CodeForces-266A-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
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0

链接：https://vjudge.net/problem/CodeForces-266A

题目大概意思是：桌子上有一排石头，每一块都可以是红色、绿色或蓝色。要从桌子上取出最少数量的石头，这样相邻的两块石头就会有不同的颜色。如果没有其他石头，一排排的石头就被认为是相邻的。

我的思路：用一个循环，将第一个与后面的比较，如果相同，则加1，依次循环下去，得到的就是应该拿掉的次数
代码如下：

#include <iostream>
using namespace std;

int main()
{
	int a,c=0;
	char b[50];
	cin >> a >> b;
	for (int i = 0; i < a; i++)
	{
		if (b[i] == b[i + 1])c++;
	}
	cout << c;
}