2021-10-13

HDU1443-java循环双链表实现

题目大意：有2k个人，前k个是好人，后k个是坏人，每隔m个人去除一个人，求k个坏人在好人前全部被去除的最小m。
代码如下：
import java.util.ArrayList;
import java.util.Scanner;

public class HDU1433{
public static void main(String []arges) {
Scanner input = new Scanner(System.in);
ArrayList ans=new ArrayList();
while (!input.hasNext(“0”)) {
int n = input.nextInt();
int m = 1;
joseph j = new joseph(n, m);

            man k = j.first;
            for (int i = 1; i <= j.jlength(); i++) {
                k = k.next;
            }
        ans.add(j.circle(n,m));
    }
    for(int i=0;i<ans.size();i++){
        System.out.println(ans.get(i));
    }
}

}
class man{
int no;
man next;
man prior;
public man(int num){
no=num;
next=null;
prior=null;
}

}
class joseph{
int n,m;
man first;

public joseph(int n1,int m1){
    man p,t;
    n=n1*2;
    m=m1;
    first=new man(1);
    t=first;
    for(int i=1;i<n;i++){
            p=new man(i+1);

        t.next=p;
        p.prior=t;
        t=p;
    }
    t.next=first;
    first.prior=t;
}
public int jlength(){
    int length=1;
    man q=first;
    while(q.next!=first){
        length++;
        q=q.next;
    }
    return length;
}
public int circle(int n,int m) {
    joseph q=this;

    man c = q.first;
    while (q.jlength()!=n) {

    for (int i = 1; i < m; i++) {
        c = c.next;
    }


        while (c.no > n  && q.jlength()>n) {
            c.prior.next = c.next;
            c.next.prior = c.prior;
            c=c.next;
            for (int i = 1; i < m; i++) {
                c = c.next;
            }

        }
        if ((c.no <= n )&& (q.jlength()>n)) {
            m=m+1;
             q=new joseph(n,m);
            c=q.first;
        }
    }
    return m;
}

}

标签：10,13,int,next,prior,joseph,2021,new,man
来源： https://blog.csdn.net/m0_62049544/article/details/120743759