ACM-簡單題之Ignatius and the Princess II——hdu1027 Ignatius and the Princess II

轉載請注明出處：

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4436    Accepted Submission(s): 2642

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

Sample Output

Author

Ignatius.L

題目：

這道題，題意就是求 N的第m種排列。

應該屬于組合數學中的一種，剛好之前做過康托展開，就感覺可以用康托展開來做。

（康拓展開詳情可戳→）

但是，我看了看資料範圍就被吓到了， N and M(1<=N<=1000, 1<=M<=10000)。

階乘，最多隻是10，怎麼N可以到1000.。。。

仔細一想就可以發現，M最大為10000，也就是說，最多也就隻有後面8個數才會動。前面不會動的。

因為1~8的階乘為：1,2,6,24,120,720,5040,40320.

8的階乘為40320>10000  10000種以内的排列序，隻能在最後8個變化。

換種說法，無論N為多少，當N>8時，前N-8是不變的，隻有後面8個在變化。

例如：N為11，那麼前面3個數為1 2 3順序一定不變，變化的永遠是後面4~11

思路想出來後，解決這道題就不算太難。

本來我用的是，邊算遍輸出，但是總是PE，可能還是有地方沒想到吧。

就直接将答案存在一個ans數組裡，最後一起輸出，就AC了。