Description

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 toalternate. 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
1
input
5
bbbbb
output
2
input
3
rbr
output
0
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.

题意:给你颜色排列,和两种操作,一种是交换颜色,一种是改变颜色,问你最少的操作,可以使得颜色交替排列

解法:首先交替排列应该是rbrbrbrbrbrbr.. 或者brbrbrbrbr 那么讨论两种情况,一种是偶数位是r 一种是偶数位是b

再求出每种情况中,哪些位置是不符合的,比如rrbbbr 中r不在符合位置的有两个,b不在符合位置的也有两个,交换就是取两个的最小值,涂色是求它们的差值

两种情况再求一次最小值

#include<bits/stdc++.h>
using namespace std;
int n, a[10000];
int dp[100000][3];
set<char>q;
map<char,int>q1,q2;
char s[100010];
int main()
{
// string s;
scanf("%d",&n);
scanf("%s",s);
int sum1,sum2;
for(int i=0; i<n; i++)
{
if(i%2==0&&s[i]!='r')
{
q1[s[i]]++;
}
else if(i%2&&s[i]!='b')
{
q1[s[i]]++;
}
}
sum1=fabs(q1['r']-q1['b'])+min(q1['r'],q1['b']);;
q1.clear();
//cout<<sum1<<endl;
for(int i=0; i<n; i++)
{
if(i%2==0&&s[i]!='b')
{
q2[s[i]]++;
}
else if(i%2&&s[i]!='r')
{
q2[s[i]]++;
}
}
//int Min2=min(q2['r'],q2['b']);
sum2=fabs(q2['r']-q2['b'])+min(q2['r'],q2['b']);
q2.clear();
printf("%d\n",min(sum1,sum2));
return 0;
}

  

Codeforces Round #373 (Div. 2) B的更多相关文章

  1. Codeforces Round #373 (Div. 1)

    Codeforces Round #373 (Div. 1) A. Efim and Strange Grade 题意 给一个长为\(n(n \le 2 \times 10^5)\)的小数,每次可以选 ...

  2. Codeforces Round #373 (Div. 2)A B

    Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 这回做的好差啊,a想不到被hack的数据,b又没有想到正确的思维 = = [题目链 ...

  3. Codeforces Round #373 (Div. 2) C. Efim and Strange Grade 水题

    C. Efim and Strange Grade 题目连接: http://codeforces.com/contest/719/problem/C Description Efim just re ...

  4. Codeforces Round #373 (Div. 2) C. Efim and Strange Grade —— 贪心 + 字符串处理

    题目链接:http://codeforces.com/problemset/problem/719/C C. Efim and Strange Grade time limit per test 1 ...

  5. Codeforces Round #373 (Div. 2)

    A,B,C傻逼题,就不说了. E题: #include <iostream> #include <cstdio> #include <cstring> #inclu ...

  6. Codeforces Round #373 (Div. 2) A B C 水 贪心 模拟(四舍五入进位)

    A. Vitya in the Countryside time limit per test 1 second memory limit per test 256 megabytes input s ...

  7. Codeforces Round #373 (Div. 2) E. Sasha and Array 线段树维护矩阵

    E. Sasha and Array 题目连接: http://codeforces.com/contest/719/problem/E Description Sasha has an array ...

  8. Codeforces Round #373 (Div. 2) B. Anatoly and Cockroaches 水题

    B. Anatoly and Cockroaches 题目连接: http://codeforces.com/contest/719/problem/B Description Anatoly liv ...

  9. Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 水题

    A. Vitya in the Countryside 题目连接: http://codeforces.com/contest/719/problem/A Description Every summ ...

  10. 线段树+矩阵快速幂 Codeforces Round #373 (Div. 2) E

    http://codeforces.com/contest/719/problem/E 题目大意:给你一串数组a,a[i]表示第i个斐波那契数列,有如下操作 ①对[l,r]区间+一个val ②求出[l ...

随机推荐

  1. 20145207 《Java程序设计》第一周学习总结

    不好意思,来晚了   别的先不说,先道个歉,放假前跟娄老师多少发生点矛盾,望原谅. 假期忙实习还有一些其他事情,为了认真对待这门课,把剩下的时间留下来,争取一天一章来弥补. 由于没选课加上另一门课没开 ...

  2. Ubuntu Firefox installs Flashplayer

    Adobe flash 下载(https://get.adobe.com/flashplayer/)  tar.gz版本(注:adobe 提供了yum,rpm,tar.gz和APT四种版本,yum和t ...

  3. mysql bin log日志

    装mysql,运行一段时间后,在mysql目录下出现一堆类似mysql-bin.000***,从mysql-bin.000001开始一直排列下来,而且占用了大量硬盘空间,高达几十个G. 对于这些超大空 ...

  4. paper 68 :MATLAB中取整函数(fix, floor, ceil, round)的使用

    MATLAB取整函数 1)fix(x) : 截尾取整. >> fix( [3.12 -3.12]) ans =      3    -3 (2)floor(x):不超过x 的最大整数.(高 ...

  5. XMLHttpRequest Level2实现跨域

    Html5提供的XMLHttpRequest Level2已经实现的跨域访问以及一些新功能 1.ie10以下版本不支持 2.在服务器端做一些小改动即可: header("Access-Con ...

  6. angular filter

    日期格式化: <span ng-bind="topShowList.sendTime|dateFormat|date:'MM-dd HH:mm'"></span& ...

  7. ajax中网页传输(一)TEXT——带有删除功能的数据库表格显示练习

    网页之间传输的三种方式:TEXT.JSON.XML. 本章将讲解带有TEXT形势的ajax网页传输 第一:body部分代码 <title>ajax中TEXT讲解并且带有删除功能的表格< ...

  8. OpenGl And 视图

    OpenGl And 视图 标签(空格分隔): game 简介 本文主要介绍坐标系的观念, 以及在openGL中的视图及其相关的变换. 大纲 视图.模型.投影变换概念 Opengl中对各种变换的支持 ...

  9. zw版【转发·台湾nvp系列Delphi例程】HALCON ClipRegion

    zw版[转发·台湾nvp系列Delphi例程]HALCON ClipRegion procedure TForm1.Button1Click(Sender: TObject);var img : HI ...

  10. SQL Server 索引视图 聚簇索引

    创建示例: 朋友的网站速度慢,让我帮忙看下,他用的SQL Server里面 有一个文章表里面有30多万条记录 还有一个用户表里面也差不多17万记录 偏偏当初设计的时候没有冗余字段 很多帖子信息需要JO ...