POJ - 2013 Symmetric Order (java)

In your job at Albatross Circus Management (yes, it's run by a bunch of clowns), you have just finished writing a program whose output is a list of names in nondescending order by length (so that each name is at least as long as the one preceding it). However, your boss does not like the way the output looks, and instead wants the output to appear more symmetric, with the shorter strings at the top and bottom and the longer strings in the middle. His rule is that each pair of names belongs on opposite ends of the list, and the first name in the pair is always in the top part of the list. In the first example set below, Bo and Pat are the first pair, Jean and Kevin the second pair, etc. 

Input

The input consists of one or more sets of strings, followed by a final line containing only the value 0. Each set starts with a line containing an integer, n, which is the number of strings in the set, followed by n strings, one per line, sorted in nondescending order by length. None of the strings contain spaces. There is at least one and no more than 15 strings per set. Each string is at most 25 characters long.

Output

For each input set print "SET n" on a line, where n starts at 1, followed by the output set as shown in the sample output.

Sample Input

7
Bo
Pat
Jean
Kevin
Claude
William
Marybeth
6
Jim
Ben
Zoe
Joey
Frederick
Annabelle
5
John
Bill
Fran
Stan
Cece
0

Sample Output

SET 1
Bo
Jean
Claude
Marybeth
William
Kevin
Pat
SET 2
Jim
Zoe
Frederick
Annabelle
Joey
Ben
SET 3
John
Fran
Cece
Stan
Bill

 

 

package 递归和回溯;

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
    // TODO Auto-generated method stub

        Scanner sc=new Scanner (System.in);
        int s=1;
        for(;;) {
            int n=sc.nextInt();
            if(n==0)
                break;
            System.out.println("SET "+s);
            s++;
            String a[]=new String [n+3];
            for(int i=1;i<=n;i++)
                a[i]=sc.next();
            if(n%2==0) {
                for(int i=1;i<=n;i=i+2) {
                    System.out.println(a[i]);
                    //System.out.println(a[n-i-1]);
                }
                for(int i=n;i>0;i=i-2) {
                    System.out.println(a[i]);
                    //System.out.println(a[n-i-1]);
                }
            }
            else {
                for(int i=1;i<=n;i=i+2){
                    System.out.println(a[i]);                    
                }
                for(int i=n-1;i>0;i=i-2) {
                    System.out.println(a[i]);
                    //System.out.println(a[n-i-1]);
                }
            }
        }
    }

}
 

这个错误是由于无法连接到本地主机的10248端口导致的。这个端口通常是kubelet进程监听的端口,用于健康检查。出现这个错误可能是由于kubelet进程没有正确启动或者配置错误导致的。 解决这个问题的方法是检查kubelet进程的状态和配置。你可以按照以下步骤进行操作: 1. 检查kubelet进程是否正在运行。你可以使用以下命令检查kubelet进程的状态: ```shell systemctl status kubelet ``` 如果kubelet进程没有运行,你可以使用以下命令启动它: ```shell systemctl start kubelet ``` 2. 检查kubelet的配置文件。你可以使用以下命令查看kubelet的配置文件路径: ```shell kubelet --kubeconfig /etc/kubernetes/kubelet.conf --config /var/lib/kubelet/config.yaml --bootstrap-kubeconfig /etc/kubernetes/bootstrap-kubelet.conf config view ``` 确保配置文件中的端口号和地址正确,并且与你的环境相匹配。 3. 检查网络连接。你可以使用以下命令检查是否可以连接到localhost10248端口: ```shell curl -sSL http://localhost:10248/healthz ``` 如果无法连接,请确保端口没有被防火墙或其他网络配置阻止。 4. 检查docker的配置。有时候,kubelet进程依赖于docker进程。你可以按照以下步骤检查docker的配置: - 创建/etc/docker目录: ```shell sudo mkdir /etc/docker ``` - 编辑/etc/docker/daemon.json文件,并添加以下内容: ```json { "exec-opts": ["native.cgroupdriver=systemd"], "log-driver": "json-file", "log-opts": { "max-size": "100m" }, "storage-driver": "overlay2", "storage-opts": [ "overlay2.override_kernel_check=true" ], "registry-mirrors": ["https://tdhp06eh.mirror.aliyuncs.com"] } ``` - 重启docker进程: ```shell systemctl restart docker ``` 请注意,以上步骤是一种常见的解决方法,但具体解决方法可能因环境而异。如果以上步骤无法解决问题,请提供更多的错误信息和环境配置,以便我们能够更好地帮助你。
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