HDOJ--1027--Ignatius and the Princess II（利用next_permutation函数去做的）

Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6405    Accepted Submission(s): 3790

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

 

题目意思：给你n,m；n代表n个数从1到n，将这个序列进行全排列，然后找到第M个排列输出结果就可以。

思路：利用next_permutation函数来对数组进行排列。找到第M个排列输出结果就好。

有关next_permutation函数请参考： http://http://blog.csdn.net/lishuhuakai/article/details/8006937

AC代码：

#include
#include
#include
#define MAX 10010
using namespace std;
int main()
{
	int n,m,i,count ;
	int seq[MAX];
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++)
			seq[i]=i;//储存要排列的数字。
		count=0; 
		while(1)
		{
			count++;//计数器，第几个排列的位置。 
			if(count==m)
			{
				for(i=1;i<=n;i++)//输出排列； 
				{
					if(i==n)
						printf("%d",seq[i]);
					else
						printf("%d ",seq[i]);			
				}
				printf("\n");
				break;//非常重要的跳出，不然的话，即使输出了结果，程序还在跑，会超时。				
			}
			if(!next_permutation(seq+1,seq+1+n))//如果都排列晚了就跳出。 
				break;
		}
	}
	return 0;
}