Python3 解法, 执行用时: <1ms, 内存消耗: 0K, 提交时间: 2023-08-12 10:13:04

import math
n,k=map(int,input().split())
f=[0]*(n+1)
f[0]=0
f[1]=0
for i in range(2,n+1):
    f[i]=int(math.log(i-1,2))+1
count=[1]*(n+1)
SUM=sum(f[1:])
if k>SUM:
    print(-1)
else:
    ret=[]
    for i in range(n,0,-1):
        if f[i]<=k:
            k-=f[i]
            ret.append(i)
            count[i]=0
    for i in range(1,n+1):
        if count[i]:
            ret.append(i)
    r=""
    for i in ret:
        r+=str(i)+" "
    print(r)

Java 解法, 执行用时: 744ms, 内存消耗: 71308K, 提交时间: 2023-08-12 10:12:41

import java.io.*;
import java.util.ArrayList;
import java.util.List;

public class Main {
    static final int N = 1000010;
    static int[] log = new int[N];
    static int[] f = new int[N];
    static int n, m;
    
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        
        String[] in = br.readLine().split(" ");
        n = Integer.parseInt(in[0]);
        m = Integer.parseInt(in[1]);
        
        // log(1) = 0, log(2) = 1, log(4) = 2,  log(8) = 3;
        log[1] = 0;
        for (int i = 2; i <= n; i++) {
            log[i] = log[i >> 1] + 1; 
        }
        
        for (int i = 2; i <= n; i++) {
            f[i] = log[i - 1] + 1;
        }
        
        List<Integer> a = new ArrayList<>();
        List<Integer> b = new ArrayList<>();
        for (int i = n; i >= 1; i--) {
            if (m >= f[i]) {
                a.add(i);
                m -= f[i];
            } else {
                b.add(i);
            }
        }
        
        if (m != 0) System.out.println("-1");
        else {
            for (int i : a) pw.print(i + " ");
            for (int i = b.size() - 1; i >= 0; i--) pw.print(b.get(i) + " ");
        }
        
        pw.flush();
    }
}