Anatoly and Cockroaches

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

如当为rbrbrbrb的时候其实可以写成两部分r_r_r_和_b_b_b_b,记录下应该为b但是却为r的个数，应该为r却为b 的个数。这样看看两个个数哪个小，那么小的那个个数一定可以通过交换使得序列符合条件，因为每次交换使得两个字母符合，所以个数多的那个也相应的完成一部分匹配，例如w1 = 5,w2 = 2；所以可以交换两个，那么交换后w1中的两个也完成了匹配，所以w1中就剩下3个，只能通过修改使得符合条件。即abs（w1-w2）+min（w1，w2）。

两种情况都判断一下取个最小就好了

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>using namespace std;int n,w1,w2,ans,ans1;string s;int main(){    scanf("%d",&n);    cin>>s;    int i;    for(i = 0; i < n; i++){        if(i%2){            if(s[i]!='r')w1++;//记录下不符合的个数        }        else{            if(s[i]!='b')w2++;        }    }    ans = abs(w1-w2)+min(w1,w2);//需要修改的加需要交换的和    w1 = w2 = 0;    for(i = 0; i < n; i++){        if(i%2){            if(s[i]!='b')w1++;        }        else{            if(s[i]!='r')w2++;        }    }    ans1 = abs(w1-w2)+min(w1,w2);//同理    ans = min(ans,ans1);    cout << ans << endl;    return 0;}

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