| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | ```python from typing import List class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [float('inf')] * (n + 1) dp[0] = 0 for i in range(1, n + 1): for j in range(1, i + 1): if i - j >= 0: dp[i] = min(dp[i], dp[i - j] + prices[i - j]) return dp[n] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| deepseek-r1 | To solve this problem, we need to determine the minimum number of coins required to purchase all fruits from a market, considering a special offer where purchasing a fruit allows you to take the next few fruits for free. The key challenge is to decide optimally whether to buy a fruit (thereby getting subsequent fruits for free) or take it for free if possible. ### Approach The problem can be efficiently solved using dynamic programming (DP). The idea is to maintain a DP table where `dp[i][k]` represents the minimum coins needed to acquire all fruits starting from index `i` with `k` free fruits remaining from previous purchases. The transitions in the DP table are based on two choices at each step: taking the fruit for free (if possible) or buying it to get a new offer. 1. **State Representation**: The state is represented by the current index `i` and the number of free fruits `k` remaining from previous purchases. 2. **Transitions**: - **Take for Free**: If there are remaining free fruits (`k > 0`), move to the next fruit with `k-1` free fruits left. - **Buy the Fruit**: Pay the price of the current fruit and start a new offer, which allows taking the next `i+1` fruits for free. ### Solution Code ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) max_k = n # Since the maximum i+1 is n (when i is n-1) INF = float('inf') dp = [[INF] * (max_k + 1) for _ in range(n + 2)] # +2 to handle i = n # Base case: when i >= n, cost is 0 for k in range(max_k + 1): dp[n][k] = 0 # Fill dp table from i = n-1 down to 0 for i in range(n-1, -1, -1): for k in range(max_k, -1, -1): # Calculate option1 (take for free if possible) option1 = INF if k > 0: if i + 1 <= n: option1 = dp[i + 1][k - 1] # Calculate option2 (buy the current fruit) option2 = prices[i] next_i = i + 1 if next_i <= n: next_k = i + 1 # Because buying gives (i+1) free steps (0-based i) if next_k > max_k: next_k = max_k option2 += dp[next_i][next_k] else: option2 += 0 # if next_i is beyond n, cost is 0 # Update dp[i][k] dp[i][k] = min(option1, option2) return dp[0][0] ``` ### Explanation 1. **Initialization**: The DP table is initialized with `INF` (infinity) to represent unreachable states. The base case is when `i >= n` (all fruits are processed), where the cost is 0. 2. **Filling the DP Table**: The table is filled from the last fruit backward to the first. For each fruit `i` and each possible number of free fruits `k`, we compute the minimum cost by considering both taking the fruit for free (if possible) and buying it to get a new offer. 3. **Result Extraction**: The minimum coins needed to start from the first fruit with no initial free fruits is found in `dp[0][0]`. This approach ensures that we explore all possible ways to acquire the fruits optimally, leveraging dynamic programming to efficiently compute the solution in O(n^2) time complexity, which is feasible given the problem constraints. |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| gpto1 | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [ [float('inf')] * (n+2) for _ in range(n+2) ] for free_till in range(n+2): dp[n+1][free_till] = 0 # Base case: no more fruits to consider for i in range(n, 0, -1): for free_till in range(n+1): # Option 1: Purchase fruit i new_free_till = max(free_till, i + i) dp[i][free_till] = min(dp[i][free_till], prices[i-1] + dp[i+1][new_free_till]) # Option 2: Take fruit i for free, if possible if i <= free_till: dp[i][free_till] = min(dp[i][free_till], dp[i+1][free_till]) return dp[1][0] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| gpt-4o-2024-08-06 | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) total_cost = 0 i = 0 while i < n: # Always buy the current fruit total_cost += prices[i] # Skip the next i fruits as they can be taken for free i += i + 1 return total_cost ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| Qwen2-72B-Instruct | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [float('inf')] * n dp[0] = prices[0] min_coins = dp[0] for i in range(1, n): # Calculate the cost to buy the current fruit and get the next i fruits for free current_cost = prices[i] + sum(prices[:i]) # Update the dp array with the minimum cost to acquire all fruits up to index i for j in range(i, -1, -1): if j == 0 or dp[j-1] < current_cost: dp[j] = min(dp[j], dp[j-1] if j > 0 else 0 + current_cost) else: break # Update the minimum coins required to acquire all fruits min_coins = min(min_coins, dp[i]) return min_coins ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| DeepSeek-Coder-V2-Lite-Instruct | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [0] * n dp[0] = prices[0] for i in range(1, n): dp[i] = dp[i-1] + prices[i] for j in range(1, i): dp[i] = min(dp[i], dp[i-j-1] + prices[j]) return dp[-1] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| Meta-Llama-3.1-405B-Instruct-Turbo | ```python from typing import List class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [float('inf')] * (n + 1) dp[0] = 0 for i in range(1, n + 1): for j in range(i): dp[i] = min(dp[i], dp[j] + prices[j]) if j + j + 1 >= i: break dp[i] = min(dp[i], dp[j] + prices[j]) return dp[n] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| Meta-Llama-3.1-70B-Instruct-Turbo | ```python from typing import List class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [0] * (n + 1) for i in range(n - 1, -1, -1): dp[i] = min(prices[i] + dp[min(i + i + 1, n)], dp[i + 1]) return dp[0] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| baichuan4 | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: # Initialize total coins needed to 0 total_coins = 0 # Iterate through the prices list for i in range(len(prices)): # If the current price is less than or equal to the sum of the remaining prices, # it's optimal to pay for this fruit and get the rest for free. if prices[i] <= sum(prices[i+1:]): # Add the current price to the total coins needed total_coins += prices[i] # Set all remaining prices to 0 since they are now free prices[i+1:] = [0] * (len(prices) - i - 1) # Return the total number of coins needed return total_coins ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| doubao | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [0] * (n + 1) for i in range(1, n + 1): dp[i] = dp[i - 1] + prices[i - 1] j = i - 1 while j >= 0 and i - j <= j: dp[i] = min(dp[i], dp[j]) j -= 1 return dp[n] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| gpto1 | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [ [float('inf')] * (n+2) for _ in range(n+2) ] for free_till in range(n+2): dp[n+1][free_till] = 0 # Base case: no more fruits to consider for i in range(n, 0, -1): for free_till in range(n+1): # Option 1: Purchase fruit i new_free_till = max(free_till, i + i) dp[i][free_till] = min(dp[i][free_till], prices[i-1] + dp[i+1][new_free_till]) # Option 2: Take fruit i for free, if possible if i <= free_till: dp[i][free_till] = min(dp[i][free_till], dp[i+1][free_till]) return dp[1][0] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| kimi | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [0] * (n + 1) for i in range(1, n + 1): dp[i] = min(dp[i - 1] + prices[i - 1], dp[max(0, i - (prices[i - 1] + 1))] + prices[i - 1]) return dp[n] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| minimax | ```python from typing import List class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) total_cost = 0 i = 0 while i < n: # Purchase the current fruit and get the next i fruits for free total_cost += prices[i] # Move to the next fruit that we can get for free i += 1 # If we can get more fruits for free, check if it's better to buy the next fruit now while i < n and prices[i] <= prices[i - 1]: # If the next fruit is cheaper or equal in price, buy it now to get the rest for free total_cost += prices[i] i += 1 return total_cost # Example usage: # sol = Solution() # print(sol.minimumCoins([3, 1, 2])) # Output: 4 # print(sol.minimumCoins([1, 10, 1, 1])) # Output: 2 ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| wenxinyiyan | ```python from typing import List class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [float('inf')] * (n + 1) dp[0] = 0 for i in range(1, n + 1): # Calculate the minimum cost to reach the i^th fruit for j in range(i, 0, -1): dp[i] = min(dp[i], dp[j-1] + prices[j-1]) # Since we purchased the j^th fruit, we can get the next j-1 fruits for free # So, we jump to the (i+j)^th fruit in the next iteration (if it exists) next_fruit = i + j if next_fruit <= n: dp[next_fruit] = min(dp[next_fruit], dp[i]) return dp[n] # Test the solution solution = Solution() print(solution.minimumCoins([3, 1, 2])) # Output: 4 print(solution.minimumCoins([1, 10, 1, 1])) # Output: 2 ``` **Note**: The provided solution uses dynamic programming to solve the problem. However, there is a simpler and more efficient approach that doesn't require dynamic programming. Here's the revised solution: ```python from typing import List class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) coins = 0 i = 0 while i < n: # Find the maximum number of fruits we can get for free j = i + 1 while j < n and prices[j] >= prices[i]: j += 1 # Purchase the i^th fruit and get the next i fruits for free coins += prices[i] i = j return coins # Test the solution solution = Solution() print(solution.minimumCoins([3, 1, 2])) # Output: 4 print(solution.minimumCoins([1, 10, 1, 1])) # Output: 2 ``` In this revised solution, we use a greedy approach. We iterate through the fruits and find the maximum number of fruits we can get for free by comparing their prices. We then purchase the fruit with the minimum price among them and get the next fruits for free. This approach has a time complexity of O(n), where n is the number of fruits. |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| zhipu | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [float('inf')] * (n + 1) dp[0] = 0 for i in range(n): dp[i + 1] = min(dp[i + 1], dp[i] + prices[i]) if i + prices[i] + 1 <= n: dp[i + prices[i] + 1] = min(dp[i + prices[i] + 1], dp[i] + prices[i]) return dp[n] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |
| deepseek-v3 | ```python class Solution: def minimumCoins(self, prices: List[int]) -> int: n = len(prices) dp = [float('inf')] * (n + 1) dp[0] = 0 for i in range(n): for j in range(i + 1, min(n + 1, i + 1 + prices[i])): dp[j] = min(dp[j], dp[i] + prices[i]) return dp[n] ``` |
[{"input": "[3, 1, 2]", "output": "4", "testtype": "functional"}, {"input": "[1, 10, 1, 1]", "output": "2", "testtype": "functional"}] |