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

本文由 Ocrosoft 于 2016-09-24 19:52:48 发表
B. Anatoly and Cockroaches

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.


Solution

题意:给出一个颜色序列,r为红,b为黑。可以任意交换两个颜色,或者改变其中一个颜色,使该序列变成颜色交替的序列。
思路:假设长度为5,则最终状态为rbrbr或者brbrb,所以只要计算当前序列与这两种状态不同的位置的数量。不同的字母需要分开统计,如果两个字母都有不同的,那就交换,如果一个没有了,那就变色。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
//#include "Osort.h"
#define ms(a) memset(a,0,sizeof(a))
typedef long long LL;
const LL LINF = 1e17;
const int INF = INT_MAX / 2;
const int MAXN = 150 + 10;
const int MOD = 100000;
int gcd(int a, int b)
{
	if (!b)return a;
	return gcd(b, a%b);
}
using namespace std;
//or2..or2..or2..or2..//
int main()
{
	string s; int len, la[2] = { 0 }, lb[2] = { 0 }; cin >> len;
	cin >> s;
	for (int i = 0; i < len; i++)
	{
 		if (i % 2 && s[i] != 'b')la[0]++;//奇数,rbrbrbrb
		if (i % 2 && s[i] != 'r')lb[0]++;//奇数,brbrbrbr
		if (!(i % 2) && s[i] != 'r')la[1]++;
		if (!(i % 2) && s[i] != 'b')lb[1]++;
	}
	int a = min(la[0], la[1]); la[0] -= a, la[1] -= a; a += la[0] + la[1];
	int b = min(lb[0], lb[1]); lb[0] -= b, lb[1] -= b; b += lb[0] + lb[1];
	int ans = min(a, b);
	printf("%d\n", ans);
	return 0;
}

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