Codeforces Beta Round #94 div2 D 优先队列
2 seconds
256 megabytes
standard input
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 < 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.
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).
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).
aa
2
a
abc
5
bc
abab
7
b
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 = , ch ;
while( !( ( ch = getchar() ) >= '' && ch <= '' ) )
{
if( ch == EOF ) return << ;
}
res = ch - '' ;
while( ( ch = getchar() ) >= '' && ch <= '' )
res = res * + ( ch - '' ) ;
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=;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-)
{
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 ;
}
Codeforces Beta Round #94 div2 D 优先队列的更多相关文章
- 图论/暴力 Codeforces Beta Round #94 (Div. 2 Only) B. Students and Shoelaces
题目传送门 /* 图论/暴力:这是个连通的问题,每一次把所有度数为1的砍掉,把连接的点再砍掉,总之很神奇,不懂:) */ #include <cstdio> #include <cs ...
- BFS Codeforces Beta Round #94 (Div. 2 Only) C. Statues
题目传送门 /* BFS:三维BFS,坐标再加上步数,能走一个点当这个地方在步数内不能落到.因为雕像最多8步就会全部下落, 只要撑过这个时间就能win,否则lose */ #include <c ...
- Codeforces Beta Round #73(Div2)
A - Chord 题意:就是环中有12个字符,给你三个字符,判断他们之间的间隔,如果第一个和第二个间隔是3并且第二个和第三个间隔是4,那么就输出minor,如果第一个和第二个间隔是4并且第二个和第三 ...
- Codeforces Beta Round #94 (Div. 1 Only)B. String sam
题意:给你一个字符串,找第k大的子字符串.(考虑相同的字符串) 题解:建sam,先预处理出每个节点的出现次数,然后处理出每个节点下面的出现次数,然后在dfs时判断一下往哪边走即可,注意一下num会爆i ...
- Codeforces Beta Round #94 div 1 D Numbers map+思路
D. Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input ou ...
- Codeforces Beta Round #94 div 2 C Statues dfs或者bfs
C. Statues time limit per test 2 seconds memory limit per test 256 megabytes input standard input ou ...
- Codeforces Beta Round #94 div 2 B
B. Students and Shoelaces time limit per test 2 seconds memory limit per test 256 megabytes input st ...
- Codeforces Beta Round #107(Div2)
B.Phone Numbers 思路:就是简单的结构体排序,只是这里有一个技巧,就是结构体存储的时候,直接存各种类型的电话的数量是多少就行,在读入电话的时候,既然号码是一定的,那么就直接按照格式%c读 ...
- Codeforces Beta Round #76 (Div. 2 Only)
Codeforces Beta Round #76 (Div. 2 Only) http://codeforces.com/contest/94 A #include<bits/stdc++.h ...
随机推荐
- vcenter web client chrome浏览器打开中文显示乱码
使用如下链接试试看https://x.x.x.x/vsphere-client/?locale=zh_CN&csp
- [转载]C#深拷贝的方法
首先了解下深拷贝和浅拷贝的定义: 浅拷贝(影子克隆):只复制对象的基本类型,对象类型,仍属于原来的引用. 深拷贝(深度克隆):不紧复制对象的基本类,同时也复制原对象中的对象.就是说完全是新对 ...
- Twitter OA prepare: Rational Sum
In mathematics, a rational number is any number that can be expressed in the form of a fraction p/q ...
- C# Bulk Operations(转)
转自http://blog.csdn.net/winnyrain/article/details/51240684 Overcome SqlBulkCopy Limitations with C# B ...
- 集合框架—HashMap
HashMap提供了三个构造函数: HashMap():构造一个具有默认初始容量 (16) 和默认加载因子 (0.75) 的空 HashMap. HashMap(int ini ...
- 命名空间“Microsoft.Office.Interop”中不存在类型或命名空间名称“Excel”。是否缺少程序集引用 的另一种解决方案
一直以来都是使用tfs进行源代码管理,系统部署也是由我本机生成后发布到服务器上,某一日,进行发布操作时,报了 [命名空间“Microsoft.Office.Interop”中不存在类型或命名空间名称“ ...
- 如何发布Maven依赖到中央仓库
平时我们都是从Maven中央仓库下载依赖,如果我们想发布我们自己写的Maven依赖到中央仓库供别人下载使用应该怎么办?这里以上传自己写的simian-maven-plugin(https://gith ...
- javashop技术培训总结,架构介绍,Eop核心机制
javashop技术培训一.架构介绍1.Eop核心机制,基于spring的模板引擎.组件机制.上下文管理.数据库操作模板引擎负责站点页面的解析与展示组件机制使得可以在不改变核心代码的情况下实现对应用核 ...
- 返回xml过长时被nginx截断的解决办法
返回xml过长时被nginx截断的解决办法 问题描述:通过网页获取数据,数据格式为xml.当xml比较短时,可以正常获取数据.但是xml长度过长时不能正常获取数据,通过观察返回数据的源代码,发现xml ...
- python之路----网络编程--黏包
黏包现象 让我们基于tcp先制作一个远程执行命令的程序(命令ls -l ; lllllll ; pwd) res=subprocess.Popen(cmd.decode('utf-8'), shell ...