codeforces 557 E Ann and Half-Palindrome

题意是要求出一个串的第k大的半回文子串

半回文串的定义是：若一个串其实位置为1，那么当所有奇数位i，且i<=(s.length+1/2)，满足s[i]=s[s.length-i+1]的时候，

那么这个串就是半回文串。

作法就是，把这个串的所有半回文子串建成一个字典树，然后查第k大就好了

#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;struct node{char val;int vis,sum;node *next[2];};string s,ans;void dfs(node *now,int k){if(now->val<'c'){ans+=now->val;k-=now->vis;if(k<1)return;}if(now->next[0]==NULL){if(now->next[1]==NULL)return;dfs(now->next[1],k);}else{if(now->next[1]==NULL)dfs(now->next[0],k);else{if(now->next[0]->sum>=k)dfs(now->next[0],k);elsedfs(now->next[1],k-now->next[0]->sum);}}}bool dp[5010][5010];void Dp(int n){for(int i=0;i<n;i++)dp[i][i]=1;for(int i=1;i<n;i++)if(s[i-1]==s[i])dp[i-1][i]=1;for(int i=2;i<n;i++)if(s[i-2]==s[i])dp[i-2][i]=1;for(int i=3;i<n;i++)if(s[i-3]==s[i])dp[i-3][i]=1;for(int i=5;i<=n;i++)for(int j=0;j+i-1<n;j++)if(s[j]==s[j+i-1]&&dp[j+2][j+i-3])dp[j][j+i-1]=1;}int bg,ed;void add(node *now,int index){for(int i=0;i<2;i++)if(s[index]=='a'+i){if(now->next[i]==NULL){now->next[i]=new node;memset(now->next[i],0,sizeof(node));now->next[i]->val=s[index];}if(index==ed){now->next[i]->vis++;now->next[i]->sum++;}else{if(dp[bg][index]){now->next[i]->vis++;now->next[i]->sum++;}add(now->next[i],index+1);}}now->sum=now->vis;for(int i=0;i<2;i++)if(now->next[i]!=NULL)now->sum+=now->next[i]->sum;}int main(){cin>>s;int k;cin>>k;int n=s.length();Dp(n);node *fs=new node;memset(fs,0,sizeof(node));fs->val='c';for(int i=0;i<n;i++)for(int j=n-1;j>=i;j--)if(dp[i][j]){bg=i;ed=j;add(fs,i);break;}dfs(fs,k);cout<<ans;}

time limit per test

1.5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Tomorrow Ann takes the hardest exam of programming where she should get an excellent mark.

On the last theoretical class the teacher introduced the notion of a half-palindrome.

String t is a half-palindrome, if for all the odd positions i () the following condition is held: ti = t|t| - i + 1, where |t| is the length of string t if positions are indexed from 1. For example, strings "abaa", "a", "bb", "abbbaa" are half-palindromes and strings "ab", "bba" and "aaabaa" are not.

Ann knows that on the exam she will get string s, consisting only of letters a and b, and number k. To get an excellent mark she has to find the k-th in the lexicographical order string among all substrings of s that are half-palyndromes. Note that each substring in this order is considered as many times as many times it occurs in s.

The teachers guarantees that the given number k doesn't exceed the number of substrings of the given string that are half-palindromes.

Can you cope with this problem?

Input

The first line of the input contains string s (1 ≤ |s| ≤ 5000), consisting only of characters 'a' and 'b', where |s| is the length of string s.

The second line contains a positive integer k —  the lexicographical number of the requested string among all the half-palindrome substrings of the given string s. The strings are numbered starting from one.

It is guaranteed that number k doesn't exceed the number of substrings of the given string that are half-palindromes.

Output

Print a substring of the given string that is the k-th in the lexicographical order of all substrings of the given string that are half-palindromes.

Sample test(s)

input

abbabaab7

output

abaa

input

aaaaa10

output

aaa

input

bbaabb13

output

bbaabb

Note

By definition, string a = a1a2... an is lexicographically less than string b = b1b2... bm, if either a is a prefix of b and doesn't coincide withb, or there exists such i, that a1 = b1, a2 = b2, ... ai - 1 = bi - 1, ai < bi.

In the first sample half-palindrome substrings are the following strings — a, a, a, a, aa, aba, abaa, abba, abbabaa, b, b, b, b, baab,bab, bb, bbab, bbabaab (the list is given in the lexicographical order).