排列-hdu_1027_Ignatius and the Princess II

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题目描述：

Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6636    Accepted Submission(s): 3926

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

6 411 8

 

Sample Output

1 2 3 5 6 41 2 3 4 5 6 7 9 8 11 10

 

Author

Ignatius.L

 

题意：现在有一个长度为N的序列：1，2,3,4，…………，N，给出N,M，求该序列的第M大的序列。

分析：如果对字典序运用比较熟悉的话，会比较容易的！！！next_permutation(a,a+n)，函数生成a[n](0<=i<n)的下一个排列，只要对1,2，…………，N，生成第M个排列就可以了，注意输出的问题。

代码：

#include<iostream>#include<stdio.h>#include<algorithm>using namespace std;const int N = 1005;int a[N],n,m;int main(){    while(~scanf("%d%d",&n,&m)){        for(int i=0;i<n;i++)            a[i]=i+1;        for(int i=0;i<m-1;i++)            next_permutation(a,a+n);        for(int i=0;i<n;i++)            if(i==n-1)            printf("%d\n",a[i]);        else printf("%d ",a[i]);    }    return 0;}