1207: Candies

Time Limit: 2 Sec  Memory Limit: 1280 MB
Submit: 223  Solved: 31
[Submit][Status][Web Board]

Description

Xiao Ming likes those N candies he collects very much. There are two kinds of candies, A and B. One day, Xiao Ming puts his candies in a row and plays games. And he can replace the Lth candy to the Rth candy with the same kind of candies. Now, he wonder that if he eats the Lth candy to Rth candy, he can eat how many B candy continuously at most. For each Xiao Ming’s query, give the number of the B candy he can eat continuously at most.

Input

In the first line, there is an integer T, indicates the number of test cases.

For each case, there are two integers N and M in the first line, indicate the number of candies and the time of Xiao Ming’s operations.

The second line is a N-length string consist of character A and B, indicates the row of candies Xiao Ming put.

The next M line is Xiao Ming’s operations. There are two kind of operations:

1. 1 L R v, indicate Xiao Ming replaces the Lth candy to the Rth candy with A candies (v==1) or B candies ( v == 2 ).

2. 2 L R, indicate Xiao Ming wonder that there are how many continuous B candies between the Lth candy to the Rth candy most.

Output

In each case, the first line is “Case #k: “, k indicates the case number.

For each query, output the number of the most continuous B candies.

Sample Input

1
5 3
ABABB
2 1 3
1 2 3 2
2 1 3

Sample Output

Case #1:
1
2

HINT

hzau 1207 Candies的更多相关文章

  1. HZAU 1207 Candies(线段树区间查询 区间修改)

    [题目链接]http://acm.hzau.edu.cn/problem.php?id=1207 [题意]给你一个字符串,然后两种操作:1,将区间L,R更新为A或者B,2,询问区间L,R最长的连续的B ...

  2. 【POJ2886】Who Gets the Most Candies?-线段树+反素数

    Time Limit: 5000MS Memory Limit: 131072K Case Time Limit: 2000MS Description N children are sitting ...

  3. poj 3159 Candies 差分约束

    Candies Time Limit: 1500MS   Memory Limit: 131072K Total Submissions: 22177   Accepted: 5936 Descrip ...

  4. Who Gets the Most Candies?(线段树 + 反素数 )

    Who Gets the Most Candies? Time Limit:5000MS     Memory Limit:131072KB     64bit IO Format:%I64d &am ...

  5. poj---(2886)Who Gets the Most Candies?(线段树+数论)

    Who Gets the Most Candies? Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 10373   Acc ...

  6. poj3159 Candies(差分约束,dij+heap)

    poj3159 Candies 这题实质为裸的差分约束. 先看最短路模型:若d[v] >= d[u] + w, 则连边u->v,之后就变成了d[v] <= d[u] + w , 即d ...

  7. HDU 5127 Dogs' Candies

    Dogs' Candies Time Limit: 30000/30000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) T ...

  8. C. Om Nom and Candies 巧妙优化枚举,将复杂度控制在10e6

    C. Om Nom and Candies 无线超大背包问题 #include <iostream> #include <cstdio> #include <cstrin ...

  9. POJ 3159 Candies (栈优化spfa)

    Candies 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/J Description During the kinderga ...

随机推荐

  1. MySQL中锁问题

    1.脏读 脏页只是在缓冲池中已经修改的页但是没有刷新到磁盘中,即数据库实例内存中的页和磁盘中的页事不一致的,当然在刷新到磁盘之前,日志都已经被写入到了重做日志文件中,而所谓的脏数据是指事务对缓冲池中行 ...

  2. Linux中的系统默认日志

    /var/log/cron 记录了系统定时任务相关的日志 /var/log/cups 记录了打印信息的日志 /var/log/dmesg 记录了系统在开机时内核自检的信息,可以通过dmesg命令直接查 ...

  3. mui 视频播放

    <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <meta name ...

  4. postgrSQL 错误ERROR: permission denied

    赋权限: GRANT ALL PRIVILEGES ON TABLE 表名 TO 用户;

  5. 【读书笔记】Java核心技术-基础知识-反射

    在网页中运行Java程序称为applet. 反射 这项功能被大量地应用于JavaBeans中,它是Java组件的体系结构. 能够分析类能力的程序称为反射(reflective).反射机制的功能及其强大 ...

  6. Android:日常学习笔记(9)———探究广播机制

    Android:日常学习笔记(9)———探究广播机制 引入广播机制 Andorid广播机制 广播是任何应用均可接收的消息.系统将针对系统事件(例如:系统启动或设备开始充电时)传递各种广播.通过将 In ...

  7. 022_Hadoop中的数据类型(Writable、WritableComparable、Comparator、RawComparator…)

    1. 在hadoop中所有的key/value都必须实现Writable接口,有两个方法,分别用于读(反序列化)和写(序列化)操作.

  8. MongoDB命令语法小用

    using System; using System.Collections.Generic; using System.Linq; using System.Text; using MongoDB; ...

  9. 非root权限的linux用法添加工作路径

    修改~目录的bashrc文件: 1.cd 到~目录. 2.ls -a ,bashrc文件是隐藏的. 3.vim .bashrc;export PATH=$PAHT:要添加的工作路径. 4.source ...

  10. iptables DNAT、SNAT和MASQUERATE

    MASQUERADE 地址伪装,和SNAT功能一样,只不过SNAT使用固定IP地址,MASQUERADE使用网卡上的地址. SNAT配置: iptables -t nat -A POSTROUTING ...