VJ第三题题解

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本文详细解析了CodeForces-266A问题，该问题要求找出最少移除多少宝石，使得桌上任意相邻的宝石颜色不同。通过分析，我们采用循环检查相邻宝石颜色的方法，利用ASCII码比较颜色，最终输出需要移除的宝石数量。

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CodeForces - 266A

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
问题简述：一行宝石，输入宝石个数以及颜色，输出至少取出多少宝石使得相邻宝石不同色。
问题分析：通过循环检查相邻宝石之间的颜色确定需要取出的宝石的个数。
程序说明：用字符串储存了宝石颜色，用ASCII码确定颜色是否相同。

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