permutation 1 多校第五场

Problem Description

 

A sequence of length n is called a permutation if and only if it's composed of the first n positive integers and each number appears exactly once.

Here we define the "difference sequence" of a permutation p1,p2,…,pn as p2−p1,p3−p2,…,pn−pn−1. In other words, the length of the difference sequence is n−1 and the i-th term is pi+1−pi

Now, you are given two integers N,K. Please find the permutation with length N such that the difference sequence of which is the K-th lexicographically smallest among all difference sequences of all permutations of length N.

 

Input

The first line contains one integer T indicating that there are T tests.

Each test consists of two integers N,K in a single line.

* 1≤T≤40

* 2≤N≤20

* 1≤K≤min(104,N!)

 

Output

For each test, please output N integers in a single line. Those N integers represent a permutation of 1 to N, and its difference sequence is the K-th lexicographically smallest.

 

Sample Input

7

3 1

3 2

3 3

3 4

3 5

3 6

20 10000

 

Sample Output

3 1 2

3 2 1

2 1 3

2 3 1

1 2 3

1 3 2

20 1 2 3 4 5 6 7 8 9 10 11 13 19 18 14 16 15 17 12

 

Source

2019 Multi-University Training Contest 5

 

题意：给你n和k  求n的全排列 使得ai+1 - ai （1<=i<=n-1）的字典序是第k小

思路：因为k的范围最大是10000 （8！=40320） 7！=5040

所以当n=9时 第一位取9 后面几位全排列 即8！种排列方法（k最大10000次不会超时）

所以分开处理 情况1：n<=8的   这种的要全部算出来 然后sort

之后就是>=9的  这种就直接第一位时n  之后的n-1位全排列

 

代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
struct ydw{
    int d[22],dd[22];
}xy[50005];
int n;
bool cmp(ydw u,ydw v)
{
    for(int i=1;i<n;i++)
    {
        if(u.dd[i]!=v.dd[i])return u.dd[i]<v.dd[i];
    }
}
int main()
{
    int i,j,m,k;
    int t;scanf("%d",&t);
    int a[25];
    while(t--)
    {
         scanf("%d%d",&n,&k);
         for(i=1;i<=n;i++)a[i]=i;
         if(n<=8)
         {
             int k2=1;
             for(i=1;i<=n;i++)
             {
                 xy[k2].d[i]=a[i];
             }
             for(i=2;i<=n;i++)
             {
                 xy[k2].dd[i-1]=a[i]-a[i-1];
             }
             k2++;
             while(next_permutation(a+1,a+1+n))
             {
                 for(i=1;i<=n;i++)
                     xy[k2].d[i]=a[i];
                 for(i=2;i<=n;i++)
                    xy[k2].dd[i-1]=a[i]-a[i-1];
                 k2++;
             }
             sort(xy+1,xy+k2,cmp);
             for(i=1;i<n;i++)
             {
                 cout<<xy[k].d[i]<<" ";
             }
             cout<<xy[k].d[n]<<endl;
             continue;
         }
         //n>9的情况
         int k1=0;
         int b[30];
         for(i=1;i<n;i++)b[k1++]=i;
         cout<<n<<" ";//第一位
         if(k==1)
         {
             for(i=0;i<k1-1;i++)cout<<b[i]<<" ";
             cout<<b[k1-1]<<endl;
             continue;
         }
         int ans=1;
         while(next_permutation(b,b+k1))
         {
             ans++;
             if(ans==k)
             {
                 for(i=0;i<k1-1;i++)cout<<b[i]<<" ";
                 cout<<b[k1-1]<<endl;
                 break;
             }
         }
    }
    return 0;
}