time, O(1). 3. **Integer Division**: The use of integer division (//) ensures that we avoid any potential floating-point inaccuracies, as the result of \(n(n+1)\) is always an even number, making the division by 2 an integer operation. This approach ensures that the function is both optimal and straightforward, providing the correct result with minimal computational overhead. |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| gpto1 | [BEGIN] 'def sum_Of_Series(n): return ((n * (n + 1) // 2) ** 2)' [DONE] |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| GPT-4o | Sure, here's the function to find the cube sum of the first `n` natural numbers: ```python def sum_Of_Series(n): return sum(i ** 3 for i in range(1, n + 1)) # Test cases assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 ``` Explanation: - The function `sum_Of_Series` takes an integer `n` as input. - It uses a generator expression to calculate the cube of each number from `1` to `n`. - The `sum()` function then adds up all these cubes to get the final result. The function should pass the given test cases. |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| llama3.1-70B-Instruct | def sum_Of_Series(n): return sum(i**3 for i in range(1, n+1)) |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| llama3.1-405B-Instruct | def sum_Of_Series(n): return sum(i**3 for i in range(1, n+1)) |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| DeepSeek-Coder-V2 | ```python def sum_Of_Series(n): return sum(i**3 for i in range(1, n+1)) ``` [DONE] |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| Qwen2-72B-Instruct | 'def sum_Of_Series(n): return (n*(n+1)//2)**2' [DONE] |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| baichuan4 | To find the cube sum of the first n natural numbers, you can use the following function: ```python def sum_Of_Series(n): return sum([i**3 for i in range(1, n + 1)]) ``` This function uses a list comprehension to calculate the cube of each number from 1 to n, and then uses the `sum()` function to calculate the sum of these cubes. Let's test the function with the given test cases: ```python assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 ``` All the test cases pass, so the function is correct. |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| doubao | 'def sum_Of_Series(n): return sum(i**3 for i in range(1, n+1))' [DONE] |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| minimax | ```python def sum_Of_Series(n): return sum(i**3 for i in range(1, n+1)) # Test cases assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 ``` |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| kimi | ```python def similar_elements(test_tup1, test_tup2): # Use set intersection to find common elements and then convert to tuple res = tuple(sorted(set(test_tup1) & set(test_tup2))) return res # Test cases 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) import math def is_not_prime(n): if n <= 1: return True # 0 and 1 are not prime numbers for i in range(2, int(math.sqrt(n)) + 1): if n % i == 0: return True return False # Test cases assert is_not_prime(2) == False assert is_not_prime(10) == True assert is_not_prime(35) == True import heapq as hq def heap_queue_largest(nums, n): return hq.nlargest(n, nums) # Test cases 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] def sum_Of_Series(n): return sum([i**3 for i in range(1, n+1)]) # Test cases assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 ``` This code defines the functions required to pass the tests provided. The `similar_elements` function finds the common elements between two tuples, `is_not_prime` checks if a number is not prime, `heap_queue_largest` uses the heap queue algorithm to find the largest integers in a list, and `sum_Of_Series` calculates the sum of cubes of the first `n` natural numbers. |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| wenxinyiyan | 'def sum_Of_Series(n): cube_sum = 0 for i in range(1, n+1): cube_sum += i**3 return cube_sum' [DONE] Explanation: The function `sum_Of_Series` takes an integer `n` as input and calculates the sum of cubes of the first `n` natural numbers. It initializes a variable `cube_sum` to 0 and then iterates from 1 to `n` (inclusive) using a `for` loop. In each iteration, it cubes the current number `i` using the exponentiation operator `**` and adds it to `cube_sum`. Finally, it returns the calculated `cube_sum`. Note: The indentation in the provided code snippet may be incorrect due to formatting issues. Ensure that the code is indented correctly when implementing it in your Python environment. Here's the corrected indentation for the code snippet: ```python def sum_Of_Series(n): cube_sum = 0 for i in range(1, n+1): cube_sum += i**3 return cube_sum ``` Now you can use the `sum_Of_Series` function to calculate the cube sum of the first `n` natural numbers. For example: ```python print(sum_Of_Series(5)) # Output: 225 print(sum_Of_Series(2)) # Output: 9 print(sum_Of_Series(3)) # Output: 36 ``` |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| zhipu | To solve this task, we need to write a function `sum_Of_Series(n)` that calculates the cube sum of the first `n` natural numbers. The cube sum of the first `n` natural numbers is given by the formula: \[ \text{Cube Sum} = 1^3 + 2^3 + 3^3 + \ldots + n^3 \] This can also be simplified using the formula: \[ \text{Cube Sum} = \left(\frac{n(n + 1)}{2}\right)^2 \] Let's implement the function using this simplified formula: ```python def sum_Of_Series(n): return (n * (n + 1) // 2) ** 2 # Test cases assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 print("All tests passed!") ``` This function calculates the sum of the first `n` natural numbers using the formula \( \frac{n(n + 1)}{2} \), and then squares the result to get the cube sum. The `//` operator is used for integer division to ensure the result is an integer. You can run this code to verify that it passes the given tests. |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| gpto1 | [BEGIN] 'def sum_Of_Series(n): return ((n * (n + 1) // 2) ** 2)' [DONE] |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |
| deepseekv3 | Here is the Python function to find the cube sum of the first `n` natural numbers: ```python def sum_Of_Series(n): sum_of_cubes = (n * (n + 1) // 2) ** 2 return sum_of_cubes ``` ### Explanation: The sum of the cubes of the first `n` natural numbers can be calculated using the formula: \[ \text{Sum of cubes} = \left(\frac{n \cdot (n + 1)}{2}\right)^2 \] This formula is derived from the fact that the sum of cubes is equal to the square of the sum of the first `n` natural numbers. ### Example: - For `n = 5`: \[ \text{Sum of cubes} = \left(\frac{5 \cdot 6}{2}\right)^2 = 15^2 = 225 \] - For `n = 2`: \[ \text{Sum of cubes} = \left(\frac{2 \cdot 3}{2}\right)^2 = 3^2 = 9 \] - For `n = 3`: \[ \text{Sum of cubes} = \left(\frac{3 \cdot 4}{2}\right)^2 = 6^2 = 36 \] This function will pass the provided test cases. |
pass 测试用例：assert sum_Of_Series(5) == 225 assert sum_Of_Series(2) == 9 assert sum_Of_Series(3) == 36 |