CodeForces-266A

最新推荐文章于 2021-03-05 21:35:52 发布

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最新推荐文章于 2021-03-05 21:35:52 发布

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本文解析了CodeForces-266A题目，介绍了一种简单的算法来解决石头颜色排列问题，确保相邻石头颜色不同。通过遍历字符串并比较相邻字符，该算法能有效找出需移除的石头数量。

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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.

这题看似比较复杂，其实不难，题目要求剩下的砖块都不同颜色，那我们每相邻是相同颜色的砖块就拿走一块就行了，即使有很多相邻颜色相同的砖块也一样，最后都会相邻的都不同颜色。
[https://vjudge.net/problem/CodeForces-266A]

#include <iostream>

using namespace std;


int main()

{

    int a,j=0;

    cin>> a;

    char b[50];

    cin>> b;

    for (int i = 1; i < a;i++)

         if (b[i - 1] == b[i])
                j++;
               cout<< j;
               return 
  }