C. Plasticine zebra（思维+猜想）

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本文解析了Codeforces C题的解决思路，通过将字符串加倍并寻找最长交替序列的方法来求解最大长度的斑马条纹问题。介绍了如何通过翻转部分序列来优化答案，并给出了实现代码。

C. Plasticine zebra

http://codeforces.com/contest/1025/problem/C

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Is there anything better than going to the zoo after a tiresome week at work? No wonder Grisha feels the same while spending the entire weekend accompanied by pretty striped zebras.

Inspired by this adventure and an accidentally found plasticine pack (represented as a sequence of black and white stripes), Grisha now wants to select several consequent (contiguous) pieces of alternating colors to create a zebra. Let's call the number of selected pieces the length of the zebra.

Before assembling the zebra Grisha can make the following operation 0

or more times. He splits the sequence in some place into two parts, then reverses each of them and sticks them together again. For example, if Grisha has pieces in the order "bwbbw" (here 'b' denotes a black strip, and 'w' denotes a white strip), then he can split the sequence as bw|bbw (here the vertical bar represents the cut), reverse both parts and obtain "wbwbb".

Determine the maximum possible length of the zebra that Grisha can produce.

Input

The only line contains a string s

(1≤|s|≤105, where |s| denotes the length of the string s

) comprised of lowercase English letters 'b' and 'w' only, where 'w' denotes a white piece and 'b' denotes a black piece.

Output

Print a single integer — the maximum possible zebra length.

Examples

Input

Copy

bwwwbwwbw

Output

Copy

Input

Copy

bwwbwwb

Output

Copy

Note

In the first example one of the possible sequence of operations is bwwwbww|bw →

w|wbwwwbwb → wbwbwwwbw, that gives the answer equal to 5

In the second example no operation can increase the answer.

这道题就是个思维题。将串加倍。然后找最长不连续的个数，再跟n比较，取最小值。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL ans,len;
int main()
{
    len=1;
    ans=1;
    string h;
    cin>>h;
    LL n=h.length();
    h=h+h;
   /// cout<<"h:"<<h<<endl;

    for(int i=1;i<h.length()-1;i++){
        if(h[i]!=h[i-1]){
            len++;
              ans=max(ans,len);
        }
        else{
            ans=max(ans,len);
            len=1;
        }
///cout<<"i:"<<i<<"    len:"<<len<<endl;
    }
    ans=min(ans,n);
    printf("%d\n",ans);
}