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This is the apex code (with result)for calculating no of perfect squares between given range . how to get total no of perfect square (in numbers) here is 6. i am unable to find please solve my problem
//Apex Programme to find the perfect squares between two numbers
public class PerfectSquare {
public static void perfectSqMethod(integer in1,integer in2){
// For every element from the range between in1 and in2
for(Integer i = in1; i<in2; i++){
double d1 = Math.sqrt(i);
double d2 = math.round(d1);
// If current number is a perfect square
if (d1==d2)
system.debug(i);
}
}
}
//calling method
PerfectSquare.perfectSqMethod(10,100);
//Result
16
25
36
49
64
81
public class PerfectSquare {
public static void perfectSqMethod(integer in1,integer in2){
// For every element from the range between in1 and in2
for(Integer i = in1; i<in2; i++){
double d1 = Math.sqrt(i);
double d2 = math.round(d1);
// If current number is a perfect square
if (d1==d2)
system.debug(i);
}
}
}
//calling method
PerfectSquare.perfectSqMethod(10,100);
//Result
16
25
36
49
64
81
The below article gives an idea of the algorithm(logic) . You will need to customize according to your requirement
https://stackoverflow.com/questions/29057462/how-to-find-number-of-perfect-squares-in-the-given-range#:~:text=There%20is%20a%20trick.,add%201%20to%20the%20difference.
Hope this helps you.Thanks
You could try using the code as below for finding the number of perfect squares and printing the squares:
public class ParkLocator {
public void countSquares(integer a, integer b)
{
integer cnt = 0; // Initialize result
// Traverse through all numbers
for (integer i = a; i <=b b; i++)
// Check if current number 'i' is perfect
// square
for (integer j = 1; j * j <= i; j++)
if (j * j == i)
{system.debug(j*j);
cnt++;}
system.debug('cnt->'+cnt);
}
}
This would be returning the number of perfect squares between given two numbers including the last number.
I hope this helps.
Regards,
Anutej