Codeforces Round #334 (Div. 2) C. Alternative Thinking 贪心

C. Alternative Thinking

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/604/problem/C

Description

Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.

However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as anot-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and{1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.

Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.

Input

The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).

The following line contains a binary string of length n representing Kevin's results on the USAICO.

Output

Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.

Sample Input

8
10000011

Sample Output

5

HINT

题意

给你一个字符串，你可以使得一个连续的01串翻转过来，然后问你最长的01相隔的子序列（不连续）的长度为多少

题解：

答案一定是max(n,maxlen+2)

maxlen是原本串的最长的01串长度

代码：

#include<iostream>
#include<stdio.h>
using namespace std;

string s;
int main()
{
    int n;
    scanf("%d",&n);
    cin>>s;
    if(n==)
    {
        printf("2\n");
        return ;
    }
    if(n==)
    {
        printf("3\n");
        return ;
    }
    int flag = ;
    for(int i=;i<s.size()-;i++)
    {
        if(s[i]==s[i+])
            flag = ;
    }
    if(flag == )
    {
        printf("%d\n",n);
        return ;
    }
    int ans = ;
    for(int i=;i<s.size()-;i++)
    {
        if(s[i]!=s[i+])
            ans++;
    }
    printf("%d\n",min(ans+,n));
}