A. Gotta Catch Em' All!
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.

Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about case — the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.

Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?

Note: uppercase and lowercase letters are considered different.

Input

Input contains a single line containing a string s (1  ≤  |s|  ≤  105) — the text on the front page of the newspaper without spaces and punctuation marks. |s| is the length of the string s.

The string s contains lowercase and uppercase English letters, i.e. .

Output

Output a single integer, the answer to the problem.

Examples
Input
Bulbbasaur
Output
1
Input
F
Output
0
Input
aBddulbasaurrgndgbualdBdsagaurrgndbb
Output
2
Note

In the first case, you could pick: Bulbbasaur.

In the second case, there is no way to pick even a single Bulbasaur.

In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur".

 #include<iostream>
using namespace std;
int f[];
int main(){
string s;
cin>>s;
for(int i=;i<s.size();i++)
f[s[i]]++;
int ans;
ans=min(f['B'],min(f['u']/,min(f['l'],min(f['b'],min(f['a']/,min(f['s'],f['r']))))));
cout<<ans;
}

757A Gotta Catch Em' All!的更多相关文章

  1. 【codeforces 757A】Gotta Catch Em' All!

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  2. CodeForces757A

    A. Gotta Catch Em' All! time limit per test 1 second memory limit per test 256 megabytes input stand ...

  3. Codeforces Round #391 A B C D E

    A. Gotta Catch Em' All! 题意 从给定的字符串中选取字符,问可构成多少个\(Bulbasaur\) // 想到柯南里一些从报纸上剪汉字拼成的恐吓信_(:з」∠)_ Code #i ...

  4. Java性能提示(全)

    http://www.onjava.com/pub/a/onjava/2001/05/30/optimization.htmlComparing the performance of LinkedLi ...

  5. JVM源码分析之javaagent原理完全解读

    概述 本文重点讲述javaagent的具体实现,因为它面向的是我们Java程序员,而且agent都是用Java编写的,不需要太多的C/C++编程基础,不过这篇文章里也会讲到JVMTIAgent(C实现 ...

  6. JVM插庄之二:Java agent基础原理

    javaagent 简介 Javaagent 只要作用在class被加载之前对其加载,插入我们需要添加的字节码. Javaagent面向的是我们java程序员,而且agent都是用java编写的,不需 ...

  7. JVM源码分析之javaagent原理完全解读--转

    原文地址:http://www.infoq.com/cn/articles/javaagent-illustrated 概述 本文重点讲述javaagent的具体实现,因为它面向的是我们Java程序员 ...

  8. JVM 源码分析之 javaagent 原理完全解读

    转载:https://infoq.cn/article/javaagent-illustrated 本文重点讲述 javaagent 的具体实现,因为它面向的是我们 Java 程序员,而且 agent ...

  9. 2016-2017 ACM-ICPC Northwestern European Regional Programming Contest (NWERC 2016)

    A. Arranging Hat $f[i][j]$表示保证前$i$个数字有序,修改了$j$次时第$i$个数字的最小值. 时间复杂度$O(n^3m)$. #include <bits/stdc+ ...

随机推荐

  1. redis 4,0 安装

    安装redis : 1,yum install wget -y 2,cd /opt: 3,wget http://download.redis.io/releases/redis-4.0.10.tar ...

  2. Jenkins 邮件发送设置(jenkins自带邮件设置)

    首先进入系统设置,找到Jenkins Location部分 这里设置 系统管理员邮件地址,然后设置邮件通知部分,这里为了方便我使用了QQ邮箱(作为发送邮件地址) 这里的 用户名 必须与上面的 系统管理 ...

  3. mybatis匹配字符串的坑

    where语句中我们经常会做一些字符串的判断,当传入的字符串参数为纯数字时,在mybatis的条件语句test里匹配全数字字符串需要注意会有如下现象: 所以里面的字符串需要加单引号,mybatis是匹 ...

  4. FDMemTable 数据集

    c++builder FDMemTable 内存表 内存数据表:现在应该首选 TFDMemTable 了(之前是 TClientDataSet) FDMemTable->CloneCursor( ...

  5. WMI 连接远程计算机并进行局域网进程扫描

    On Error Resume Next Dim myArray(231) myArray(0)="smss.exe"myArray(1)="csrss.exe" ...

  6. VB6 创建控制台应用程序

    ' 功能:为VB程序创建一个consolewindow.Private Declare Function AllocConsole Lib "kernel32" () As Lon ...

  7. vue基础——vue实例

    创建一个vue实例 每个vue应用都是通过Vue函数创建一个新的Vue实例开始的 var vm = new Vue({ //选项 }) 一个Vue应用由一个通过new Vue创建的根Vue实例,以及可 ...

  8. playbook相关

    ansible-playbook site.yml  -f 10 ansible-playbook常用参数说明: -f  10          启用10个并发进程数执行playbook -u  RM ...

  9. iKcamp|基于Koa2搭建Node.js实战(含视频)☞ 处理静态资源

    视频地址:https://www.cctalk.com/v/15114923882788 处理静态资源 无非花开花落,静静. 指定静态资源目录 这里我们使用第三方中间件: koa-static 安装并 ...

  10. (转)游戏引擎中三大及时光照渲染方法介绍(以unity3d为例)

    重要:在目前市面上常见的游戏引擎中,主要采用以下三种灯光实现方式: 顶点照明渲染路径细节 Vertex Lit Rendering Path Details 正向渲染路径细节 Forward Rend ...