编写以上出列三代码py

在Python中，我们可以编写一个简单函数来模拟"数到三出列"的游戏。下面是一个基本的示例，假设我们将1到41的数字分为3组，每组轮流移出3个数字：

def count_to_three(numbers):
    # 初始化三组数字
    groups = [numbers[i:i+3] for i in range(0, len(numbers), 3)]
    
    while numbers:
        # 对于每组，先移出第一个数字
        for group in groups:
            if group:
                numbers.remove(group.pop(0))
                
        # 更新剩余数字的分组
        new_groups = [numbers[i:i+3] for i in range(0, len(numbers), 3)]

        # 检查是否有新生成的空组，如果有，则合并相邻的非空组
        merged_groups = []
        for group in zip(new_groups[:-1], new_groups[1:]):
            if all(g == [] for g in group):
                continue
            merged_groups.extend(group)
        
        groups = merged_groups

    return groups

# 使用1到41的数字测试
numbers = list(range(1, 42))
result = count_to_three(numbers)
print("最终出列后的组:", result)