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. 问题总结——window平台下grunt\bower安装后无法运行的问题

    一.问题: 安装grunt或者bower后,在cmd控制台运行grunt -version 或者 bower -v会出现:“xxx不是内部或外部命令,也不是可运行的程序或批处理文件”,

  2. DOM常用对象

    一.select对象 HEML中的下拉列表 属性: 1.options 获得当前select下所有option 2.options[i] 获得当前select下i位置的option 3.selecte ...

  3. LeetCode189:Rotate Array

    Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array ...

  4. python排序sorted与sort比较

    Python list内置sort()方法用来排序,也可以用python内置的全局sorted()方法来对可迭代的序列排序生成新的序列. sorted(iterable,key=None,revers ...

  5. Net Core 控制台程序使用Nlog 输出到log文件

    using CoreImportDataApp.Common; using Microsoft.Extensions.Configuration; using Microsoft.Extensions ...

  6. Stack的三种含义 ----超级经典 明白了 栈 的三种含义

    来自:http://www.ruanyifeng.com/blog/2013/11/stack.html ----------------------------------------------- ...

  7. HBuilder开发移动App——manifest.json文件解析

    以前做过Android App开发,对于各项配置都是在AndroidManifest.xml文件中完成的,包括权限的设定.图标.标签.App的名字.Activity注册等等 使用HBuilder开发移 ...

  8. VS2010编写C++程序出现error C1010: 在查找预编译头时遇到意外的文件结尾。是否忘记了向源中添加“#include "StdAfx.h"”?

    用VS2010编写C++程序,编译时出现如下错误: 修改方法: 右击项目,选择属性 点击确定,重新编译,错误解决.

  9. Go语言,用原子函数atomic避免资源竞争

    下一步应该是互斥锁了. package main import ( "fmt" "runtime" "sync" "sync/at ...

  10. paramiko 使用总结(SSH 操作远端机器)

    1.用户名.密码登陆方式 import paramikoparamiko.util.log_to_file('paramiko.log') # 记录日志文件ssh = paramiko.SSHClie ...