Codeforces 719B Anatoly and Cockroaches

最新推荐文章于 2019-08-07 00:17:29 发布

转载最新推荐文章于 2019-08-07 00:17:29 发布 · 47 阅读

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原文链接：http://www.cnblogs.com/ECJTUACM-873284962/p/6415496.html

本文介绍了一个关于交替颜色蟑螂排列的问题，并提供了一种有效的算法解决方案。通过统计不同模式下的操作次数来确定最少的操作步骤。

Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are n cockroaches living in Anatoly's room.

Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line to alternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.

Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of cockroaches.

The second line contains a string of length n, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively.

Output

Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate.

Examples

Input

5

rbbrr

Output

Input

5

bbbbb

Output

Input

3

rbr

Output

Note

In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.

In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.

In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0.

题目链接：http://codeforces.com/problemset/problem/719/B

分析：细想只有两种模式，一种brbrbr... 另一种rbrbrb... 只需要统计这两种模式下，需要的两种操作数中最小的一个，即是答案。

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 100005;
 4 char a[MAXN];
 5 int main()
 6 {
 7     int n;
 8     while(cin>>n)
 9     {
10         scanf("%s",a);
11         int m = 0;
12         int t=0;
13         int u=0;
14         int v=0;
15         for(int i=0; i<n; i++)
16         {
17             if(i%2==0)
18             {
19                 if(a[i]=='r')
20                     m++;
21                 if(a[i]=='b')
22                     t++;
23             }
24             else
25             {
26                 if(a[i]=='r')
27                     u++;
28                 if(a[i]=='b')
29                     v++;
30             }
31         }
32         int x=max(t,u);
33         int y=max(m,v);
34         int z=min(x,y);
35         printf("%d\n",z);
36     }
37     return 0;
38 }

转载于:https://www.cnblogs.com/ECJTUACM-873284962/p/6415496.html