Reduce Time complexity of the following program

问题

import java.util.Scanner;
class Special_Pairs{

private static Scanner scan;


public static void main(String [] args) {
    byte t;
    int n;
    scan = new Scanner(System.in);
    t=scan.nextByte();
    int[] a=new int[100000];
    while(t>0)
    {
        int i,j,count=0;
        n=scan.nextInt();
    for(i=0;i<n;i++)
    {
        a[i]=scan.nextInt();
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(((a[i]&a[j])==0)||((a[j]&a[i])==0))
            {
                count++;
            }
        }
    }
    t--;
    System.out.println(count);
    }       
}
}

Help me reduce time complexity of this program

Question :

You have been given an integer array A on size N. You must report the number of ordered pairs (i,j) such that A[i] & A[j]=0. Here & denotes the BITWISE AND (i,j) and (j,i) are considered different.

Input: First line contains T-Number of Test cases. First line of each test contains N. Next line contains N integers - the i'th integer A[i].

Output: Output the number of such pairs for each test case.

Constraints: T ≤ 10; N ≤ 100000; A[i] ≤ 1000000

Sample Input(Plaintext Link)

1 5 41 47 34 40 29

Sample Output(Plaintext Link)

2

Explanation: These are the required pairs (3 5) (5 3)

回答1:

I would suggest three optimization for this. I have modified the code as well.

1. You need not to always start from 0 for each iteration of outer loop. The second loop can start from current+1 of the first loop. So will not be comparing elements which you have already compared.
2. You don't need to check for both pairs (i,j) and (j,i). If one is zero then other will always be zero.
3. You need not to initialize the array with fix size. You can always initialize it reading the value of n.

import java.util.Scanner;

public class Pairs {
  public static void main(String [] args) {
    Scanner scan = new Scanner(System.in);
    int t = scan.nextInt();
    while(t > 0) {
      t--;
      int count = 0;
      int n = scan.nextInt();
      int a[] = new int[n];
      for(int i = 0; i<n; i++) {
        a[i]=scan.nextInt();
      }
      for(int i = 0; i<n-1; i++) {
        for(int j = i+1; j<n; j++) {
          if((a[i] & a[j])==0)
          {
            count += 2;
          }
        }
      }
      System.out.println(count);
    }
  }
}

回答2:

If you are competing on a programming contest (like ICPC or something like this), maybe you shouldn't use Scanner. It's too slow for reading from the keyboard. I've already competed at ICPC, but I used to use C++. Maybe you should try BufferedReader instead of Scanner.

来源：https://stackoverflow.com/questions/32405298/reduce-time-complexity-of-the-following-program