C. Plasticine zebra
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||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.
这个题我也是很迷，赛后队友告诉我，不管怎么翻得到串一定是 s + s的子串
例如 bwwwbwwbw 翻转之后就是 bwwwbwwbwbwwwbwwbw的一个子串，所以直接在这个串里面找答案就可以了，
坑点只有一个字母答案是 1 &_&
#include<iostream>
#include<vector>
#include<set>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long LL;
const int MOD=1e9+7;
const int maxn=1e5+7;
const double ESP=1e-8;
int n;
char a[maxn<<1];
int main()
{
scanf("%s",a+1);
int len=strlen(a+1);
for(int i=1;i<=len;++i)
a[i+len]=a[i];
int ans=-2e9;
int cnt=1;
for(int i=2;i<=len*2;++i)
if(a[i]!=a[i-1])
{
++cnt;
}
else
{
ans=max(ans,cnt);
cnt=1;
}
ans=max(ans,cnt);
if(ans%len==0) ans=len;
cout<<ans<<endl;
}