CodeForces21C - Stripe 2 解题报告

博客探讨了一种算法问题,即如何找出一个序列可以被分割成三个连续子序列,使得它们的和相等。首先,通过判断序列总和是否为3的倍数排除不可能情况。接着,找到序列中前缀和等于总和三分之一和三分之二的位置,通过双指针方法计算符合条件的分割方案数。最后,给出了AC代码实现。

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原题链接

题目大意

给定一个序列,求将该序列分割为三段连续的子序列,其和相等的方案数.

给定数据

第一行为 n n n,表示序列长度.
第二行给定的序列 A A A,每个元素为 A i A_i Ai.
对于每个 A i A_i Ai, ∣ A i ∣ ≤ 1 0 4 |A_i| \leq 10^4 Ai104 A i A_i A

### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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