C - Stones on the Table

标签：AC   NPU   single   保留   and   意思   input   mes   names   

Problem description

There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.

Input

The first line contains integer n (1?≤?n?≤?50) — the number of stones on the table.

The next line contains string s, which represents the colors of the stones. We‘ll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it‘s green and "B", if it‘s blue.

Output

Print a single integer — the answer to the problem.

Examples

3
RRG

1

5
RRRRR

4

4
BRBG

0
解题思路：题目的意思是输入的字符串中如果有相同的字符，保留一个就好，去掉其他相同的字符，保证序列中不出现相同的字符，问一共去掉多少个字符。简单处理字符串。
AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     int n,i=0,sum=0,num=0;char str[55];
 5     cin>>n;getchar();//吃掉回车符对字符串的影响
 6     cin>>str;
 7     while(str[i]!=‘\0‘){
 8         if(str[i+1]==str[i]){
 9             while(str[i+1]==str[i]){num++;i++;}
10         }
11         else {sum+=num;num=0;i++;}
12     }
13     cout<<sum<<endl;
14     return 0;
15 }

C - Stones on the Table

标签：AC   NPU   single   保留   and   意思   input   mes   names   

原文地址：https://www.cnblogs.com/acgoto/p/9112415.html