2019 Multi-University Training Contest 5 permutation 1 hdu6628 （排列，枚举）-优快云博客

本文探讨了一个特定的算法挑战，即找到长度为N的排列，其差分序列在字典序中是第K小的。通过分类讨论和枚举排序策略，文章提供了一种解决方案，适用于N不超过20的情况。

permutation 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 524 Accepted Submission(s): 212

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

思路：因为k最大为1e4，所以可以以根据是否大于8来分类讨论，当当于8时，直接按照全排列依次取数即可，而小于8时，则可以将所有情况枚举，然后排序得到答案。

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
typedef long long ll;
typedef vector<ll> vec;
//typedef vector<vec> mat;
const ll MOD = 1000000007;//根据需要更改
const int MAXN = 50;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

struct P{
    int a[11];
    int b[11];
}v[500000];
int fac[MAXN];
int ans[MAXN];
int T, n, k;
bool cmp(P a, P b){
    for(int i = 0; i < n - 1; i++) {
        if(a.b[i] < b.b[i]) return true;
        if(a.b[i] > b.b[i]) return false;
    }
    return false;
}
int main()
{
	//std::ios::sync_with_stdio(false);
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
//	fac[0] = 1;
//	rep(i, 1, 20){
//	    if(i > 8)   fac[i] = INF;
//        else        fac[i] = fac[i-1]*i;
//        printf("%d\n", fac[i]);
//	}
	scanf("%d", &T);
	while(T--){
        scanf("%d%d", &n, &k);
        if(n > 8){
            ans[0] = n;
            rep(i, 1, n-1){
                ans[i] = i;
            }
            int cnt = 0;
            do{
                cnt++;
                if(cnt == k) break;
            }while(next_permutation(ans, ans+n));
            rep(i, 0, n-1){
                if(i!=0) putchar(' ');
                printf("%d", ans[i]);
            }
            putchar('\n');
        }else{
            rep(i, 0, n-1){
                ans[i] = i+1;
            }
            int cnt = 0;
            rep(i, 0, INF){
                rep(j, 0, n-1){
                    v[i].a[j] = ans[j];
                    if(j>0) v[i].b[j-1] = v[i].a[j] - v[i].a[j-1];
                }
                v[i].b[n-1] = -INF;
                cnt++;
                if(!next_permutation(ans, ans+n)) break;
            }
            sort(v, v+cnt, cmp);
            k--;
            rep(i, 0, n-1){
                if(i!=0) putchar(' ');
                printf("%d", v[k].a[i]);
            }
            putchar('\n');
        }
	}
    return 0;
}
/*
5
3
7 7 7
0 7 5
*/