Finding largest square less than a given number in Java

Question

I want to write a java program to find the largest square number which is less than or equal to a given number.

For example when I insert 10 as the inputted number, the answer is 9 because 9is the largest square up to 10

I understand loops, but I can't figure out the logic behind this one.

Please read [the tour](http://stackoverflow.com/tour) and [*How do I ask a good question?*](http://stackoverflow.com/help/how-to-ask). — Biffen, Jan 24 '17 at 10:44

score 2 · Answer 1 · answered Jan 24 '17 at 10:52

2

How about something like the following to get you going ...

double x = 20;
double root = Math.sqrt(x);
int t = (int)root;
System.out.println("Root is:" + root + " and answer is :" + t*t);

answered Jan 24 '17 at 10:52

Stoat

117
9

This is the way to go IMHO. – Ole V.V. Jan 24 '17 at 11:16

score 0 · Accepted Answer · answered Jan 24 '17 at 11:00

0

Here is how to do this with loops

int n=10; //This is your Number
int i = 0;
for(i=n;i>=1;i--)
    if((int)Math.sqrt(i)*(int)Math.sqrt(i)==i)
        break;
System.out.println(i);

Below is how it works:-

The Loop runs from the n, which is your number, to 1. It then checks whether, the square root of i, which is running through n to 1, is a perfect square. If it is, it breaks the loop, and prints the value of i on the screen.

answered Jan 24 '17 at 11:00

dumbPotato21

5,669
5
21
34

Thank you that was very helpful – kago Jan 24 '17 at 11:10
@kago If it helped, please select it as the answer. – dumbPotato21 Jan 24 '17 at 11:11
i just did thank you – kago Jan 24 '17 at 11:28

score 0 · Answer 3 · answered Jan 24 '17 at 11:15

    public static void main(String argv[]){
        System.out.println(largestSquareLessOrEqualTo(145)); 
    } 

    public static int largestSquareLessOrEqualTo(int limit){
    int i = 0;
        for (; i <(int)Math.sqrt(limit); i++){
        }
        return(i*i);
    }

Finding largest square less than a given number in Java

3 Answers3