Ignatius and the Princess II

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 2   Accepted Submission(s) : 1
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

 
Author
Ignatius.L
 
 
 
#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
int str[1001];
int n,m,i;
while (scanf("%d%d",&n,&m)!=EOF)
{
for (i=0;i<n;i++)
{
str[i]=i+1;
}
for (i=1;i<m;i++) next_permutation(str,str+n);
printf("%d",str[0]);
for (i=1;i<n;i++)
{
printf(" %d",str[i]);
}
puts("");
}
return 0;
}

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

  1. HDU Ignatius and the Princess II 全排列下第K大数

    #include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include& ...

  2. (全排列)Ignatius and the Princess II -- HDU -- 1027

    链接: http://acm.hdu.edu.cn/showproblem.php?pid=1027 Ignatius and the Princess II Time Limit: 2000/100 ...

  3. HDU 1027 Ignatius and the Princess II(求第m个全排列)

    传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1027 Ignatius and the Princess II Time Limit: 2000/10 ...

  4. 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 ( ...

  5. HDU - 1027 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 ( ...

  6. HDU 1027 Ignatius and the Princess II[DFS/全排列函数next_permutation]

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

  7. 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 ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. npm获取配置值的两种方式

    命令行标记 在命令行上放置--foo bar设置foo配置参数为bar. 一个 -- 参数(argument)告诉cli解析器停止读取flags.一个 在命令行结尾的--flag参数(paramete ...

  2. 搭建eova开发环境

    1.安装好maven 2.下载Eova项目解压到文件夹eova下 3.dos命令到eova文件夹下执行mvn eclipse:eclipse(注:构建eclipse项目命令) 4.修改*.tag文件错 ...

  3. [洛谷P1074] 靶形数独

    洛谷题目链接:靶形数独 题目描述 小城和小华都是热爱数学的好学生,最近,他们不约而同地迷上了数独游戏,好胜的他们想用数独来一比高低.但普通的数独对他们来说都过于简单了,于是他们向 Z 博士请教,Z 博 ...

  4. 创建ipadWEB应用程序到主屏幕

    1.webkit内核中的一些私有的meta标签,这些 meta标签在开发webapp时起到非常重要的作用 (1)<meta content="width=device-width; i ...

  5. 微信小程序登录状态

    我们知道,WEB服务器通过浏览器携带的cookie获取session来判断是否是同一用户(或浏览器):Restful服务通过客户端传过来唯一ID,来识别调用用户. >为什么需要维护登录态? 有自 ...

  6. cube中的判断类型

    import { createAddAPI } from '../util' const DATE_RE = /^(1|2)\d{3}[.\-/]\d{1,2}[.\-/]\d{1,2}$/ cons ...

  7. JVM指令的使用深入详解

    原文地址:https://www.jb51.net/article/155293.htm 一.未归类系列A 此系列暂未归类. 指令码    助记符                            ...

  8. JSP(1) - JSP简介、原理、语法 - 小易Java笔记

    1.JSP简介 (1)JSP的全称是Java Server Pages(运行在服务器端的页面),实际就是Servlet(学习JSP的关键就是时刻联想到Servlet) (2)JSP.Servlet各自 ...

  9. 编译opencv2.4.11时出现错误:error: ‘NppiGraphcutState’ has not been declared

    安装cuda之后再安装opencv时出现错误: /data/opencv-2.4.11/modules/gpu/src/graphcuts.cpp:120:54: error: ‘NppiGraphc ...

  10. C#格式化时间,货币

    1.格式化货币(跟系统的环境有关,中文系统默认格式化人民币,英文系统格式化美元) string.Format("{0:C}",0.2) 结果为:¥0.20 (英文操作系统结果:$0 ...