C. Plasticine zebra
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 00 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 ss (1≤|s|≤1051≤|s|≤105, where |s||s| denotes the length of the string ss) 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
5
input
Copy
bwwbwwb
output
Copy
3
Note
In the first example one of the possible sequence of operations is bwwwbww|bw →→ w|wbwwwbwb →→ wbwbwwwbw, that gives the answer equal to 55.
In the second example no operation can increase the answer.
同时翻转其实相当于把字符串看成是首尾连接的环
所以先把字符串后面接一个与自己一样的字符串
从头到尾【原字符串】枚举最长的长度 当然不可以超过原字符串的长度
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1e5 + 5;
string s;
int main()
{
while(cin>>s){
s += s;
int maxlen = 0;
for(int i = 0; i < s.length() / 2; ){
char now = s[i];
for(int j = i + 1; j - i <= s.length() / 2; j++){
if(s[j] != now){
now = s[j];
if(j - i == s.length() / 2){
maxlen = max(maxlen, j - i);
}
}
else{
maxlen = max(maxlen, j - i);
i = j;
break;
}
}
if(maxlen == s.length() / 2){
break;
}
}
printf("%d
", maxlen);
}
return 0;
}