CODEFORCES Rockethon 2015 B. Permutations

You are given a permutation p of numbers 1,?2,?…,?n. Let’s define f(p) as the following sum:

Find the lexicographically m-th permutation of length n in the set of permutations having the maximum possible value of f(p).
Input

The single line of input contains two integers n and m (1?≤?m?≤?cntn), where cntn is the number of permutations of length n with maximum possible value of f(p).

The problem consists of two subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows.

In subproblem B1 (3 points), the constraint 1?≤?n?≤?8 will hold.
In subproblem B2 (4 points), the constraint 1?≤?n?≤?50 will hold. 

Output

Output n number forming the required permutation.
Sample test(s)
Input

2 2

Output

2 1

Input

3 2

Output

1 3 2

Note

In the first example, both permutations of numbers {1, 2} yield maximum possible f(p) which is equal to 4. Among them, (2,?1) comes second in lexicographical order.

先给我们一个f(p), 求出n个数中f(p)最大的那些排列里面字典序是m的那个排列

可以发现
f(1) =1
f(2) = 2
f(3) = 4
f(4) = 8
f(5) = 16
….
所以f(n) = (2 ^ (n - 1))
而且在这些排列中，1一定在第1或者最后，2一定在第2或者倒数第二 ….. ，根据这些，我们可以递归求出排列

/*************************************************************************
    > File Name: B.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月09日 星期一 14时42分08秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int n, cnt;
LL m;
int ans[55];

void work (int l, int r)
{
    if (l == r)
    {
        ans[l] = cnt++;
        return;
    }
    LL tmp = (1LL << (r - l - 1));
    if (tmp >= m)
    {
        ans[l] = cnt++;
        work (l + 1, r);
    }
    else
    {
        ans[r]  = cnt++;
        m -= tmp;
        work (l, r - 1);
    }
}

int main ()
{
    while (cin >> n >> m)
    {
        cnt = 1;
        work (1, n);
        printf("%d", ans[1]);
        for (int i = 2; i <= n; ++i)
        {
            printf(" %d", ans[i]);
        }
        printf("\n");
    }
    return 0;
}