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”.
贪心,在[l + 1, l +m-1]里选出最小,字典序最下,把比选中的最大的字符小的都加上?
AC代码:
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int K = 1e5 + 10;bool vis[K];char s[K],st[K],ss ='a';int main(){ int N; scanf("%d %s",&N,s); int nl = strlen(s),pl = 0,kl = 0; while(pl + N - 1 < nl){ int ml = pl; for(int i = pl; i <= pl + N - 1; i++) if(s[i] <= s[ml]) ml = i; st[++kl] = s[ml],ss = max(ss,s[ml]),pl = ml + 1,vis[ml] = 1; } for(int i = 0 ; i < nl ; i++) if(!vis[i] && s[i] < ss) st[++kl] = s[i]; sort(st + 1, st + 1 + kl); printf("%s/n",st + 1); return 0;}新闻热点
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