Codeforces Round #373 (Div. 2) Anatoly and Cockroaches —

Codeforces Round #373 (Div. 2) Anatoly and Cockroaches —— 贪心

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

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
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.

题解：

1.遍历一遍，找出b放错的个数和r放错的个数， min（b，r）即为交换的次数， max（b，r）-min（b，r）即为涂色的次数，所以总的次数为max（b，r）

2.由于序列只能是rbrbrb……或者brbrbr……，所以分两次遍历，然后取最小值，即ans = min（max（b1，r1）， max（b2，r2））

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <sstream>
#include <algorithm>
using namespace std;
#define pb push_back
#define mp make_pair
#define ms(a, b)  memset((a), (b), sizeof(a))
#define eps 0.0000001
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 100000+10;
char s[maxn],a[maxn];

int main()
{
    int n, t1,t2, ans;
    scanf("%d",&n);
    scanf("%s",s);

    t1 = t2 = 0;
    for(int i = 0; i<n; i++)
    {
        if((i&1) && s[i]=='r')
            t1++;

        if(!(i&1) && s[i]=='b')
            t2++;
    }

    ans = max(t1,t2);
    t1 = t2 = 0;
    for(int i = 0; i<n; i++)
    {
        if((i&1) && s[i]=='b')
            t1++;

        if(!(i&1) && s[i]=='r')
            t2++;
    }

    ans = min(  ans, max(t1,t2)  );
    printf("%d
",ans);
}