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 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.

Sample Input

5
rbbrr

Sample Output

1

Hint

题意

有n个字符，要么是r,要么是b，现在你想让他变成交替的

每次你可以修改一个字符，或者交换两个字符的位置。

问你最少花费是多少

题解：

水题，要么是rbrbrbrbrb这样，要么是brbrbrbrb这样

都判断一下取个最小就好了、

代码

#include<bits/stdc++.h>
using namespace std;

string s;
int n,w1,w2,ans,ans1,ans2;
int main()
{
    scanf("%d",&n);
    cin>>s;
    for(int i=0;i<n;i++)
    {
        if(i%2){if(s[i]!='r')w1++;}
        else{if(s[i]!='b')w2++;}
    }
    ans=abs(w1-w2)+min(w1,w2);
    w1=w2=0;
    for(int i=0;i<n;i++)
    {
        if(i%2){if(s[i]!='b')w1++;}
        else{if(s[i]!='r')w2++;}
    }
    ans=min(ans,abs(w1-w2)+min(w1,w2));
    cout<<ans<<endl;
}