Codeforce 719B

Posted 2021-02-10 fangxiaoqi

B. Anatoly and Cockroaches

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

　　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 1

　　5

　　rbbrr

　　Output 1

　　1

 

　　Input 2

　　5

　　bbbbb

　　Output 2

　　2

 

　　Input 3

　　3

　　rbr

　　Output 3

　　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.

 

思路：

　　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

 

代码：

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     int n;
 5     while(cin>>n){
 6         char a[10005];
 7         int m=0,t=0,u=0,v=0;
 8         scanf("%s",a);
 9         for(int i=0;i<n;i++){
10             if(i%2==0){
11                 if(a[i]==‘r‘)m++;
12                 if(a[i]==‘b‘)t++;
13             }else{
14                 if(a[i]==‘r‘)u++;
15                 if(a[i]==‘b‘)v++;
16             }
17         }
18         int x=max(t,u);
19         int y=max(m,v);
20         int z=min(x,y);
21         cout<<z<<endl;
22     }
23     return 0;
24 }