Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) -- D. Dense Subsequence （技巧枚举）

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大体题意：

给你一个字符串，和一个数m ，要求在这个字符串中选出一些位置，使得任意连续的长度为m 的字符串中至少包括一个这些位置的字符！输出这些字符组成字符串的最小字典序！

思路：

借鉴了学长的博客！

感觉很巧妙：

枚举最后一个字母id！

然后开始扫这个字符串，如果这个字符小于 id 加入到ans里

如果等于id 先用栈记录一下，为什么用栈呢？因为最后用的时候肯定是用最远的一个，栈最后加入的就是最远的！

如果跑了长度为m 了还是没找到比id 小的，就处理等于id的。

这个栈里第一个元素，如果位置在那个长度为m 的区间内就是合法的！

然后输出记录的结果即可！

#include <bits/stdc++.h>using namespace std;const int maxn = 100000 + 10;char s[maxn];int len,m;set<int>ans;stack<int>sk;bool ok(int id){    ans.clear();    while(!sk.empty())sk.pop();    int sum = 0;    for (int i = 0; i < len; ++i){        if (s[i] < id+'a')ans.insert(i),sum = 0;        else if (s[i] == id+'a')sk.push(i),++sum;        else ++sum;        if (sum == m){            if (sk.empty())return 0;            int pos = sk.top();            if (pos >= i-m+1 && pos <= i)ans.insert(pos);            else return 0;            sum = i-pos;//            if (i == 1)printf("sum = %d\n",sum);        }    }    return 1;}int main(){    scanf("%d",&m);    scanf("%s",s);    len = strlen(s);    for (int i = 0; i < 26; ++i){        if (ok(i)) break;    }    string t = "";    while(!ans.empty()){        t += s[*(ans.begin())];        ans.erase(ans.begin());    }    sort(t.begin(),t.end());    printf("%s\n",t.c_str());    return 0;}

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
3cbabc
output
a
input
2abcab
output
aab
input
3bcabcbaccba
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".

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