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

题意：t 组数据，每组给出 n、k 两个数，定义差异排列为 $p_i=a_{i+1}-a_i$ ，求 n 的全排列，求出 n 的所有全排列中差异序列字典序第 k 小的排列

思路：

k 的范围是 1 到 min(1E4,n!)，而 n 最大是 20，当 n=8 时，n!=40320，因此 k 最大是 1E4

故而对 n 分情况讨论：

n<=8 时：直接求出所有差异序列的全排列，然后 sort 排序后输出即可
n>=9 时：第一位直接取 n，然后再取剩下的 n-1 位的全排列，到 k 为止即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<unordered_map>
#include<bitset>
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
#define Pair pair<int,int>
LL quickPow(LL a,LL b){ LL res=1; while(b){if(b&1)res*=a; a*=a; b>>=1;} return res; }
LL multMod(LL a,LL b,LL mod){ a%=mod; b%=mod; LL res=0; while(b){if(b&1)res=(res+a)%mod; a=(a<<=1)%mod; b>>=1; } return res%mod;}
LL quickPowMod(LL a, LL b,LL mod){ LL res=1,k=a; while(b){if((b&1))res=multMod(res,k,mod)%mod; k=multMod(k,k,mod)%mod; b>>=1;} return res%mod;}
LL getInv(LL a,LL mod){ return quickPowMod(a,mod-2,mod); }
LL GCD(LL x,LL y){ return !y?x:GCD(y,x%y); }
LL LCM(LL x,LL y){ return x/GCD(x,y)*y; }
const double EPS = 1E-10;
const int MOD = 998244353;
const int N = 10000+5;
const int dx[] = {-1,1,0,0,1,-1,1,1};
const int dy[] = {0,0,-1,1,-1,1,-1,1};
using namespace std;struct Node {int pos[22], p[22];
} node[50005];
int n,k;
bool cmp(Node x, Node y) {for (int i = 1; i < n; i++) {if (x.p[i] != y.p[i])return x.p[i] < y.p[i];}
}
int a[25];
int main() {int t;scanf("%d", &t);while (t--) {scanf("%d%d", &n, &k);if (n <= 8) { // n<=8时for (int i = 1; i <= n; i++)a[i] = i;for (int i = 1; i <= n; i++)//记录位置node[1].pos[i] = a[i];for (int i = 2; i <= n; i++)//求差异排列node[1].p[i - 1] = a[i] - a[i - 1];int num = 2;while (next_permutation(a + 1, a + 1 + n)) {//生成全排列for (int i = 1; i <= n; i++)node[num].pos[i] = a[i];for (int i = 2; i <= n; i++)node[num].p[i - 1] = a[i] - a[i - 1];num++;}sort(node + 1, node + num, cmp);for (int i = 1; i <= n - 1; i++)printf("%d ", node[k].pos[i]);printf("%d\n", node[k].pos[n]);} else { // n>=9时printf("%d ",n); //第一位一定是nfor(int i=1;i<=n-1;i++) //剩余的n-1位a[i]=i;if (k == 1) { //特判k=1for(int i=1;i<=n-2;i++)printf("%d ",a[i]);printf("%d\n",a[n-1]);}else{int num = 1;while (next_permutation(a + 1, a + 1 + (n - 1))) { //输出前k个num++;if (num == k) {for (int i = 1; i <= n - 2; i++)printf("%d ", a[i]);printf("%d\n", a[n - 1]);break;}}}}}return 0;
}

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permutation 1（HDU-6628）

Problem Description

Input

Output

Sample Input

Sample Output

Source Program

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