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最新推荐文章于 2020-03-27 13:23:17 发布

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最新推荐文章于 2020-03-27 13:23:17 发布

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分类专栏： codeforces

本文链接：https://blog.youkuaiyun.com/qq_36711003/article/details/80866592

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本文介绍了一个简单的算法，该算法按照字母顺序从给定字符串中删除指定数量的字符，并提供了实现这一功能的C++代码示例。

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问题描述：

You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ( $k \leq n$ ) from the string $s$ . Polycarp uses the following algorithm $k$ times:

if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
...
remove the leftmost occurrence of the letter 'z' and stop the algorithm.

This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly $k$ times, thus removing exactly $k$ characters.

Help Polycarp find the resulting string.

Input

The first line of input contains two integers $n$ and $k$ ( $1 \leq k \leq n \leq 4 \cdot 10^{5}$ ) — the length of the string and the number of letters Polycarp will remove.

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

Output

Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ times.

If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).

问题分析：

从a到z顺序删除，一直到删除了k个字符为止。输出删除后的字符。

AC代码：

#include <iostream>
#include<iomanip>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<set>
#include<stack>
#include<queue>
using namespace std;

const int N = 400005;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n,k;
	string s;
	cin>>n>>k>>s;
	int flag[N];
	memset(flag,1,sizeof(flag));
	for(int i = 0;i < 26;i++){
		for(int j = 0;j < s.size();j++){
			//从a开始删除 
			if(s[j] == 'a'+i && k > 0){
				k--;
				//标记已经删除了的字符 
				flag[j] = 0;
			}
		}
	}
	for(int i = 0;i < s.size();i++){
		if(flag[i]){
			cout<<s[i];
		}
	}
	cout<<endl;
	return 0;
}