AtCoder Beginner Contest 083

D - Wide Flip

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

You are given a string S consisting of 0 and 1. Find the maximum integer K not greater than |S| such that we can turn all the characters of S into 0 by repeating the following operation some number of times.

• Choose a contiguous segment [l,r] in S whose length is at least K (that is, r−l+1≥K must be satisfied). For each integer i such that l≤i≤r, do the following: if Si is 0, replace it with 1; if Si is 1, replace it with 0.

Constraints

• 1≤|S|≤105
• Si(1≤i≤N) is either 0 or 1.

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Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum integer K such that we can turn all the characters of S into 0 by repeating the operation some number of times.

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Sample Input 1

Copy

010

Sample Output 1

Copy

2

We can turn all the characters of S into 0 by the following operations:

• Perform the operation on the segment S[1,3] with length 3. S is now 101.
• Perform the operation on the segment S[1,2] with length 2. S is now 011.
• Perform the operation on the segment S[2,3] with length 2. S is now 000.

题意：

找出最大的k，满足题意。哈哈哈哈哈，实在不想翻译，偷懒一波

POINT：

如果k是x，

那么比如11101000,从左开始数，第k+1开始到末尾都可以任意变0,1。 比如变个[1,k]和[1,k+1]，那么k+1这个位置就变了。

从右开始数，那么[1,len-k]都可以任意变0，1。所以我们只要遍历x，然后找到他们从左从右都不能任意变的重合位置，判断这些位置上的值是不是相同的，若相同

则这个x可以，不相同就绝对不能成功。

感觉做的有点蠢。

#include <set>#include <map>#include <math.h>#include <vector>#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define  LL long longconst double pi = acos(-1);const double g = -9.8;const int maxn = 1e5+555; int main(){char s[maxn];scanf("%s",s);int ll=strlen(s);int ans=ll/2;int l,r;while(1){l=ll-ans;r=ans+1;int flag=1;int now=s[l-1];for(int i=l;i<r;i++){if(s[i]!=now){flag=0;break;}}if(!flag){break;}ans++;}printf("%d\n",ans);}