Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) D. Dense Subsequence 暴力

D. Dense Subsequence

题目连接：

http://codeforces.com/contest/724/problem/D

Description

You are given a string s, consisting of lowercase English letters, and the integer m.

One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.

Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.

Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < ... < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j,  j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.

Then we take any permutation p of the selected indices and form a new string sip1sip2... sipt.

Find the lexicographically smallest string, that can be obtained using this procedure.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).

The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s.

Output

Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.

Sample Input

3
cbabc

Sample Output

a

Hint

题意

给你一个m和一个字符串s

一开始你的now为0，你需要在[now,now+m-1]里面选一个字符，假设选的位置为p，则now=p+1，然后重复

直到now+m-1>=s.size()，并且不能选择使得字典序变得更小的时候，就不选了。

你需要找到字典序最小的答案，答案最后可以排序，选出来的东西。

题解：

暴力枚举最大的字符是什么，然后再贪心的去放，小于这个字符的全选，其他的贪心的去放，使得能够覆盖所有区域。

肯定是放覆盖当前区间最后面的那个。

代码

#include<bits/stdc++.h>
using namespace std;

set<int>S[26];
int m;
string s;
int cnt[26];
int main()
{
    cin>>m>>s;
    for(int i=0;i<26;i++)
    {
        int last=-1,lastcur=-1e5,flag=1;
        memset(cnt,0,sizeof(cnt));
        for(int j=0;j<s.size();j++)
        {
            if(s[j]==i+'a')
                lastcur=j;
            else if(s[j]<i+'a'){
                last=j;
                cnt[s[j]-'a']++;
            }
            if(j-last>=m){
                if(lastcur<=last){
                    flag=0;
                    break;
                }
                last=lastcur;
                cnt[i]++;
            }
        }
        if(flag==0)continue;
        for(int ii=0;ii<26;ii++)
            for(int jj=0;jj<cnt[ii];jj++)
                printf("%c",ii+'a');
        cout<<endl;
        return 0;
    }
}