codeforces 266A Stones on the Table（模拟水题）

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本文介绍了一道编程题，任务是计算在一行不同颜色的石头中，为了确保任意相邻的两颗石头颜色不同所需移除的最少石头数量。文中提供了一个示例代码，展示了如何通过遍历字符串并比较相邻字符来解决问题。

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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.

Sample test(s)

Input

3
RRG

Output

Input

5
RRRRR

Output

Input

4
BRBG

Output

0

不多说就是找有几个相邻的是不同的：
代码：
#include <iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int t,ans,i;
    char c[51];
    while(cin>>t)
    {
        cin>>c;
        ans=0;
        for(i=0;i<(t-1);i++)
        {
            if(c[i]!=c[i+1])
                ans++;
        }
        cout<<(t-1-ans)<<endl;
    }
    return 0;
}