couple hints: 1. don't compute all fibonacci numbers, the even ones are sufficient 2. store a fine selection of even fibonacci numbers and their pre sums for later reference 3. you can even speed up the search for the matching even fibonacci number by using a matching exponential factor and some log functions

import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
   for(int a0 = 0; a0 < t; a0++){
        long n = in.nextLong();  // Read the limit N for this test case
        long sum = 0;

        // Starting the Fibonacci sequence
        long a = 1, b = 2;

        // Calculate the sum of even Fibonacci numbers not exceeding N
        while (b <= n) {
            if (b % 2 == 0) {
                sum += b;
            }

            // Generate the next Fibonacci number
            long next = a + b;
            a = b;
            b = next;
        }

        // Output the result for this test case
        System.out.println(sum);
    }

    in.close(); 
}

}

Easier than #1

Python 3 solution. Thoughts?

import sys

t = int(input().strip())

for a0 in range(t):
    n = int(input().strip())
    
    a, b, s = 1, 2, 0
    
    while b <= n:
        if b % 2 == 0:
            s += b
            
        a, b = b, a + b
        
    print(s)

public static long fibonacci(long count){ long i= 0, k = 0, j= 1; long sum = 0; while(i

    }
    return sum;
}