hdu1027 Ignatius and the Princess II （全排列 & STL中的神器）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?

pid=1027

Ignatius and the Princess II 

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

题意：给定n和m n表示1……n的数字排列。求第m个排列。

通过这题发现了STL中的神器：next_permutation（后一个）和prev_permutation（前一个）函数

依照STL文档的描写叙述，next_permutation函数将按字母表顺序生成给定序列的下一个较大的序列。直到整个序列为减序为止。prev_permutation函数与之相反。是生成给定序列的上一个较小的序列。

二者原理同样。仅遍例顺序相反

代码例如以下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define N 1047
int num[N];
int main()
{
	int n , m , i ,k;
	while(~scanf("%d%d",&n,&m))
	{
		memset(num,0,sizeof(num));
		for(i = 0 ; i < n ; i++)
		{
			num[i] = i+1;
		}
		for(i = 1 ; i < m ; i++)
		{
			next_permutation(num,num+n);
		}
		for(i = 0 ; i < n-1 ; i++)
			printf("%d ",num[i]);
		printf("%d
",num[n-1]);
	}
	return 0;
}