数组错误在给定条件下有多少奇数和偶数

Question

1. I want to output it like so
2. my compiler doesn't prompt errors in code I use netbeans apache
3. when i put how many numbers ======================================================

        public class NewClass 
        {
          public static void main(String[] args)
          {
            int n = 0;
            int EvenNo;
            int OddNo;
            Scanner sc = new Scanner(System. in);
     
            System.out.println("How many numbers will be entered?");
            n = sc.nextInt();
            int a[]=new int[n];
        
            //I think Im missing some code here

            System.out.println("Enter " + n + " Elements Separated by Space >> " );
            String input;
            input = sc. nextLine();
            String[] tokens = input.split(" ");
            int[] inputNumbers = new int[tokens.length];
 
            EvenNo = OddNo = 0;

            for(int i = 0; i < tokens.length; i++)
            {
               inputNumbers[i] = Integer.parseInt(tokens[i]);

               if(inputNumbers[i] % 2 == 0 && inputNumbers[i] != 0)
               EvenNo++;
 
               else if(inputNumbers[i] % 2 != 0)
               OddNo++;
            }
               System.out.print("\n");
               System.out.println("The number of Even Numbers >> " + EvenNo);
               System.out.print("The number of Odd Numbers >> " + OddNo);
          }
        }    

Answer 1

You should convert the string format of scanner input to an integer, using Integer.parseInt()

....
System.out.println("How many numbers will be entered?");
n = Integer.parseInt(sc.nextLine());

...

Answer 2

Here is the problem. You have this...

System.out.println("How many numbers will be entered?");
n = sc.nextInt();

...which is later followed by this...

input = sc. nextLine();

Long story short, nextLine() does not play well with the other next***() methods of the scanner. You can read here for more info - https://stackoverflow.com/a/13102066/10118965

The easiest way to solve your problem would be to do this

System.out.println("How many numbers will be entered?");
n = sc.nextInt();
sc.nextLine();

That's probably the least painful way to solve this. But in the future, you can avoid this problem entirely if you only use nextLine() . Those convenience methods are nice, but as you see, they can cause you problems if you don't know their particular nuances.