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 10000Sample 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 ，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n
for(int i=1;i<=n-1;i++) //剩余的n-1位
a[i]=i;
if (k == 1) { //特判k=1
for(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;
}