Codeforces Beta Round #94 (Div. 1 Only)String

B. String
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 < i) xj = 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”.

题意：
给一个字符串，要求求出它的字典序为第k小的那个子串。

思路：
模拟这整个过程，先求出字典序最小的那个串，再求出字典序为第二小的那个串….最后求出字典序为k 的那个串

#include<cstdio>#include<string>#include<set>#include<iostream>using namespace std;typedef pair<string,int> data;set<data>se;string ss;int k;int main(){    while(cin>>ss>>k)    {        long long  n=ss.size();        if(k>n*(n+1)/2)        {            puts("No such line.");            continue;        }        se.clear();        string a[26]={"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};        for(int i=0;i<ss.size();i++)        {            se.insert(data(a[ss[i]-'a'],i+1));        }        k--;        while(k--)        {           data now=*se.begin();            se.erase(se.begin());            if(now.second==n)                continue;            now.first+=ss[now.second];            now.second++;            se.insert(now);        }        cout<<se.begin()->first<<endl;    }    return 0;}