Codeforces 604 C Alternative Thinking【思维】

本文探讨了一个有趣的问题:如何通过对一个二进制字符串进行一次区间翻转操作,来最大化该字符串中最长交替子序列的长度。文章提供了详细的解决思路及AC代码。

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C. Alternative Thinking

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 a not-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.

Examples

Input

8
10000011

Output

5

Input

2
01

Output

2

Note

In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.

In the second sample, Kevin can flip the entire string and still have the same score.

 

题目大意:

给你一个长度为n的01串,我们可以进行一次区间翻转(起点终点自己定),得到一个新的01串,使得得到的新的01串中的一个子串最长,且这个子串是01间隔的。

e.g:

1010 4

1011 3

1101101 5


思路:


1、通过枚举发现,在原串中如果有一段长度大于3的连续相等子串:10000000001-------------->10100000001,那么其解ans=原串最长子串长度+2.


2、再通过枚举发现,在原串中有一段长度为2的连续相等的子串:110101---------->010101,那么其解ans=原串最长子串长度+1.


以上两种方案都是通过改变原串某一个字符增加长度的,接下来我们再枚举一下通过区间翻转的情况带来增加的情况:


3、再通过枚举发现,在原串中假如有两段以及两段以上的长度为2的连续相等子串:1100---------------->1010,那么其解ans=原串最长子串长度+2.


4、思路构建完毕,剩下的工作就是敲代码、


Ac代码:


#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char a[1000000];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int output=1;
        scanf("%s",a);
        int maxn=0;
        int contz=1;
        int cont2=0;
        for(int i=0;i<n-1;i++)
        {
            if(a[i]==a[i+1])
            {
                contz++;
            }
            else
            {
                if(contz==2)cont2++;
                contz=1,output++;
            }
            maxn=max(contz,maxn);
        }
        if(contz==2)cont2++;
        if(maxn>=3)output+=2;
        if(maxn==2&&cont2>1)output+=2;
        if(maxn==2&&cont2==1)output+=1;
        printf("%d\n",output);
    }
}





### Codeforces 思维题解题思路和技巧 #### 预处理的重要性 对于许多竞赛编程问题而言,预处理能够显著提高效率并简化后续操作。通过提前计算某些固定的数据结构或模式匹配表,可以在实际求解过程中节省大量时间。例如,在字符串处理类题目中预先构建哈希表来加速查找过程[^1]。 #### 算法优化策略 针对特定类型的输入数据设计高效的解决方案至关重要。当面对大规模测试案例时,简单的暴力破解往往无法满足时限要求;此时则需考虑更高级别的算法改进措施,比如动态规划、贪心算法或是图论中的最短路径算法等。此外,合理利用空间换取时间也是一种常见的优化手段[^2]。 #### STL库的应用价值 C++标准模板库提供了丰富的容器类型(vector, deque)、关联式容器(set,map)以及各种迭代器支持,极大地便利了程序开发工作。熟练掌握这些工具不仅有助于快速实现功能模块,还能有效减少代码量从而降低出错几率。特别是在涉及频繁插入删除场景下,优先选用双向队列deque而非单向链表list可获得更好的性能表现。 ```cpp #include <iostream> #include <deque> using namespace std; int main(){ deque<int> dq; // 向两端添加元素 dq.push_back(5); dq.push_front(3); cout << "Front element is: " << dq.front() << endl; cout << "Back element is : " << dq.back() << endl; return 0; } ``` #### 实际应用实例分析 以一道具体题目为例:给定一系列查询指令,分别表示往左端/右端插入数值或者是询问某个指定位置到边界之间的最小距离。此题目的关键在于如何高效地追踪最新状态而无需重复更新整个数组。采用双指针技术配合静态分配的一维数组即可轻松解决上述需求,同时保证O(n)级别的总运行成本[^4]。
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