AtCoder Beginner Contest 083 - D - Wide Flip（思维）

最新推荐文章于 2021-07-30 19:59:19 发布

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本文介绍了一个有趣的编程挑战，目标是在给定字符串S由0和1组成的情况下，寻找最大整数K，使得通过一系列操作可以使所有字符变为0。文章详细解释了解决方案的思路，并提供了完整的代码实现。

D - Wide Flip

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.

Input

Input is given from Standard Input in the following format:

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.

Sample Input 1

Copy

Sample Output 1

Copy

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 long
const 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);
}