DeCantor Expansion (逆康托展开)
Background\text{Background}Background
The \text{The }The Listen&Say
Test will be hold on May 11, so I decided to fill my blog \text{Test will be hold on May 11, so I decided to fill my blog }Test will be hold on May 11, so I decided to fill my blog
with English words until that day.\text{with English words until that day.}with English words until that day.
Problem\text{Problem}Problem
There goes a problem.\text{There goes a problem.}There goes a problem.
You’ve got 2 intergers N,k. Please calculate the kth permutation of ∀k∈[i,n].\text{You've got 2 intergers }N,k\text{. Please calculate the }k\text{th permutation of }\forall k\in[i,n].You’ve got 2 intergers N,k. Please calculate the kth permutation of ∀k∈[i,n].
Solution\text{Solution}Solution
It’s easy to know that we can got it by Depth-first Search, \text{It's easy to know that we can got it by Depth-first Search, }It’s easy to know that we can got it by Depth-first Search, but its Time complexity is O(n!).\text{but its Time complexity is }O(n!).but its Time complexity is O(n!).
DeCantor Expansion
is a algorithm which can solve problems like these calculating the kth permutation\text{is a algorithm which can solve problems like these calculating the }k\text{th permutation}is a algorithm which can solve problems like these calculating the kth permutationin O(nlogn) with heap optimization.\text{in }O(n\log n)\text{ with heap optimization.}in O(nlogn) with heap optimization.
Let’s explain how it works in a simple example. Set N=5,k=61, the answer is a[].\text{Let's explain how it works in a simple example. Set }N=5,k=61,\text{ the answer is }a[].Let’s explain how it works in a simple example. Set N=5,k=61, the answer is a[].
1.Let 61 / 4! = 2 ... 13, it shows that there’re 2 numbers behind a[1] are smaller than a[1].\text{1.\quad Let 61 / 4! = 2 ... 13, it shows that there're 2 numbers behind }a[1]\text{ are smaller than a[1].}1.Let 61 / 4! = 2 ... 13, it shows that there’re 2 numbers behind a[1] are smaller than a[1].
Therefore, a[1]=3;\text{Therefore, }a[1]=3;Therefore, a[1]=3;
2.Let 13 / 3! = 2 ... 1, it shows that there’re 2 numbers behind a[2] are smaller than a[2].\text{2.\quad Let 13 / 3! = 2 ... 1, it shows that there're 2 numbers behind }a[2]\text{ are smaller than a[2].}2.Let 13 / 3! = 2 ... 1, it shows that there’re 2 numbers behind a[2] are smaller than a[2].
Therefore, a[2]=4;\text{Therefore, }a[2]=4;Therefore, a[2]=4;
3.Let 1 / 2! = 0 ... 1, it shows that there’re 0 number behind a[3] are smaller than a[3].\text{3.\quad Let 1 / 2! = 0 ... 1, it shows that there're 0 number behind }a[3]\text{ are smaller than a[3].}3.Let 1 / 2! = 0 ... 1, it shows that there’re 0 number behind a[3] are smaller than a[3].
Therefore, a[3]=1;\text{Therefore, }a[3]=1;Therefore, a[3]=1;
4.Let 1 / 1! = 1 ... 0, it shows that there’re 1 number behind a[4] are smaller than a[4].\text{4.\quad Let 1 / 1! = 1 ... 0, it shows that there're 1 number behind }a[4]\text{ are smaller than a[4].}4.Let 1 / 1! = 1 ... 0, it shows that there’re 1 number behind a[4] are smaller than a[4].
Therefore, a[4]=5;\text{Therefore, }a[4]=5;Therefore, a[4]=5;
Therefore, a[5]=2,a[]={3,4,1,5,2}.\text{Therefore, }a[5]=2, a[]=\{3,4,1,5,2\}.Therefore, a[5]=2,a[]={3,4,1,5,2}.
Summary\text{Summary}Summary
∀i∈[1,n−1], let k / (n−1)!, the answer you’ve got is the number of interger j∈[i+1,n] which has a[j]<a[i]. And let k equals to the remainder.\forall i\in[1,n-1],\text{ let }k\ /\ (n-1)!\text{, the answer you've got is the number of interger }\newline j\in[i+1,n]\text{ which has }a[j]<a[i].\text{ And let }k\text{ equals to the remainder.}∀i∈[1,n−1], let k / (n−1)!, the answer you’ve got is the number of interger j∈[i+1,n] which has a[j]<a[i]. And let k equals to the remainder.
The End\text{The End}The End
Reference material\text{Reference material}Reference material
DeCantor Expansion (逆康托展开)的更多相关文章
- LightOJ1060 nth Permutation(不重复全排列+逆康托展开)
一年多前遇到差不多的题目http://acm.fafu.edu.cn/problem.php?id=1427. 一开始我还用搜索..后来那时意外找到一个不重复全排列的计算公式:M!/(N1!*N2!* ...
- nyoj 139——我排第几个|| nyoj 143——第几是谁? 康托展开与逆康托展开
讲解康托展开与逆康托展开.http://wenku.baidu.com/view/55ebccee4afe04a1b071deaf.html #include<bits/stdc++.h> ...
- 题解报告:NYOJ 题目143 第几是谁?(逆康托展开)
描述 现在有"abcdefghijkl”12个字符,将其按字典序排列,如果给出任意一种排列,我们能说出这个排列在所有的排列中是第几小的.但是现在我们给出它是第几小,需要你求出它所代表的序列. ...
- HDU1027 Ignatius and the Princess II( 逆康托展开 )
链接:传送门 题意:给出一个 n ,求 1 - n 全排列的第 m 个排列情况 思路:经典逆康托展开,需要注意的时要在原来逆康托展开的模板上改动一些地方. 分析:已知 1 <= M <= ...
- 康托展开&逆康托展开学习笔记
啊...好久没写了...可能是最后一篇学习笔记了吧 题目大意:给定序列求其在全排列中的排名&&给定排名求排列. 这就是康托展开&&逆康托展开要干的事了.下面依次介绍 一 ...
- Codeforces-121C(逆康托展开)
题目大意: 给你两个数n,k求n的全排列的第k小,有多少满足如下条件的数: 首先定义一个幸运数字:只由4和7构成 对于排列p[i]满足i和p[i]都是幸运数字 思路: 对于n,k<=1e9 一眼 ...
- hdoj 1027 Ignatius and the Princess II 【逆康托展开】
Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K ( ...
- 康托展开与逆康托展开模板(O(n^2)/O(nlogn))
O(n2)方法: namespace Cantor { ; int fac[N]; void init() { fac[]=; ; i<N; ++i)fac[i]=fac[i-]*i; } in ...
- 【HDU - 1043】Eight(反向bfs+康托展开)
Eight Descriptions: 简单介绍一下八数码问题:在一个3×3的九宫格上,填有1~8八个数字,空余一个位置,例如下图: 1 2 3 4 5 6 7 8 在上图中,由于右下角位置是空的 ...
随机推荐
- Guava的RateLimiter实现接口限流
最近开发需求中有需要对后台接口进行限流处理,整理了一下基本使用方法. 首先添加guava依赖: <dependency> <groupId>com.google.guava&l ...
- Jenkins把GitHub项目做成Docker镜像
本文是<Jenkins流水线(pipeline)实战>系列的第三篇,前面已对Jenkins流水线有了基本认识,也试过从GitHub下载pipeline脚本并执行,今天的实战是编写一段pip ...
- Sping学习笔记(一)----Spring源码阅读环境的搭建
idea搭建spring源码阅读环境 安装gradle Github下载Spring源码 新建学习spring源码的项目 idea搭建spring源码阅读环境 安装gradle 在官网中下载gradl ...
- Vue学习之vue属性绑定和双向数据绑定
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- jenkins自动化部署项目3 --设置用户
我直接设置的admin ,jenkins可以新建多个用户,并赋予不同的权限(TODO) 等后续需要严格规范操作人的时候再补充
- JAVA设计模式---总述篇
一.设计模式(Design Pattern): 1.设计模式的概念 是前辈们对代码开发经验的总结,是解决特定问题的一系列套路.它不是语法规定,而是一套用来提高代码可复用性.可维护性.可读性.稳健性以及 ...
- golang1.13中重要的新特新
本文索引 语言变化 数字字面量 越界索引报错的完善 工具链改进 GOPROXY GOSUMDB GOPRIVATE 标准库的新功能 判断变量是否为0值 错误处理的革新 Unwrap Is As gol ...
- 09.Django基础七之Ajax
一 Ajax简介 1.简介 AJAX(Asynchronous Javascript And XML)翻译成中文就是"异步的Javascript和XML".即使用Javascrip ...
- thymeleaf 将后端绑定数据直接传递js变量
根据自我需求,thymeleaf可以直接将后端数据传递给js中进行使用,例如: 1.后端接口数据: @Controllerpublic class TestController { @RequestM ...
- CSS 换行
默认情况下,元素的属性是 white-space:normal:自动换行:(不把单词截断,会把单词看作一个整体) -----但是但是但是但是..当元素中的内容是一对没有空格的字符/数字时,超过容器宽度 ...