华中农业大学第五届程序设计大赛网络同步赛-L

本文介绍了一个简单的编程问题解决方案:如何计算鸡兄从盒子中取出特定组合的饼干获得的幸福感。通过遍历字符串并检查相邻饼干类型来实现。

L.Happiness

Chicken brother is very happy today, because he attained N pieces of biscuits whose tastes are A or B. These biscuits are put into a box. Now, he can only take out one piece of biscuit from the box one time. As we all know, chicken brother is a creative man. He wants to put an A biscuit and a B biscuit together and eat them. If he take out an A biscuit from the box and then he take out a B biscuit continuously, he can put them together and eat happily. Chicken brother’s happiness will plus one when he eat A and B biscuit together one time. Now, you are given the arrangement of the biscuits in the box(from top to bottom) ,please output the happiness of Chicken Brother when he take out all biscuit from the box.

Input Description

The first line is an integer indicates the number of test cases. In each case, there is one line includes a string consists of characters ‘A’ and ‘B’. The length of string is not more than 1000000.

Output Description

For each test case: The first line output “Case #k:", k indicates the case number. The second line output the answer.

Sample Input

1

ABABBBA

Sample Output

Case #1:

2

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 
 8 int main()
 9 {
10     int T;
11     string str;
12     cin>>T;
13     for(int kase = 1; kase <= T; kase++)
14     {
15         cin>>str;
16         int len = str.length(), ans = 0;
17         for(int i = 0; i < len-1; i++)
18             if(str[i] == 'A' && str[i+1] == 'B')
19                   ans++;
20         cout<<"Case #"<<kase<<":"<<endl;
21         cout<<ans<<endl;
22     }
23 
24     return 0;
25 }

 

转载于:https://www.cnblogs.com/Penn000/p/6756257.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值