Which is the most efficient way of taking input in Java?

问题

I am solving this question.

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 submit it, it get's timed out. Please help me optimize this to as much as is possible.

Example

Input:

Output:

They say :

You are expected to be able to process at least 2.5MB of input data per second at runtime.

Modified CODE

Thank you all...I modified 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);
    }

regards

shahensha

回答1:

This could be slightly faster, based on limc's solution, BufferedReader should be faster still 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);

    }
}

回答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);

回答3:

You may consider reading big chunks of input and then get the numbers from there.

Other change is, you may use Integer.parseInt() instead of Scanner.nextInt() although I don't know the details of each one, somethings tells me Scanner version performs a bit more computation to know if the input is correct. Another alternative is to convert the number yourself ( although Integer.parseInt should be fast enough )

Create a sample input, and measure your code, change a bit here and there and see what the difference is.

Measure, measure!

回答4:

BufferedReader is supposed to be faster than Scanner. You will need to parse everything yourself though and depending on your implementation it could be worse.

来源：https://stackoverflow.com/questions/4960736/which-is-the-most-efficient-way-of-taking-input-in-java

标签

java

optimization

user-input

inputstream