重排数列和代码碎片的ac代码，求魔幻王国的代码

1.代码碎片

package nowcoder.netease0909;

import java.text.DecimalFormat;
import java.util.Scanner;

public class Q1 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            String str = sc.next();
            System.out.println(getAVG(str));
        }
    }

    private static String getAVG(String str) {
        char[] c = str.toCharArray();
        int sum = 0;
        int time = 0;
        int temp = 1;
        int p = 0, q = 0;
        while (q < c.length) {
            if (c[q] == c[p]) {
                q++;
            } else {
                sum += (q - p);
                time++;
                p = q;
            }
        }
        if (p == q) {
            sum++;
            time++;
        } else {
            sum += (q - p);
            time++;
        }
        DecimalFormat df = new DecimalFormat("0.00");
        return df.format((sum * 1.0) / time);
    }
}

2.重排数列

package nowcoder.netease0909;

import java.util.Scanner;

public class Q2 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int t = sc.nextInt();
            String[] strArr = new String[t];
            int index = 0;
            while (index < t) {
                int n = sc.nextInt();
                int[] arr = new int[n];
                for (int i = 0; i < n; i++) {
                    arr[i] = sc.nextInt();
                }
                strArr[index] = canDo(arr);
                index++;
//                System.out.println(canDo(arr));
            }
            for (String s :
                    strArr) {
                System.out.println(s);
            }
        }
    }

    private static String canDo(int[] arr) {
        int count4 = 0, count2 = 0, countOther = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] % 4 == 0) {
                count4++;
                continue;
            } else if (arr[i] % 2 == 0) {
                count2++;
                continue;
            } else
                countOther++;
        }

        if (countOther + 1 <= count4 || count2 == arr.length || count4 == countOther)
            return "Yes";
        else
            return "No";

    }
}

求魔幻王国的代码， 题目都没读懂，真难受。目测已挂。