Codeforces Round #163 (Div. 2) / 226A Stones on the Table (字符串处理)

最新推荐文章于 2024-07-15 17:55:50 发布

原创最新推荐文章于 2024-07-15 17:55:50 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces #acm

acm之路--数据结构同时被 3 个专栏收录

111 篇文章

订阅专栏

Codeforces

83 篇文章

订阅专栏

acm之路--字符串

42 篇文章

订阅专栏

本文介绍了一道来自CodeForces平台的编程题266A的解决方案，题目要求从一列三种颜色的石子中移除最少数量的石子，使得任意相邻的石子颜色不同。文章提供了完整的C++代码实现。

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
   
1

     input
   
5
RRRRR

     output
   
4

     input
   
4
BRBG

     output
   
0

题意：一列的石子有三种颜色，取出x个，剩下的石子相邻间颜色不同，求min(x)。

完整代码：

/*30ms,0KB*/

#include <cstdio>

int main()
{
	int n, count = 0;
	scanf("%d", &n);
	getchar();
	char now = getchar(), ch;
	while (--n)
	{
		ch = getchar();
		if (ch == now)
			++count;
		else
			now = ch;
	}
	printf("%d", count);
	return 0;
}