(513B2)codeforce

B1. Permutations
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.


#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<set>
#include<algorithm>
#include<cmath>


#define ll __int64
#define MAX 1000009
using namespace std;


/*
看题解做的题,这个题的规律真是难看····为什么我就是看不出来嘛····
让m-1与1,2,3,4,5,....,n-1的二进制进行与运算,看是否为0,如果为0,从左到右填,否则从右到左,看不懂这个规律


1ll 这个是长整形long long 下的1


左移n位等于乘以2的n次方 <<
右移n位等于整除2的n次方 >>


我觉得这题的规律是···m-1与2的k(n-2>=k>=1)次幂进行与运算,看是否为0,如果为0,从左到右填,否则从右到左,这是前n-1位,第n位,是m-1与2的n-1次幂,进行运算,填的规则,与前者相同,n的时候单独拿出来
*/
int a[MAX];


int main()
{
    ll n,m;
    while(~scanf("%I64d%I64d",&n,&m))
    {
        int l = 0;
        int r = n - 1;
        for(int i = 1; i<=n; i++)
        {
            // cout<<i<<" "<<(1ll<<n-i-1)<<endl;
            //cout<<((m-1)&(1ll<<n-i-1))<<endl;
            if(i==n)
            {
                //cout<<i<<" "<<((1ll<<n)-1)<<endl;
                //cout<<((m-1)&((1ll<<n)-1))<<endl;
                if((m-1)&((1ll<<n)-1))
                {
                    a[r--] = i;
                }
                else
                {
                    a[l++] = i;
                }
            }
            else
            {
                if((m-1)&(1ll<<(n-i-1)))
                {
                    a[r--] = i;
                }
                else
                {
                    a[l++] = i;
                }
//            }
            }
        }
        for(int i=0; i<n; i++)
        {
            if(i!=0)
            {
                cout<<" ";
            }
            cout<<a[i];
        }
        cout<<endl;
    }
    return 0;
}

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