Codeforces Beta Round #94 div2 D 优先队列

B. String
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

One day in the IT lesson Anna and Maria learned about the lexicographic order.

String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < ixj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages​​.

The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.

Input

The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).

Output

Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes).

Examples
input
aa
2
output
a
input
abc
5
output
bc
input
abab
7
output
b
Note

In the second sample before string "bc" follow strings "a", "ab", "abc", "b".

思路:优先队列;被c++卡死。。我是c++11过的

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#define true ture
#define false flase
using namespace std;
#define ll __int64
#define inf 0xfffffff
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF )  return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
struct is
{
    string a;
    int st;
    bool operator <(const is &x) const
    {
        return a>x.a;
    }
};
string a;
int len;
priority_queue<is>q;
void k_big(int k)
{
    is s;
    for(ll i=0;i<len;i++)
    {
        s.a=a[i];
        s.st=i;
        q.push(s);
    }
    while(!q.empty())
    {
        s=q.top();
        //cout<<s.a<<" "<<s.st<<endl;
        q.pop();
        k--;
        if(!k)
        {
            printf("%s\n",s.a.c_str());
            return;
        }
        if(s.st<len-1)
        {
            s.a+=a[++s.st];
            q.push(s);
        }
    }
    printf("No such line.\n");
}
int main()
{
    int x,y,z,i,t;
    cin>>a;
    scanf("%d",&x);
    len=a.size();
    k_big(x);
    return 0;
}
View Code

 

posted @ 2016-05-03 20:22  jhz033  阅读(222)  评论(0)    收藏  举报