Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1862 Accepted Submission(s): 1119

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

發現 7! < 10000 < 8!

如果 n > 8，第一個排序是 1, 2, ..., n - 1, n，而且前 10000 個排序只有最後的 8 個數可能發生變動，所以可以只對後面 8 個數模擬排序，若模擬次數是 X，那麼時間複雜度是 O(X*8)。

接下來是模擬次數 X：

n 個數的排序，他的最後一個排序一定是 n, n - 1, ..., 2, 1，所以我們可以預先知道的若干個排序是第1!, 2!, 3! ... 個。

比如 1~7 的第 1!, 2!, 3! ... 個排序分別是：

1, 2, 3, 4, 5, 6, 7

1, 2, 3, 4, 5, 7, 6

1, 2, 3, 4, 7, 6, 5

1, 2, 3, 7, 6, 5, 4

1, 2, 7, 6, 5, 4, 3

1, 7, 6, 5, 4, 3, 2

7, 6, 5, 4, 3, 2, 1

這樣模擬次數大約可以下降到 O(m / 8)。

總時間複雜度大約是 O(m / 8 * 8)

#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

int main () {
    int n, m;
    int fac [9] = {1, 1, 2, 6, 24, 120, 720, 720*7, 720*56};
    while (scanf ("%d%d", &n, &m) == 2) {
        assert (m < fac [8]);
        vector <int> perm;
        while ((int)perm.size() < 8 && (int)perm.size() < n) {
            perm.push_back (perm.size() + 1);
        }
        int i = 0;
        while (fac [++i] <= m) {}
        --i;
        assert (i <= (int)perm.size());
        reverse (perm.end() - i, perm.end());
        int p = fac [i];
        while (p++ < m) {
#ifdef _DEBUG
            bool b =
#endif
            next_permutation (perm.begin(), perm.end());
#ifdef _DEBUG
            assert (b);
#endif
        }
        if (n <= 8) {
            for (int k=0; k<(int)perm.size(); ++k) {
                printf ("%d%s", perm[k], k==(int)perm.size()-1 ? "\n" : " ");
            }
        } else {
            for (int k=0; k<n-8; ++k) {
                printf ("%d ", k + 1);
            }
            for (int k=0; k<8; ++k) {
                printf ("%d%s", perm[k] + n - 8, k==7 ? "\n" : " ");
            }
        }
    }
    return 0;
}