ACM新手训练题（CodeForces - 266A）

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#简单编程问题

探讨了一道编程题，桌上有一排n个石头，颜色为红、绿、蓝三种，目标是最小化拿掉的石头数量，确保任意相邻石头颜色不同。通过遍历石头颜色并比较相邻石头，实现最优解。

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Stones on the Table
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

Input

3
RRG

Output

Input

5
RRRRR

Output

Input

4
BRBG

Output

0
问题链接：https://vjudge.net/problem/36572/origin
问题简述：
桌子上有n（用户输入 ,1<=n<=50）个石头,每个石头有不同的颜色，颜色有三种，红（R)绿(G)蓝(B)。n个石头排列成一排，计算需要拿走多少个石头才能让相邻两个石头的颜色不一样。最终输出这个数。
问题分析：
用一个字符数组接收所有石头的颜色，通过循环一个个判断颜色是否相同，相同则令计数加一，最终输出计数。
AC代码如下：

#include<iostream>
using namespace std;
int main()
{
 char a[51];
 int n,i,count=0;
 cin >> n;
 cin >> a;
 for (i = 0; i < n ; i++)
 {
  if (a[i] == a[i + 1])
   count++;
 }
 cout << count << endl;
 return 0;
}