C. Good String Codeforces Round #560 (Div. 3)

最新推荐文章于 2023-02-04 15:01:53 发布

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最新推荐文章于 2023-02-04 15:01:53 发布

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本文详细解析了C.GoodString问题，探讨了如何通过最少的字符删除使字符串符合特定条件，即长度为偶数且奇数位置字符不等于其后的偶数位置字符。文章提供了完整的代码实现及样例解释，帮助读者理解并解决此类字符串操作问题。

C. Good String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's call (yet again) a string good if its length is even, and every character in odd position of this string is different from the next character (the first character is different from the second, the third is different from the fourth, and so on). For example, the strings good, stringand xyyx are good strings, and the strings bad, aa and aabc are not good. Note that the empty string is considered good.

You are given a string ss, you have to delete minimum number of characters from this string so that it becomes good.

Input

The first line contains one integer n (1≤n≤2⋅1051≤n≤2⋅105) — the number of characters in s.

The second line contains the string s, consisting of exactly n lowercase Latin letters.

Output

In the first line, print one integer k (0≤k≤n0≤k≤n) — the minimum number of characters you have to delete from s to make it good.

In the second line, print the resulting string s. If it is empty, you may leave the second line blank, or not print it at all.

Examples

input

Copy

4
good

output

Copy

0
good

input

Copy

4
aabc

output

Copy

2
ab

input

Copy

3
aaa

output

Copy

题意：题目中规定合格的字符串的标准：1，长度为偶数 2，字符串奇数位和下一位即偶数位数字不能相等。求在原字符串中最少删除几个字符使其成为符合要求的字符串。

我的代码：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <stack>

using namespace std;
stack<char>s;
char a[400005];
stack<char>ch;

void prtstk(){
	while(!s.empty()){
		ch.push(s.top());
		s.pop();
	}
	while(!ch.empty()){
		cout<<ch.top();
		s.push(ch.top());
		ch.pop();
	}
	cout<<endl;
}

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		while(!s.empty())
		{
			s.pop(); 
		}
		while(!ch.empty())
		{
			ch.pop(); 
		}
		scanf("%s",a);
		int len = strlen(a);
		for(int i=0;i<len;i++)
		{
			if(i == 0)
			{
				s.push(a[i]);
			}
			else
			{
				if(s.size()%2==0)
				{
					s.push(a[i]);
				}
				else
				{
					if(s.top()!=a[i])
					{
						s.push(a[i]);
					}
				}
			}
		}
		if(s.size()%2==1)
		{
			printf("%d\n",n-s.size()+1);
			s.pop();
			if(s.size()!=0)
			{
				prtstk();
			}
		}
		else
		{
			printf("%d\n",n-s.size());
			if(s.size()!=0)
			{
				prtstk(); 
			}
		}
	}
	return 0;
}

容易错的样例：

输入

gdoo

aaaaa

aabba

输出

（为空串）

abba