解析答案如下:
import java.util.Arrays;
import java.util.Scanner;
/*
本题的关键:设置一个特大的数组,通过数组下标来记录所被覆盖的数!
如记录5这个数,就是在数组下标为5的位置上标记一下,
*/
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[5000];
int[] ioarr = new int[n];
for (int i = 0; i < n; i++) {
ioarr[i] = sc.nextInt();
}
int[] key = new int[n];
//复制一下输入的数组
for (int i = 0; i < n; i++) {
key[i] = ioarr[i];
}
int[] ey = new int[n];
for (int j = 0; j < n; j++) {
while (ioarr[j]!=1){
if (ioarr[j] % 2 == 0) {
ioarr[j] = ioarr[j] / 2;
} else {
ioarr[j] = (3 * ioarr[j] + 1) / 2;
}
arr[ioarr[j]] = 1;}
}
for(int i=0;i<key.length;i++){
if(arr[key[i]]!=1){
ey[i]=key[i];
}
}
Arrays.sort(ey);
for(int i=ey.length-1;i>0;i--){
if (ey[i]!=0&&ey[i-1]!=0){
System.out.print(ey[i]+" ");
}else if (ey[i]!=0&&ey[i-1]==0){
System.out.print(ey[i]);
}
}
}
}
Java|PTA乙级|1005 继续(3n+1)猜想 (25分)|解析
原文:https://www.cnblogs.com/niujiajun/p/14348934.html