Plasticine zebratime 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.
Examplesinput
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.
题意：没有连续的b和w的字符串为zebra length，求一个字符串的zebra length最长长度，字符串可变化：将字符串分成两部分然后倒置每部分组成新的字符串
分析：考虑字符串的变化：将字符串分为前缀和后缀两部分，字符串的变化就是将后缀倒置前缀倒置
考虑将两个一样的字符串连接在一起得到新字符串，则新字符串的子串包含了原来一个字符串的所有变化
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
map<string,ll> mp;
string s;
int main() {
ios::sync_with_stdio(0);
cin >> s;
string ts = "";
ts = s + s;
ll ans = 1, maxans = -1;
for( ll i = 0; i < ts.length(); i ++ ) {
if( i == 0 ) {
continue;
}
if( ts[i] != ts[i-1] ) {
ans ++;
if( i == ts.length()-1 ) {
maxans = max(maxans,ans);
}
} else {
maxans = max(maxans,ans);
ans = 1;
}
}
if( maxans > s.length() ) {
maxans = maxans - s.length();
}
cout << maxans << endl;
return 0;
}
