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

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

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".

Source

Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined)

My Solution

ST表+贪心

ST表，用O(nlogn)的预处理，然后O(1)查询，找出ans中出现的最大的字母，

然后把比它小的字母都从小到大push到ans里，

然后贪心的去那个需要的最大字母的个数，

每次if(maxv == query_min(i, i + m - 1)) 则在这个 [ i, i + m -1 ]区间内找出最右端的一个maxv + 'a'  字母，//如果优先选左边的，则右边的还会取到，所以取区间最右边的maxv+'a'

就好。

复杂度 O(nlogn)

#include <iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;typedef long long LL;const int maxn = 1e5 + 8;const int MAXN = 1e5 + 8;int stTable[MAXN][32], preLog2[MAXN], arr[MAXN];inline void st_prepare(const int &n){    preLog2[1] = 0;    for(int i = 2; i <= n; i++){        preLog2[i] = preLog2[i-1];        if((1 << (preLog2[i] + 1)) == i){            preLog2[i]++;        }    }    for(int i = n - 1; i >= 0; i--){        stTable[i][0] = arr[i];        for(int j = 1; (i + (1 << j) - 1) < n; j++){            stTable[i][j] = min(stTable[i][j - 1], stTable[i + (1 << j - 1)][j - 1]);        }    }}inline int query_min(const int &l, const int r){    int len = r - l + 1, k = preLog2[len];    return min(stTable[l][k], stTable[r - (1 << k) + 1][k]);}int cnt[26];string s, ans;//bool flag[maxn];int main(){    #ifdef LOCAL    freopen("d.txt", "r", stdin);    //freopen("d.out", "w", stdout);    int T = 4;    while(T--){    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    for(int i = 0; i < maxn; i++){        arr[i] = 1e9 + 7;    }    int m, sz;    cin >> m;    cin >> s;    sz = s.size();    ans.clear();    for(int i = 0; i < sz; i++){        cnt[s[i] - 'a']++;        arr[i] = s[i] - 'a';        //cout << arr[i+1];    }    st_prepare(sz);    int maxv = -1;    for(int i = 0; i < sz; i++){        if(i + m - 1 < sz){            maxv = max(maxv, query_min(i, i + m - 1));        }        else break;    }    for(int i = 0; i < maxv; i++){        while(cnt[i]){            ans += 'a' + i;            cnt[i]--;        }    }    for(int i = 0; i < sz; i++){        //cout << i << endl;        if(i + m - 1 < sz){            if(maxv == query_min(i, i + m - 1)){                for(int j = i + m - 1; j >= i; j--){         //在区域内找到最右边的maxv, 比赛的时候找的最左边的 maxv 尴尬                    if(s[j] - 'a' == maxv){                        ans += 'a' + maxv;                        i = j;                        //i += m - 1;      //不能跳跃，看样例                        //cout << i << endl;                        break;                    }                }            }        }        else break;    }    cout << ans << endl;    #ifdef LOCAL    memset(cnt, 0, sizeof cnt);    cout << endl;    }    #endif // LOCAL    return 0;}

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