【codeforces 724D】【贪心】 Dense Subsequence 【一个字符串，按照一定的区间要求从中选出一些字符，使得这串字符的sort后字典序最小】

传送门：D. Dense Subsequence

描述：

D. Dense Subsequence

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

input

3
cbabc

output

a

input

2
abcab

output

aab

input

3
bcabcbaccba

output

aaabb

Note

In the first sample, one can choose the subsequence {3} and form a string "a".

In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a',
'b' and 'a') and rearrange the chosen symbols to form a string "aab".

题意:

给你一个字符串，让你从中选出一串字符，使得这串字符的sort后字典序最小，选字符的要求是如果上一个选的是i 那么[i+1,i+m]中至少选一个,[1,m]中必须选一个

思路：

贪心，先考虑只由'a'-c的字符组成该字符串，首先有‘a’的话肯定是拿'a'越多越好,拿'b'也是越多越好(aaabbXX肯定比aaabXXX字典序小)，也就c字符需要特判一下,因为c字符不是越多越好，而是越少越好，只有在满足前面的越多越好的前提下，不得不拿c才拿。

有了上面的策略我们具体做的时候就先满足区间长度不超过m的条件扫一遍字符串，再按字典序小的加入生成的字符串中策略扫一遍字符串，然后就好了。

比赛的时候贪心策略少考虑了点_(:зゝ∠)_ ，贪心没想象中这么简单！！！

代码：
#include <bits/stdc++.h>
#define pr(x) cout << #x << "= " << x << " " ；
#define pl(x) cout << #x << "= " << x << endl;
#define ll __int64
#define mod 1000000007
using namespace std;

#define mst(ss,b) memset(ss,b,sizeof(ss));
#define rep(i,k,n) for(int i=k;i<=n;i++)

const int N=1e5+10;

char s
,a,mx='a';
int vis
,top=0,m;

int main(){
scanf("%d",&m);
scanf("%s",s+1);
int n=strlen(s+1);
int now=1;
while(now+m-1<=n){
int res=now;
rep(i, now+1, now+m-1){
if(s[i]<=s[res])res=i;
}a[++top]=s[res];
vis[res]=1;
mx=max(mx, a[top]);
now=res+1;
}
rep(i, 1, n)if(!vis[i] && s[i]<mx)a[++top]=s[i];//贪心把字典序小的加入生成的字符串中
sort(a+1, a+top+1);
printf("%s\n", a+1);
return 0;
}