Problem Description
Now our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is about to kill our pretty Princess. But now the BEelzebub has to beat our hero first. feng5166 says, "I have three question for you, if you can work them out, I will release the Princess, or you will be my dinner, too." Ignatius says confidently, "OK, at last, I will save the Princess."

"Now I will show you the first problem." feng5166 says, "Given a sequence of number 1 to N, we define that 1,2,3...N-1,N is the smallest sequence among all the sequence which can be composed with number 1 to N(each number can be and should be use only once in this problem). So it's easy to see the second smallest sequence is 1,2,3...N,N-1. Now I will give you two numbers, N and M. You should tell me the Mth smallest sequence which is composed with number 1 to N. It's easy, isn't is? Hahahahaha......"
Can you help Ignatius to solve this problem?

 
Input
The input contains several test cases. Each test case consists of two numbers, N and M(1<=N<=1000, 1<=M<=10000). You may assume that there is always a sequence satisfied the BEelzebub's demand. The input is terminated by the end of file.
 
Output
For each test case, you only have to output the sequence satisfied the BEelzebub's demand. When output a sequence, you should print a space between two numbers, but do not output any spaces after the last number.
 
Sample Input
6 4
11 8
 
Sample Output
1 2 3 5 6 4

1 2 3 4 5 6 7 9 8 11 10

 
Code:
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime> using namespace std; int a[]; int main()
{
int n,m,i;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=;i<n;i++)
a[i]=i+; int cnt=; do
{
cnt++;
if(cnt==m)
break; }while(next_permutation(a,a+n));
//next_permutation(first,last)按字典序升序全排列,prev_permutation(first,last)按字典序降序全排列 for(i=;i<n;i++)
{
printf("%d",a[i]);
if(i!=n-)
printf(" ");
}
printf("\n");
}
return ;
}

HDU1027 Ignatius and the Princess II的更多相关文章

  1. HDU1027 Ignatius and the Princess II 【next_permutation】【DFS】

    Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K ( ...

  2. hdu1027 Ignatius and the Princess II (全排列 &amp; STL中的神器)

    转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=1027 Ignatiu ...

  3. 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 <= ...

  4. ACM-简单题之Ignatius and the Princess II——hdu1027

    转载请注明出处:http://blog.csdn.net/lttree Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Othe ...

  5. ACM-简单的主题Ignatius and the Princess II——hdu1027

    转载请注明出处:http://blog.csdn.net/lttree Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Othe ...

  6. HDU 1027 Ignatius and the Princess II(康托逆展开)

    Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K ( ...

  7. Ignatius and the Princess II(全排列)

    Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K ( ...

  8. Ignatius and the Princess II

    Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Jav ...

  9. (next_permutation)Ignatius and the Princess II hdu102

    Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K ( ...

随机推荐

  1. Linux(UBUNTU) 下安装Eclipse

    到目前为止,Eclipse 的官方最新版本为 Eclipse Kepler (4.3.2),我们可以使用如下步骤在 Ubuntu 14.04 或其它 Ubuntu 版本中进行快速安装. 1.安装Ope ...

  2. python 解析docx文档的方法,以及利用Python从docx文档提取插入的文本对象和图片

    首先安装docx模块,通过pip install docx或者在docx官方链接上下载安装都可以 下面来看下如何解析docx文档:文档格式如下 有3个部分组成 1 正文:text文档 2 一个表格. ...

  3. Jexus部署.Net Core项目

    Jexus Jexus 即 Jexus Web Server,简称JWS,是Linux平台上 的一款ASP.NET WEB服务器.它是 Linux.Unix.FreeBSD 等非Windows系统架设 ...

  4. final用法

    1.修饰类 如果一个类被定义为final类型,那么该类无法被其他类继承,该类中的所有方法都是final类型的,字段是否是final类型取决于字段自身的定义. 2.修饰方法 一个方法被定义为final类 ...

  5. Intellij IDEA 2017集成MyBatis三剑客

    MyBatis三剑客指的是:MyBatis-Generate.Mybatis Plus.MyBatis-PageHelper MyBatis-Generate 使用 Mybatis Generator ...

  6. CoolBlog开发笔记第3课:创建Django应用

    教程目录 1.1 CoolBlog开发笔记第1课:项目分析 1.2 CoolBlog开发笔记第2课:搭建开发环境 前言 经过上一节我们已经创建了CoolBlog工程,但是关于CoolBlog的功能代码 ...

  7. cygwin vi编辑器左右上下键和删除键乱码错误

    安装cygwin后使用其中的vi编辑器时发现上下左右键和删除键乱码,搜索了中文的帮助方案,没有解决,最后搜索了英文的网站,找到了解决方案.参考链接如下:http://superuser.com/que ...

  8. maven打包 bat自动化打包

    maven打包,首先cd到项目根目录,如果想跳过测试阶段,可用:mvn package -DskipTests bat命令,说明start是打开文件夹的意思:e: cd E:\workspace\it ...

  9. validateform正则表达式 datatype验证数字

    正则表达式验证正数负数 浮点数/^\-?[0-9]+(.[0-9]+)?$/ datatype="/^\-?[0-9]+(.[0-9]+)?$/"

  10. PHP加密解密数字

    <?php /** * 加密解密类,PHP加密解密数字,适用于URL加密 * 该算法仅支持加密数字.比较适用于数据库中id字段的加密解密,以及根据数字显示url的加密. * @version a ...