CodeForces 961-A Tetris(模拟)

这篇博客介绍了CodeForces 961-A题目的模拟过程,讨论了类似俄罗斯方块的问题,其中玩家在n列平台上放置m个方块,每列至少有一个方块时,底部一行会消除并得分。博客内容包括问题描述、输入输出格式、样例输入输出以及解题思路和代码实现。通过模拟找到最小相同数的个数来计算得分。

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Problem Description

You are given a following process.

There is a platform with n columns. 1×1 squares are appearing one after another in some columns on this platform. If there are no squares in the column, a square will occupy the bottom row. Otherwise a square will appear at the top of the highest square of this column.

When all of the n columns have at least one square in them, the bottom row is being removed. You will receive 1 point for this, and all the squares left will fall down one row.

You task is to calculate the amount of points you will receive.

 

Input 

The first line of input contain 2 integer numbers n and m (1≤n,m≤1000) — the length of the platform and the number of the squares.

The next line contain mm integer numbers c1,c2,…,cm (1≤ci≤n) — column in which i-th square will appear

### 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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