CodeForces - 266A Stones on the Table

最新推荐文章于 2020-03-12 19:12:38 发布

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最新推荐文章于 2020-03-12 19:12:38 发布

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本文链接：https://blog.youkuaiyun.com/weixin_43979544/article/details/84887192

本文探讨了一道算法题，目标是最小化调整使一排红、绿、蓝三种颜色的石头任意相邻两颗不同色，通过计算相邻同色石头对数确定移除数量。

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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.

Examples

Input
3
RRG
Output
1

Input
5
RRRRR
Output
4

Input
4
BRBG
Output
0

题目解析：
即判断相邻的两个石头颜色相同的对数。
AC通过的c++程序如下：

#include<iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
	int n = 0, k = 0;
	cin >> n;
	string a;
	cin >> a;
	vector<int>obj;
	for (int i = 0; i<n; i++)
	{
		obj.push_back(a[i]);
	}
	for (int i = 0; i<n-1; i++)
	{
		if (obj[i] == obj[i + 1])
			k += 1;
	}
	cout << k;
}