欢迎来到起遇信息学
起遇信息学正处于上线筹建阶段,以下功能已全部开放免费体验: ✅ 完整题库浏览与代码提交评测(C / C++ / Python / Java 等) ✅ 入门到进阶的系列课程试读、作业与考试 ✅ AI 提示、AI 作业分析等智能助教功能 ✅ 赛事模拟与个人能力报告 ✅ 邮箱注册开放 ⏳ 付费课程订阅与微信/支付宝支付通道 ⏳ 手机号登录,微信扫码登录、微信公众号绑定 使用中如遇任何问题,欢迎通过页面底部 **"联系我们"** 与我们沟通。
CF2173A.Sleeping Through Classes
Sleeping Through Classes
You have classes today, which are numbered from to .
The classes are described by a binary string of length . We call class important if and only if . For each important class, you must stay awake and listen to it.
You are very tired and wish to sleep through as many classes as possible. However, falling asleep takes time. If you listen to an important class , then you cannot fall asleep for the next classes, i.e., you must also stay awake in classes (or until the end of the day, if fewer than classes remain).
For classes that are not important, you may sleep through them unless the rule above forces you to stay awake.
Your task is to find out the maximum number of classes you can sleep through today.
A binary string is a string where each character is either 0 or 1.
Input
Each test contains multiple test cases. The first line contains the number of test cases (). The description of the test cases follows.
The first line of each test case contains two integers and ().
The second line of each test case contains the string of length ( or ).
Output
For each test case, output a single integer — the maximum number of classes you can sleep through today.
Note
In the first test case, you must listen to class and class . After listening to class , you cannot fall asleep in class . So the only class you can sleep through is class .
In the second test case, you can sleep through all the classes.
In the fourth test case, you can only sleep through classes and .
Samples
4
4 1
1001
3 3
000
3 1
001
8 2
01000101
1
3
2
2
在线编程 IDE
建议全屏模式获得最佳体验
| 进入全屏编程 | Alt+E |
| 递交评测 | Ctrl+Enter |
| 注释/取消注释 | Ctrl+/ |
| 缩放字体 | Ctrl+滚轮 |