Gym 100269G-Garage

本文探讨了如何在限定的空间内放置尽可能少的矩形车库,以满足合同要求的最大数量限制。通过算法计算最少所需的车库数量,确保无法再添加额外的车库。

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Problem G. Garage
Input file: garage.in
Output file: garage.out
Time limit: 2 seconds
Memory limit: 256 megabytes
Wow! What a lucky day! Your company has just won a social contract for building a garage complex.
Almost all formalities are done: contract payment is already transferred to your account.
So now it is the right time to read the contract. Okay, there is a sandlot in the form of W × H rectangle
and you have to place some garages there. Garages are w × h rectangles and their edges must be parallel
to the corresponding edges of the sandlot (you may not rotate garages, even by 90◦
). The coordinates of
garages may be non-integer.
You know that the economy must be economical, so you decided to place as few garages as possible.
Unfortunately, there is an opposite requirement in the contract: placing maximum possible number of
garages.
Now let’s see how these requirements are checked… The plan is accepted if it is impossible to add a new
garage without moving the other garages (the new garage must also have edges parallel to corresponding
sandlot edges).
这里写图片描述
Time is money, find the minimal number of garages that must be ordered, so that you can place them
on the sandlot and there is no place for an extra garage.
Input
The only line contains four integers: W, H, w, h — dimensions of sandlot and garage in meters. You may
assume that 1 ≤ w ≤ W ≤ 30 000 and 1 ≤ h ≤ H ≤ 30 000.
Output
Output the optimal number of garages.
Examples
garage.in garage.out
11 4 3 2 2
10 8 3 4 2
15 7 4 2 4
The plan on the first picture is accepted and optimal for the first example. Note that a rotated (2 × 3)
garage could be placed on the sandlot, but it is prohibited by the contract.

题意:问能放几个停车位将沙地占满,使得所放的停车位最少。

代码:

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;
int main (void)
{
    //freopen
    //freopen
    int X,Y,x,y;
    scanf("%d %d %d %d",&X,&Y,&x,&y);
    int xx,yy;
    xx=2*x;
    yy=2*y;
    int cntx=X/xx;
    int cnty=Y/yy;
//  printf("%d %d__\n",cntx,cnty);
    if(X%xx>=x)
        cntx++;
//      printf("cntx=%d\n",cntx);
    if(Y%yy>=y)
        cnty++;
//  printf("cnty=%d\n",cnty);
    printf("%d\n",cntx*cnty);
    return 0;
}
### 使用PPO算法在gym-super-mario-bros环境中的实现 为了在 `gym-super-mario-bros` 游戏环境中应用近端策略优化 (Proximal Policy Optimization, PPO),可以按照以下方法构建模型并训练代理。以下是详细的说明: #### 安装依赖库 首先,确保安装必要的 Python 库来支持 `gym-super-mario-bros` 和强化学习框架 Stable Baselines3。 ```bash pip install nes-py gym-super-mario-bros stable-baselines3[extra] ``` 上述命令会安装 `nes-py`, `gym-super-mario-bros` 以及用于实现 PPO 的强化学习工具包 `Stable-Baselines3`[^1]。 --- #### 创建超级马里奥环境 通过导入 `SuperMarioBros-v0` 或其他变体创建游戏环境,并设置动作空间和观察空间。 ```python import gym_super_mario_bros from nes_py.wrappers import JoypadSpace from gym.spaces import Box from gym.wrappers import FrameStack from stable_baselines3.common.env_checker import check_env from stable_baselines3 import PPO # 初始化 Super Mario Bros 环境 env = gym_super_mario_bros.make('SuperMarioBros-v0') # 设置简化操作集 env = JoypadSpace(env, [['right'], ['right', 'A']]) # 将帧堆叠到一起以提供时间序列数据给神经网络 env = FrameStack(env, num_stack=4) # 验证环境是否兼容稳定基线的要求 check_env(env) ``` 此部分代码定义了一个简单的控制方案(右移或跳跃),并通过 `FrameStack` 提供连续四帧作为输入状态。 --- #### 训练PPO模型 使用 `stable-baselines3.PPO` 来初始化和训练代理。 ```python model = PPO( policy="CnnPolicy", env=env, verbose=1, tensorboard_log="./mario_ppo_tensorboard/" ) # 开始训练过程 model.learn(total_timesteps=int(1e6)) # 保存训练好的模型 model.save("ppo_mario") ``` 在此配置中: - **policy**: 使用卷积神经网络 (`CnnPolicy`) 处理图像型观测值。 - **total_timesteps**: 总共执行 $1 \times 10^6$ 时间步数进行训练。 - **tensorboard_log**: 可视化日志路径以便监控训练进展。 --- #### 测试已训练的模型 加载先前保存的模型并对环境运行推理测试。 ```python del model # 删除旧模型以防冲突 # 加载预训练模型 model = PPO.load("ppo_mario") state = env.reset() done = False while not done: action, _states = model.predict(state) state, reward, done, info = env.step(action) env.render() env.close() ``` 这段脚本展示了如何利用训练完成后的模型在游戏中做出决策。 --- ### 注意事项 1. 超参数调整对于性能至关重要。例如,更改学习率、批量大小或其他超参数可能显著影响收敛速度与最终效果。 2. 如果希望扩展功能,可考虑引入更复杂的奖励机制或者自定义环境封装器。 3. 对于更高难度级别(如世界 1-2 或以上),建议增加训练时间和样本数量。 ---
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