What is the most efficient way to input data in Java?

Question

I solve this issue.

This is my code:

import java.io.IOException;
import java.util.Scanner;


public class Main {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] t = new int[n];
        int count = 0;
        for (int i = 0; i < n; i++) {
            t[i] = sc.nextInt();
            if (t[i] % k == 0) {
                count++;
            }
        }
        System.out.println(count);

    }
}

But when I send it, it will expire. Please help me optimize this as much as possible.

Example

Input:

7 3
1
51
966369
7
9
999996
11

Conclusion:

4

They say:

It is expected that you will be able to process at least 2.5 MB of input for the second at run time.

Modified CODE

Thanks to everyone ... I changed my code and it worked ... here it is ....

 public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] input = br.readLine().split(" ");
        int n = Integer.parseInt(input[0]);
        int k = Integer.parseInt(input[1]);
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (Integer.parseInt(br.readLine()) % k == 0) {
                count++;
            }
        }
        System.out.println(count);
    }

considers

shahensha

Answer 1

It can be a little faster, based on the limc solution, it BufferedReadershould be faster though.

import java.io.IOException;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int count = 0;
        while (true) {
            try {
                if (sc.nextInt() % k == 0) {
                    count++;
                }
            } catch (NoSuchElementException e) {
                break;
            }
        }
        System.out.println(count);

    }
}

Answer 2

How about this?

Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int count = 0;
for (int i = 0; i < n; i++) {
    if (sc.nextInt() % k == 0) {
        count++;
    }
}
System.out.println(count);

Answer 3

, .

- Integer.parseInt() Scanner.nextInt(), , somethings , , , . ( Integer.parseInt )

, , .

, !

Answer 4

BufferedReader , Scanner. , .