acm 2 1024 Sequence one

1.1024

2.Subset sequence

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

Total Submission(s): 2339 Accepted Submission(s): 1242

Problem Description

Consider the aggregate An= { 1, 2, …, n }. For example, A1={1}, A3={1,2,3}. A subset sequence is defined as a array of a non-empty subset. Sort all the subset sequece of An in lexicography order. Your task is to find the m-th one.

Input

The input contains several test cases. Each test case consists of two numbers n and m ( 0< n<= 20, 0< m<= the total number of the subset sequence of An ).

Output

For each test case, you should output the m-th subset sequence of An in one line.

Sample Input

1 1 

2 1 

2 2 

2 3 

2 4 

3 10

Sample Output

1 

1 

1 2 

2 

2 1 

2 3 1

3.一个从1...n的数组，对这个数组进行规则排序,给出一个m，输出该数组第几个的表达形式

4.例如n=2

1 

1 

1 2 

2 

2 1 

2 3 1

5.#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

struct numberr

{

    int number;

    int postion;

};

numberr b[1005];

int a[1005];

int n,p,cou,dep;

int flag;

int isok(int ss,int ee)

{

    for(int i=ss;i<ee;i++)

    {

        if(a[i]==a[ee])return 0;

    }

    return 1;

}

void dfs(int nowdep,int pos)

{

    if(cou>=p)

        return ;

    if(nowdep==dep)

    {

        cou++;

        flag=1;

        for(int i=0;i<nowdep-1;i++)

        {

            cout<<b[i].number<<" ";

        }

        cout<<b[nowdep-1].number<<endl;

        return ;

    }

    for(int i=pos;i<n;i++)

    {

        if((nowdep!=0&&b[nowdep-1].number<=a[i])||nowdep==0)

        {

            if(nowdep!=0)

            {

                if(!isok(b[nowdep-1].postion+1,i))

                    continue;

            }

            else

            {

                if(nowdep==0&&!isok(0,i))

                    continue;

            }

            b[nowdep].number=a[i];

            b[nowdep].postion=i;

            dfs(nowdep+1,i+1);

        }

    }

}

int main()

{

    while(cin>>n>>p)

    {

        for(int i=0;i<n;i++)

            cin>>a[i];

        cou=0;

        for(int i=1;i<n;i++)

        {

            flag=0;

            dep=i;

            dfs(0,0);

            if(cou>=p||(!flag))break;

        }

        cout<<endl;

    }

    return 0;

}