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. 说明exit()函数作用的程序

    // algo1-4.cpp 说明exit()函数作用的程序 #include"c1.h" int a(int i) { if(i==1) { printf("退出程序的 ...

  2. 动态规划:树形DP

    典型例题有三道: 没有上司的舞会 选课 景点中心 我们可以把动态规划的状态和转移描述成DAG 对于有根树来说,如果我们规定边的方向由父节点指向叶子节点 或者是由叶子节点指向父节点(奇葩) 那么它也是一 ...

  3. laravel 获得各个根文件夹路径的方法及路由的一些使用

    各个根文件夹路径的方法 APP目录: app_path(); config目录: config_path(); public目录: public_path(); storage目录: storage_ ...

  4. kuangbin带你飞 并查集 题解

    做这套题之前一直以为并查集是很简单的数据结构. 做了才发现自己理解太不深刻.只看重片面的合并集合.. 重要的时发现每个集合的点与这个根的关系,这个关系可以做太多事情了. 题解: POJ 2236 Wi ...

  5. 虚拟机vmware10.0.0里设置Suse Linux Enterprise 11系统静态IP上网

    http://blog.csdn.net/usbdrivers/article/details/50035615 首次在虚拟机里安装Suse Linux Enterprise 11,采用NET方式能够 ...

  6. [ Python - 7 ] 简单的省份查询系统

    主要是练习while 循环和 if 条件判断的使用 #!_*_coding:utf-8_*_ # Author: hkey def options(list): for i, v in enumera ...

  7. MATLAB作图方法与技巧(一)

    下面从折线图.曲面图和图形说明与定制三个方面说明MATLAB作图的一些方法与技巧,注意,文中出现的单引号在MATLAB中实际运行的时候,需要改为MATLAB认可的单引号. 一. 作折线图 1.plot ...

  8. 2014年spark开发者大赛火热进行中!

    “发现最有正能量的网络达人”,Spark开发者大赛火热进行! 2014年9月30日,2014 Spark开发者大赛在北京正式启动.本次大赛由Spark亚太研究院联合国内领先的IT技术创新与发展的互联网 ...

  9. ubantu的python2与python3的相关兼容更新问题

    Ubuntu14.04, 系统内同时装了Python3.3 和 2.7用sudo apt-get install python-pipsudo apt-get install python3-pip分 ...

  10. ZCMU Problem H: Crixalis's Equipment(贪心,排序)

    #include<stdio.h> #include<stdlib.h> struct node { int a,b; }c[1002]; int cmpxy(const st ...