在线判断Codevita的运行时错误

Question

下面是问题 主要问题是要获得反演的总数

Count Bits
Problem Description
Given a sequence of distinct numbers a1, a2, ….. an, an inversion occurs if there are indices i<j such that ai > aj .

For example, in the sequence 2 1 4 3 there are 2 inversions (2 1) and (4 3).

The input will be a main sequence of N positive integers. From this sequence, a Derived derived sequence will be obtained using the following rule. The output is the number of inversions in the derived sequence.

Rule for forming derived sequence

The derived sequence is formed by counting the number of 1s bits in the binary representation of the corresponding number in the input sequence.

Thus, if the input is 3,4,8, the binary representation of the numbers are 11,100 and 1000. The derived sequence is the number of 1sin the binary representation of the numbers in the input sequence, and is 2,1,1

Constraints
N <= 50

Integers in sequence <= 107

Input Format
The first line of the input will have a single integer, which will give N.

The next line will consist of a comma separated string of N integers, which is the main sequence

Output
The number of inversions in the derived sequence formed from the main sequence.


Explanation
Example 1

Input

5

55, 53, 88, 27, 33

Output

8

Explanation

The number of integers is 5, as specified in the first line. The given sequence is 55, 53, 88, 27, 33.

The binary representation is 110111, 110101, 1011000, 11011, 100001and 100001 . The derived sequence is 5,4,3,4,2, 4,3,4,2 (corresponding to the number of 1s bits). The number of inversions in this is 8, namely (5,4),(5,3),(5,4),(5,2),(4,3),(4,2),(3,2),(4,2). Hence the output is 8.

Example 2

Input

8

120,21,47,64,72,35,18,98

Output

15

Explanation

The number of integers is 8. The given sequence is 120,21,47,64,72,35,18,98. The corresponding binary sequence is 1111000,10101,101111,1000000,1001000,100011, 10010,1100010. The derived sequence (number of 1s) is 4,3,5,1,2,3,2,3. The number of inversions is 15, namely (4,3),(4,1),(4,2),(4,3),(4,2),(4,3),(3,1),(3,2),(3,2),(5,1),(5,2),(5,3), (5,2), (5,3),(3,2). Hence the output is 15.

这是我的解决方法

    package SudoPlacement;

import java.util.Scanner;

public class Inversion {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String line = null;
        int n = sc.nextInt();
        line = sc.next();
        String[] parts = line.split(",");
        int[] arr = new int[n];
        for (int i=0;i<parts.length;i++)
            arr[i]=Integer.parseInt(parts[i]);
        int[] countArray = new int[n];
        for (int i=0;i<n;i++){

            int value=0;
            while (arr[i]!=0) {
                value+= arr[i]%2;
                arr[i]=arr[i]/2;
            }
            countArray[i]=value;
        }

        int noOfInversions = 0;
        for (int i=0;i<countArray.length;i++) {
            for (int j=0;j<countArray.length;j++){
                if (countArray[i]>countArray[j] && i<j)
                    noOfInversions++;
            }
        }

        System.out.println(noOfInversions);
    }


}

使用本地ide（Intellij）时，我会按要求获取正确的输出，但是当我尝试提交代码时，却遇到了运行时错误。 有人可以帮我吗？

这是在线ide针对运行时错误显示的唯一内容

这是我的想法输出

Answer 1

尝试注释此行package SudoPlacement;  从您的解决方案。在将解决方案上传到在线法官时，我已经多次犯过这个错误，并且遇到了相同的运行时错误。