Find all perfect square root of n digit
Here given code implementation process.
// C program
// Find all perfect square root of n digit
#include <stdio.h>
#include <math.h>
int isSquareRoot(int sum)
{
int value = (int) sqrt(sum);
if (value *value == sum)
{
return 1;
}
return 0;
}
// Display Calculated result
void show(int output[], int n, int sum)
{
for (int i = 0; i < n; ++i)
{
if (i != 0)
{
printf("+");
}
printf("%d²", output[i]);
}
printf(" = %d(%d²)\n ", sum, (int) sqrt(sum));
}
// Find all perfect square root of n digit sum
void perfectSquare(int output[], int n, int counter, int num, int sum)
{
if (counter == n)
{
if (isSquareRoot(sum))
{
show(output, n, sum);
}
return;
}
for (int x = num; x <= 9; ++x)
{
output[counter] = x;
sum += x *x;
perfectSquare(output, n, counter + 1, x, sum);
sum -= x *x;
}
}
// Handles the request of finding digit square sum
void findDigitSquare(int digit)
{
if (digit <= 0)
{
// When digit is less than 1
return;
}
// Assume digit is not exceed the length of maximum numbers
int output[digit];
printf("\n Perfect Square of %d digits are \n ", digit);
perfectSquare(output, digit, 0, 1, 0);
}
int main()
{
// Test case
findDigitSquare(3);
findDigitSquare(4);
return 0;
}Output
Perfect Square of 3 digits are
1²+2²+2² = 9(3²)
1²+4²+8² = 81(9²)
2²+3²+6² = 49(7²)
2²+4²+4² = 36(6²)
2²+6²+9² = 121(11²)
3²+6²+6² = 81(9²)
4²+4²+7² = 81(9²)
4²+8²+8² = 144(12²)
6²+6²+7² = 121(11²)
Perfect Square of 4 digits are
1²+1²+1²+1² = 4(2²)
1²+1²+3²+5² = 36(6²)
1²+1²+7²+7² = 100(10²)
1²+2²+2²+4² = 25(5²)
1²+3²+3²+9² = 100(10²)
1²+4²+4²+4² = 49(7²)
1²+5²+5²+7² = 100(10²)
2²+2²+2²+2² = 16(4²)
2²+2²+3²+8² = 81(9²)
2²+2²+4²+5² = 49(7²)
2²+2²+7²+8² = 121(11²)
2²+4²+4²+8² = 100(10²)
2²+4²+5²+6² = 81(9²)
2²+8²+8²+8² = 196(14²)
3²+3²+3²+3² = 36(6²)
3²+5²+9²+9² = 196(14²)
4²+4²+4²+4² = 64(8²)
4²+4²+5²+8² = 121(11²)
4²+5²+8²+8² = 169(13²)
4²+6²+6²+9² = 169(13²)
4²+8²+8²+9² = 225(15²)
5²+5²+5²+5² = 100(10²)
6²+6²+6²+6² = 144(12²)
7²+7²+7²+7² = 196(14²)
8²+8²+8²+8² = 256(16²)
9²+9²+9²+9² = 324(18²)
/*
Java Program for
Find all perfect square root of n digit number
*/
public class DigitSquare
{
public boolean isSquareRoot(int sum)
{
int value = (int) Math.sqrt(sum);
if (value * value == sum)
{
return true;
}
return false;
}
// Display Calculated result
public void show(int[] output, int n, int sum)
{
for (int i = 0; i < n; ++i)
{
if (i != 0)
{
System.out.print("+");
}
System.out.print(" "+output[i] + "²");
}
System.out.print(" = " + sum + "(" + (int) Math.sqrt(sum) + "²)\n");
}
// Find all perfect square root of n digit sum
public void perfectSquare(int[] output, int n, int counter, int num, int sum)
{
if (counter == n)
{
if (isSquareRoot(sum))
{
show(output, n, sum);
}
return;
}
for (int x = num; x <= 9; ++x)
{
output[counter] = x;
sum += x * x;
perfectSquare(output, n, counter + 1, x, sum);
sum -= x * x;
}
}
// Handles the request of finding digit square sum
public void findDigitSquare(int digit)
{
if (digit <= 0)
{
// When digit is less than 1
return;
}
// Assume digit is not exceed the length of maximum numbers
int[] output = new int[digit];
System.out.print("\n Perfect Square of " + digit + " digits are \n");
perfectSquare(output, digit, 0, 1, 0);
}
public static void main(String[] args)
{
DigitSquare task = new DigitSquare();
// Test case
task.findDigitSquare(3);
task.findDigitSquare(4);
}
}Output
Perfect Square of 3 digits are
1²+ 2²+ 2² = 9(3²)
1²+ 4²+ 8² = 81(9²)
2²+ 3²+ 6² = 49(7²)
2²+ 4²+ 4² = 36(6²)
2²+ 6²+ 9² = 121(11²)
3²+ 6²+ 6² = 81(9²)
4²+ 4²+ 7² = 81(9²)
4²+ 8²+ 8² = 144(12²)
6²+ 6²+ 7² = 121(11²)
Perfect Square of 4 digits are
1²+ 1²+ 1²+ 1² = 4(2²)
1²+ 1²+ 3²+ 5² = 36(6²)
1²+ 1²+ 7²+ 7² = 100(10²)
1²+ 2²+ 2²+ 4² = 25(5²)
1²+ 3²+ 3²+ 9² = 100(10²)
1²+ 4²+ 4²+ 4² = 49(7²)
1²+ 5²+ 5²+ 7² = 100(10²)
2²+ 2²+ 2²+ 2² = 16(4²)
2²+ 2²+ 3²+ 8² = 81(9²)
2²+ 2²+ 4²+ 5² = 49(7²)
2²+ 2²+ 7²+ 8² = 121(11²)
2²+ 4²+ 4²+ 8² = 100(10²)
2²+ 4²+ 5²+ 6² = 81(9²)
2²+ 8²+ 8²+ 8² = 196(14²)
3²+ 3²+ 3²+ 3² = 36(6²)
3²+ 5²+ 9²+ 9² = 196(14²)
4²+ 4²+ 4²+ 4² = 64(8²)
4²+ 4²+ 5²+ 8² = 121(11²)
4²+ 5²+ 8²+ 8² = 169(13²)
4²+ 6²+ 6²+ 9² = 169(13²)
4²+ 8²+ 8²+ 9² = 225(15²)
5²+ 5²+ 5²+ 5² = 100(10²)
6²+ 6²+ 6²+ 6² = 144(12²)
7²+ 7²+ 7²+ 7² = 196(14²)
8²+ 8²+ 8²+ 8² = 256(16²)
9²+ 9²+ 9²+ 9² = 324(18²)// Include header file
#include <iostream>
#include <math.h>
using namespace std;
/*
C++ Program for
Find all perfect square root of n digit number
*/
class DigitSquare
{
public: bool isSquareRoot(int sum)
{
int value = (int) sqrt(sum);
if (value *value == sum)
{
return true;
}
return false;
}
// Display Calculated result
void show(int output[], int n, int sum)
{
for (int i = 0; i < n; ++i)
{
if (i != 0)
{
cout << "+";
}
cout << " " << output[i] << "²";
}
cout << " = " << sum << "(" << (int) sqrt(sum) << "²)\n";
}
// Find all perfect square root of n digit sum
void perfectSquare(int output[], int n, int counter, int num, int sum)
{
if (counter == n)
{
if (this->isSquareRoot(sum))
{
this->show(output, n, sum);
}
return;
}
for (int x = num; x <= 9; ++x)
{
output[counter] = x;
sum += x *x;
this->perfectSquare(output, n, counter + 1, x, sum);
sum -= x *x;
}
}
// Handles the request of finding digit square sum
void findDigitSquare(int digit)
{
// When digit is less than 1
if (digit <= 0)
{
return;
}
// Assume digit is not exceed the length of maximum numbers
int output[digit];
cout << "\n Perfect Square of " << digit << " digits are \n";
this->perfectSquare(output, digit, 0, 1, 0);
}
};
int main()
{
DigitSquare task = DigitSquare();
// Test case
task.findDigitSquare(3);
task.findDigitSquare(4);
return 0;
}Output
Perfect Square of 3 digits are
1²+ 2²+ 2² = 9(3²)
1²+ 4²+ 8² = 81(9²)
2²+ 3²+ 6² = 49(7²)
2²+ 4²+ 4² = 36(6²)
2²+ 6²+ 9² = 121(11²)
3²+ 6²+ 6² = 81(9²)
4²+ 4²+ 7² = 81(9²)
4²+ 8²+ 8² = 144(12²)
6²+ 6²+ 7² = 121(11²)
Perfect Square of 4 digits are
1²+ 1²+ 1²+ 1² = 4(2²)
1²+ 1²+ 3²+ 5² = 36(6²)
1²+ 1²+ 7²+ 7² = 100(10²)
1²+ 2²+ 2²+ 4² = 25(5²)
1²+ 3²+ 3²+ 9² = 100(10²)
1²+ 4²+ 4²+ 4² = 49(7²)
1²+ 5²+ 5²+ 7² = 100(10²)
2²+ 2²+ 2²+ 2² = 16(4²)
2²+ 2²+ 3²+ 8² = 81(9²)
2²+ 2²+ 4²+ 5² = 49(7²)
2²+ 2²+ 7²+ 8² = 121(11²)
2²+ 4²+ 4²+ 8² = 100(10²)
2²+ 4²+ 5²+ 6² = 81(9²)
2²+ 8²+ 8²+ 8² = 196(14²)
3²+ 3²+ 3²+ 3² = 36(6²)
3²+ 5²+ 9²+ 9² = 196(14²)
4²+ 4²+ 4²+ 4² = 64(8²)
4²+ 4²+ 5²+ 8² = 121(11²)
4²+ 5²+ 8²+ 8² = 169(13²)
4²+ 6²+ 6²+ 9² = 169(13²)
4²+ 8²+ 8²+ 9² = 225(15²)
5²+ 5²+ 5²+ 5² = 100(10²)
6²+ 6²+ 6²+ 6² = 144(12²)
7²+ 7²+ 7²+ 7² = 196(14²)
8²+ 8²+ 8²+ 8² = 256(16²)
9²+ 9²+ 9²+ 9² = 324(18²)// Include namespace system
using System;
/*
C# Program for
Find all perfect square root of n digit number
*/
public class DigitSquare
{
public Boolean isSquareRoot(int sum)
{
int value = (int) Math.Sqrt(sum);
if (value * value == sum)
{
return true;
}
return false;
}
// Display Calculated result
public void show(int[] output, int n, int sum)
{
for (int i = 0; i < n; ++i)
{
if (i != 0)
{
Console.Write("+");
}
Console.Write(" " + output[i] + "²");
}
Console.Write(" = " + sum + "(" + (int) Math.Sqrt(sum) + "²)\n");
}
// Find all perfect square root of n digit sum
public void perfectSquare(int[] output, int n, int counter, int num, int sum)
{
if (counter == n)
{
if (isSquareRoot(sum))
{
show(output, n, sum);
}
return;
}
for (int x = num; x <= 9; ++x)
{
output[counter] = x;
sum += x * x;
perfectSquare(output, n, counter + 1, x, sum);
sum -= x * x;
}
}
// Handles the request of finding digit square sum
public void findDigitSquare(int digit)
{
// When digit is less than 1
if (digit <= 0)
{
return;
}
// Assume digit is not exceed the length of maximum numbers
int[] output = new int[digit];
Console.Write("\n Perfect Square of " + digit + " digits are \n");
perfectSquare(output, digit, 0, 1, 0);
}
public static void Main(String[] args)
{
DigitSquare task = new DigitSquare();
// Test case
task.findDigitSquare(3);
task.findDigitSquare(4);
}
}Output
Perfect Square of 3 digits are
1²+ 2²+ 2² = 9(3²)
1²+ 4²+ 8² = 81(9²)
2²+ 3²+ 6² = 49(7²)
2²+ 4²+ 4² = 36(6²)
2²+ 6²+ 9² = 121(11²)
3²+ 6²+ 6² = 81(9²)
4²+ 4²+ 7² = 81(9²)
4²+ 8²+ 8² = 144(12²)
6²+ 6²+ 7² = 121(11²)
Perfect Square of 4 digits are
1²+ 1²+ 1²+ 1² = 4(2²)
1²+ 1²+ 3²+ 5² = 36(6²)
1²+ 1²+ 7²+ 7² = 100(10²)
1²+ 2²+ 2²+ 4² = 25(5²)
1²+ 3²+ 3²+ 9² = 100(10²)
1²+ 4²+ 4²+ 4² = 49(7²)
1²+ 5²+ 5²+ 7² = 100(10²)
2²+ 2²+ 2²+ 2² = 16(4²)
2²+ 2²+ 3²+ 8² = 81(9²)
2²+ 2²+ 4²+ 5² = 49(7²)
2²+ 2²+ 7²+ 8² = 121(11²)
2²+ 4²+ 4²+ 8² = 100(10²)
2²+ 4²+ 5²+ 6² = 81(9²)
2²+ 8²+ 8²+ 8² = 196(14²)
3²+ 3²+ 3²+ 3² = 36(6²)
3²+ 5²+ 9²+ 9² = 196(14²)
4²+ 4²+ 4²+ 4² = 64(8²)
4²+ 4²+ 5²+ 8² = 121(11²)
4²+ 5²+ 8²+ 8² = 169(13²)
4²+ 6²+ 6²+ 9² = 169(13²)
4²+ 8²+ 8²+ 9² = 225(15²)
5²+ 5²+ 5²+ 5² = 100(10²)
6²+ 6²+ 6²+ 6² = 144(12²)
7²+ 7²+ 7²+ 7² = 196(14²)
8²+ 8²+ 8²+ 8² = 256(16²)
9²+ 9²+ 9²+ 9² = 324(18²)<?php
/*
Php Program for
Find all perfect square root of n digit number
*/
class DigitSquare
{
public function isSquareRoot($sum)
{
$value = (int) sqrt($sum);
if ($value * $value == $sum)
{
return true;
}
return false;
}
// Display Calculated result
public function show($output, $n, $sum)
{
for ($i = 0; $i < $n; ++$i)
{
if ($i != 0)
{
echo "+";
}
echo " ". $output[$i] ."²";
}
echo " = ". $sum ."(". (int) sqrt($sum) ."²)\n";
}
// Find all perfect square root of n digit sum
public function perfectSquare($output, $n, $counter, $num, $sum)
{
if ($counter == $n)
{
if ($this->isSquareRoot($sum))
{
$this->show($output, $n, $sum);
}
return;
}
for ($x = $num; $x <= 9; ++$x)
{
$output[$counter] = $x;
$sum += $x * $x;
$this->perfectSquare($output, $n, $counter + 1, $x, $sum);
$sum -= $x * $x;
}
}
// Handles the request of finding digit square sum
public function findDigitSquare($digit)
{
// When digit is less than 1
if ($digit <= 0)
{
return;
}
// Assume digit is not exceed the length of maximum numbers
$output = array_fill(0, $digit, 0);
echo "\n Perfect Square of ". $digit ." digits are \n";
$this->perfectSquare($output, $digit, 0, 1, 0);
}
}
function main()
{
$task = new DigitSquare();
// Test case
$task->findDigitSquare(3);
$task->findDigitSquare(4);
}
main();Output
Perfect Square of 3 digits are
1²+ 2²+ 2² = 9(3²)
1²+ 4²+ 8² = 81(9²)
2²+ 3²+ 6² = 49(7²)
2²+ 4²+ 4² = 36(6²)
2²+ 6²+ 9² = 121(11²)
3²+ 6²+ 6² = 81(9²)
4²+ 4²+ 7² = 81(9²)
4²+ 8²+ 8² = 144(12²)
6²+ 6²+ 7² = 121(11²)
Perfect Square of 4 digits are
1²+ 1²+ 1²+ 1² = 4(2²)
1²+ 1²+ 3²+ 5² = 36(6²)
1²+ 1²+ 7²+ 7² = 100(10²)
1²+ 2²+ 2²+ 4² = 25(5²)
1²+ 3²+ 3²+ 9² = 100(10²)
1²+ 4²+ 4²+ 4² = 49(7²)
1²+ 5²+ 5²+ 7² = 100(10²)
2²+ 2²+ 2²+ 2² = 16(4²)
2²+ 2²+ 3²+ 8² = 81(9²)
2²+ 2²+ 4²+ 5² = 49(7²)
2²+ 2²+ 7²+ 8² = 121(11²)
2²+ 4²+ 4²+ 8² = 100(10²)
2²+ 4²+ 5²+ 6² = 81(9²)
2²+ 8²+ 8²+ 8² = 196(14²)
3²+ 3²+ 3²+ 3² = 36(6²)
3²+ 5²+ 9²+ 9² = 196(14²)
4²+ 4²+ 4²+ 4² = 64(8²)
4²+ 4²+ 5²+ 8² = 121(11²)
4²+ 5²+ 8²+ 8² = 169(13²)
4²+ 6²+ 6²+ 9² = 169(13²)
4²+ 8²+ 8²+ 9² = 225(15²)
5²+ 5²+ 5²+ 5² = 100(10²)
6²+ 6²+ 6²+ 6² = 144(12²)
7²+ 7²+ 7²+ 7² = 196(14²)
8²+ 8²+ 8²+ 8² = 256(16²)
9²+ 9²+ 9²+ 9² = 324(18²)/*
Node Js Program for
Find all perfect square root of n digit number
*/
class DigitSquare
{
isSquareRoot(sum)
{
var value = parseInt(Math.sqrt(sum));
if (value * value == sum)
{
return true;
}
return false;
}
// Display Calculated result
show(output, n, sum)
{
for (var i = 0; i < n; ++i)
{
if (i != 0)
{
process.stdout.write("+");
}
process.stdout.write(" " + output[i] + "²");
}
process.stdout.write(" = " + sum + "(" + parseInt(Math.sqrt(sum)) + "²)\n");
}
// Find all perfect square root of n digit sum
perfectSquare(output, n, counter, num, sum)
{
if (counter == n)
{
if (this.isSquareRoot(sum))
{
this.show(output, n, sum);
}
return;
}
for (var x = num; x <= 9; ++x)
{
output[counter] = x;
sum += x * x;
this.perfectSquare(output, n, counter + 1, x, sum);
sum -= x * x;
}
}
// Handles the request of finding digit square sum
findDigitSquare(digit)
{
// When digit is less than 1
if (digit <= 0)
{
return;
}
// Assume digit is not exceed the length of maximum numbers
var output = Array(digit).fill(0);
process.stdout.write("\n Perfect Square of " + digit + " digits are \n");
this.perfectSquare(output, digit, 0, 1, 0);
}
}
function main()
{
var task = new DigitSquare();
// Test case
task.findDigitSquare(3);
task.findDigitSquare(4);
}
main();Output
Perfect Square of 3 digits are
1²+ 2²+ 2² = 9(3²)
1²+ 4²+ 8² = 81(9²)
2²+ 3²+ 6² = 49(7²)
2²+ 4²+ 4² = 36(6²)
2²+ 6²+ 9² = 121(11²)
3²+ 6²+ 6² = 81(9²)
4²+ 4²+ 7² = 81(9²)
4²+ 8²+ 8² = 144(12²)
6²+ 6²+ 7² = 121(11²)
Perfect Square of 4 digits are
1²+ 1²+ 1²+ 1² = 4(2²)
1²+ 1²+ 3²+ 5² = 36(6²)
1²+ 1²+ 7²+ 7² = 100(10²)
1²+ 2²+ 2²+ 4² = 25(5²)
1²+ 3²+ 3²+ 9² = 100(10²)
1²+ 4²+ 4²+ 4² = 49(7²)
1²+ 5²+ 5²+ 7² = 100(10²)
2²+ 2²+ 2²+ 2² = 16(4²)
2²+ 2²+ 3²+ 8² = 81(9²)
2²+ 2²+ 4²+ 5² = 49(7²)
2²+ 2²+ 7²+ 8² = 121(11²)
2²+ 4²+ 4²+ 8² = 100(10²)
2²+ 4²+ 5²+ 6² = 81(9²)
2²+ 8²+ 8²+ 8² = 196(14²)
3²+ 3²+ 3²+ 3² = 36(6²)
3²+ 5²+ 9²+ 9² = 196(14²)
4²+ 4²+ 4²+ 4² = 64(8²)
4²+ 4²+ 5²+ 8² = 121(11²)
4²+ 5²+ 8²+ 8² = 169(13²)
4²+ 6²+ 6²+ 9² = 169(13²)
4²+ 8²+ 8²+ 9² = 225(15²)
5²+ 5²+ 5²+ 5² = 100(10²)
6²+ 6²+ 6²+ 6² = 144(12²)
7²+ 7²+ 7²+ 7² = 196(14²)
8²+ 8²+ 8²+ 8² = 256(16²)
9²+ 9²+ 9²+ 9² = 324(18²)import math
# Python 3 Program for
# Find all perfect square root of n digit number
class DigitSquare :
def isSquareRoot(self, sum) :
value = int(math.sqrt(sum))
if (value * value == sum) :
return True
return False
# Display Calculated result
def show(self, output, n, sum) :
i = 0
while (i < n) :
if (i != 0) :
print("+", end = "")
print("{0}²".format(output[i]),end="")
i += 1
print(" = {0}({1})² ".format(sum,int(math.sqrt(sum))))
# Find all perfect square root of n digit sum
def perfectSquare(self, output, n, counter, num, sum) :
if (counter == n) :
if (self.isSquareRoot(sum)) :
self.show(output, n, sum)
return
x = num
while (x <= 9) :
output[counter] = x
sum += x * x
self.perfectSquare(output, n, counter + 1, x, sum)
sum -= x * x
x += 1
# Handles the request of finding digit square sum
def findDigitSquare(self, digit) :
# When digit is less than 1
if (digit <= 0) :
return
# Assume digit is not exceed the length of maximum numbers
output = [0] * (digit)
print("\n Perfect Square of ", digit ," digits are ")
self.perfectSquare(output, digit, 0, 1, 0)
def main() :
task = DigitSquare()
# Test case
task.findDigitSquare(3)
task.findDigitSquare(4)
if __name__ == "__main__": main()Output
Perfect Square of 3 digits are
1²+2²+2² = 9(3)²
1²+4²+8² = 81(9)²
2²+3²+6² = 49(7)²
2²+4²+4² = 36(6)²
2²+6²+9² = 121(11)²
3²+6²+6² = 81(9)²
4²+4²+7² = 81(9)²
4²+8²+8² = 144(12)²
6²+6²+7² = 121(11)²
Perfect Square of 4 digits are
1²+1²+1²+1² = 4(2)²
1²+1²+3²+5² = 36(6)²
1²+1²+7²+7² = 100(10)²
1²+2²+2²+4² = 25(5)²
1²+3²+3²+9² = 100(10)²
1²+4²+4²+4² = 49(7)²
1²+5²+5²+7² = 100(10)²
2²+2²+2²+2² = 16(4)²
2²+2²+3²+8² = 81(9)²
2²+2²+4²+5² = 49(7)²
2²+2²+7²+8² = 121(11)²
2²+4²+4²+8² = 100(10)²
2²+4²+5²+6² = 81(9)²
2²+8²+8²+8² = 196(14)²
3²+3²+3²+3² = 36(6)²
3²+5²+9²+9² = 196(14)²
4²+4²+4²+4² = 64(8)²
4²+4²+5²+8² = 121(11)²
4²+5²+8²+8² = 169(13)²
4²+6²+6²+9² = 169(13)²
4²+8²+8²+9² = 225(15)²
5²+5²+5²+5² = 100(10)²
6²+6²+6²+6² = 144(12)²
7²+7²+7²+7² = 196(14)²
8²+8²+8²+8² = 256(16)²
9²+9²+9²+9² = 324(18)²# Ruby Program for
# Find all perfect square root of n digit number
class DigitSquare
def isSquareRoot(sum)
value = (Math.sqrt(sum)).to_i
if (value * value == sum)
return true
end
return false
end
# Display Calculated result
def show(output, n, sum)
i = 0
while (i < n)
if (i != 0)
print("+")
end
print(" ", output[i] ,"²")
i += 1
end
print(" = ", sum ,"(", (Math.sqrt(sum)).to_i ,"²)\n")
end
# Find all perfect square root of n digit sum
def perfectSquare(output, n, counter, num, sum)
if (counter == n)
if (self.isSquareRoot(sum))
self.show(output, n, sum)
end
return
end
x = num
while (x <= 9)
output[counter] = x
sum += x * x
self.perfectSquare(output, n, counter + 1, x, sum)
sum -= x * x
x += 1
end
end
# Handles the request of finding digit square sum
def findDigitSquare(digit)
# When digit is less than 1
if (digit <= 0)
return
end
# Assume digit is not exceed the length of maximum numbers
output = Array.new(digit) {0}
print("\n Perfect Square of ", digit ," digits are \n")
self.perfectSquare(output, digit, 0, 1, 0)
end
end
def main()
task = DigitSquare.new()
# Test case
task.findDigitSquare(3)
task.findDigitSquare(4)
end
main()Output
Perfect Square of 3 digits are
1²+ 2²+ 2² = 9(3²)
1²+ 4²+ 8² = 81(9²)
2²+ 3²+ 6² = 49(7²)
2²+ 4²+ 4² = 36(6²)
2²+ 6²+ 9² = 121(11²)
3²+ 6²+ 6² = 81(9²)
4²+ 4²+ 7² = 81(9²)
4²+ 8²+ 8² = 144(12²)
6²+ 6²+ 7² = 121(11²)
Perfect Square of 4 digits are
1²+ 1²+ 1²+ 1² = 4(2²)
1²+ 1²+ 3²+ 5² = 36(6²)
1²+ 1²+ 7²+ 7² = 100(10²)
1²+ 2²+ 2²+ 4² = 25(5²)
1²+ 3²+ 3²+ 9² = 100(10²)
1²+ 4²+ 4²+ 4² = 49(7²)
1²+ 5²+ 5²+ 7² = 100(10²)
2²+ 2²+ 2²+ 2² = 16(4²)
2²+ 2²+ 3²+ 8² = 81(9²)
2²+ 2²+ 4²+ 5² = 49(7²)
2²+ 2²+ 7²+ 8² = 121(11²)
2²+ 4²+ 4²+ 8² = 100(10²)
2²+ 4²+ 5²+ 6² = 81(9²)
2²+ 8²+ 8²+ 8² = 196(14²)
3²+ 3²+ 3²+ 3² = 36(6²)
3²+ 5²+ 9²+ 9² = 196(14²)
4²+ 4²+ 4²+ 4² = 64(8²)
4²+ 4²+ 5²+ 8² = 121(11²)
4²+ 5²+ 8²+ 8² = 169(13²)
4²+ 6²+ 6²+ 9² = 169(13²)
4²+ 8²+ 8²+ 9² = 225(15²)
5²+ 5²+ 5²+ 5² = 100(10²)
6²+ 6²+ 6²+ 6² = 144(12²)
7²+ 7²+ 7²+ 7² = 196(14²)
8²+ 8²+ 8²+ 8² = 256(16²)
9²+ 9²+ 9²+ 9² = 324(18²)
/*
Scala Program for
Find all perfect square root of n digit number
*/
class DigitSquare
{
def isSquareRoot(sum: Int): Boolean = {
var value: Int = (Math.sqrt(sum)).toInt;
if (value * value == sum)
{
return true;
}
return false;
}
// Display Calculated result
def show(output: Array[Int], n: Int, sum: Int): Unit = {
var i: Int = 0;
while (i < n)
{
if (i != 0)
{
print("+");
}
print(" " + output(i) + "²");
i += 1;
}
print(" = " + sum + "(" + (Math.sqrt(sum)).toInt + "²)\n");
}
// Find all perfect square root of n digit sum
def perfectSquare(output: Array[Int], n: Int, counter: Int, num: Int, sum: Int): Unit = {
if (counter == n)
{
if (this.isSquareRoot(sum))
{
this.show(output, n, sum);
}
return;
}
var x: Int = num;
while (x <= 9)
{
output(counter) = x;
this.perfectSquare(output, n, counter + 1, x, sum + x * x);
x += 1;
}
}
// Handles the request of finding digit square sum
def findDigitSquare(digit: Int): Unit = {
// When digit is less than 1
if (digit <= 0)
{
return;
}
// Assume digit is not exceed the length of maximum numbers
var output: Array[Int] = Array.fill[Int](digit)(0);
print("\n Perfect Square of " + digit + " digits are \n");
this.perfectSquare(output, digit, 0, 1, 0);
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: DigitSquare = new DigitSquare();
// Test case
task.findDigitSquare(3);
task.findDigitSquare(4);
}
}Output
Perfect Square of 3 digits are
1²+ 2²+ 2² = 9(3²)
1²+ 4²+ 8² = 81(9²)
2²+ 3²+ 6² = 49(7²)
2²+ 4²+ 4² = 36(6²)
2²+ 6²+ 9² = 121(11²)
3²+ 6²+ 6² = 81(9²)
4²+ 4²+ 7² = 81(9²)
4²+ 8²+ 8² = 144(12²)
6²+ 6²+ 7² = 121(11²)
Perfect Square of 4 digits are
1²+ 1²+ 1²+ 1² = 4(2²)
1²+ 1²+ 3²+ 5² = 36(6²)
1²+ 1²+ 7²+ 7² = 100(10²)
1²+ 2²+ 2²+ 4² = 25(5²)
1²+ 3²+ 3²+ 9² = 100(10²)
1²+ 4²+ 4²+ 4² = 49(7²)
1²+ 5²+ 5²+ 7² = 100(10²)
2²+ 2²+ 2²+ 2² = 16(4²)
2²+ 2²+ 3²+ 8² = 81(9²)
2²+ 2²+ 4²+ 5² = 49(7²)
2²+ 2²+ 7²+ 8² = 121(11²)
2²+ 4²+ 4²+ 8² = 100(10²)
2²+ 4²+ 5²+ 6² = 81(9²)
2²+ 8²+ 8²+ 8² = 196(14²)
3²+ 3²+ 3²+ 3² = 36(6²)
3²+ 5²+ 9²+ 9² = 196(14²)
4²+ 4²+ 4²+ 4² = 64(8²)
4²+ 4²+ 5²+ 8² = 121(11²)
4²+ 5²+ 8²+ 8² = 169(13²)
4²+ 6²+ 6²+ 9² = 169(13²)
4²+ 8²+ 8²+ 9² = 225(15²)
5²+ 5²+ 5²+ 5² = 100(10²)
6²+ 6²+ 6²+ 6² = 144(12²)
7²+ 7²+ 7²+ 7² = 196(14²)
8²+ 8²+ 8²+ 8² = 256(16²)
9²+ 9²+ 9²+ 9² = 324(18²)import Foundation
/*
Swift 4 Program for
Find all perfect square root of n digit number
*/
class DigitSquare
{
func isSquareRoot(_ sum: Int)->Bool
{
let value: Int = Int(sqrt(Double(sum)));
if (value * value == sum)
{
return true;
}
return false;
}
// Display Calculated result
func show(_ output: [Int], _ n: Int, _ sum: Int)
{
var i: Int = 0;
while (i < n)
{
if (i != 0)
{
print(terminator: "+");
}
print("\(output[i])²",terminator: "");
i += 1;
}
print(" = \(sum) (\(Int(sqrt(Double(sum)))))²");
}
// Find all perfect square root of n digit sum
func perfectSquare(_ output: inout[Int], _ n: Int, _ counter: Int, _ num: Int, _ sum: Int)
{
if (counter == n)
{
if (self.isSquareRoot(sum))
{
self.show(output, n, sum);
}
return;
}
var x: Int = num;
while (x <= 9)
{
output[counter] = x;
self.perfectSquare(&output, n, counter + 1, x, sum + x * x);
x += 1;
}
}
// Handles the request of finding digit square sum
func findDigitSquare(_ digit: Int)
{
// When digit is less than 1
if (digit <= 0)
{
return;
}
// Assume digit is not exceed the length of maximum numbers
var output: [Int] = Array(repeating: 0, count: digit);
print("\n Perfect Square of ", digit ," digits are ");
self.perfectSquare(&output, digit, 0, 1, 0);
}
}
func main()
{
let task: DigitSquare = DigitSquare();
// Test case
task.findDigitSquare(3);
task.findDigitSquare(4);
}
main();Output
Perfect Square of 3 digits are
1²+2²+2² = 9 (3)²
1²+4²+8² = 81 (9)²
2²+3²+6² = 49 (7)²
2²+4²+4² = 36 (6)²
2²+6²+9² = 121 (11)²
3²+6²+6² = 81 (9)²
4²+4²+7² = 81 (9)²
4²+8²+8² = 144 (12)²
6²+6²+7² = 121 (11)²
Perfect Square of 4 digits are
1²+1²+1²+1² = 4 (2)²
1²+1²+3²+5² = 36 (6)²
1²+1²+7²+7² = 100 (10)²
1²+2²+2²+4² = 25 (5)²
1²+3²+3²+9² = 100 (10)²
1²+4²+4²+4² = 49 (7)²
1²+5²+5²+7² = 100 (10)²
2²+2²+2²+2² = 16 (4)²
2²+2²+3²+8² = 81 (9)²
2²+2²+4²+5² = 49 (7)²
2²+2²+7²+8² = 121 (11)²
2²+4²+4²+8² = 100 (10)²
2²+4²+5²+6² = 81 (9)²
2²+8²+8²+8² = 196 (14)²
3²+3²+3²+3² = 36 (6)²
3²+5²+9²+9² = 196 (14)²
4²+4²+4²+4² = 64 (8)²
4²+4²+5²+8² = 121 (11)²
4²+5²+8²+8² = 169 (13)²
4²+6²+6²+9² = 169 (13)²
4²+8²+8²+9² = 225 (15)²
5²+5²+5²+5² = 100 (10)²
6²+6²+6²+6² = 144 (12)²
7²+7²+7²+7² = 196 (14)²
8²+8²+8²+8² = 256 (16)²
9²+9²+9²+9² = 324 (18)²/*
Kotlin Program for
Find all perfect square root of n digit number
*/
class DigitSquare
{
fun isSquareRoot(sum: Int): Boolean
{
var value: Int = Math.sqrt(sum.toDouble()).toInt();
if (value * value == sum)
{
return true;
}
return false;
}
// Display Calculated result
fun show(output: Array <Int> , n: Int, sum: Int): Unit
{
var i: Int = 0;
while (i < n)
{
if (i != 0)
{
print("+");
}
print("" + output[i] + "²");
i += 1;
}
print(" = " + sum + "(" + Math.sqrt(sum.toDouble()).toInt() + "²)\n ");
}
// Find all perfect square root of n digit sum
fun perfectSquare(output: Array < Int > , n: Int, counter: Int, num: Int, sum: Int): Unit
{
if (counter == n)
{
if (this.isSquareRoot(sum))
{
this.show(output, n, sum);
}
return;
}
var x: Int = num;
while (x <= 9)
{
output[counter] = x;
this.perfectSquare(output, n, counter + 1, x, sum + x * x);
x += 1;
}
}
// Handles the request of finding digit square sum
fun findDigitSquare(digit: Int): Unit
{
// When digit is less than 1
if (digit <= 0)
{
return;
}
// Assume digit is not exceed the length of maximum numbers
var output: Array < Int > = Array(digit)
{
0
};
print("\n Perfect Square of " + digit + " digits are \n ");
this.perfectSquare(output, digit, 0, 1, 0);
}
}
fun main(args: Array < String > ): Unit
{
var task: DigitSquare = DigitSquare();
// Test case
task.findDigitSquare(3);
task.findDigitSquare(4);
}Output
Perfect Square of 3 digits are
1²+2²+2² = 9(3²)
1²+4²+8² = 81(9²)
2²+3²+6² = 49(7²)
2²+4²+4² = 36(6²)
2²+6²+9² = 121(11²)
3²+6²+6² = 81(9²)
4²+4²+7² = 81(9²)
4²+8²+8² = 144(12²)
6²+6²+7² = 121(11²)
Perfect Square of 4 digits are
1²+1²+1²+1² = 4(2²)
1²+1²+3²+5² = 36(6²)
1²+1²+7²+7² = 100(10²)
1²+2²+2²+4² = 25(5²)
1²+3²+3²+9² = 100(10²)
1²+4²+4²+4² = 49(7²)
1²+5²+5²+7² = 100(10²)
2²+2²+2²+2² = 16(4²)
2²+2²+3²+8² = 81(9²)
2²+2²+4²+5² = 49(7²)
2²+2²+7²+8² = 121(11²)
2²+4²+4²+8² = 100(10²)
2²+4²+5²+6² = 81(9²)
2²+8²+8²+8² = 196(14²)
3²+3²+3²+3² = 36(6²)
3²+5²+9²+9² = 196(14²)
4²+4²+4²+4² = 64(8²)
4²+4²+5²+8² = 121(11²)
4²+5²+8²+8² = 169(13²)
4²+6²+6²+9² = 169(13²)
4²+8²+8²+9² = 225(15²)
5²+5²+5²+5² = 100(10²)
6²+6²+6²+6² = 144(12²)
7²+7²+7²+7² = 196(14²)
8²+8²+8²+8² = 256(16²)
9²+9²+9²+9² = 324(18²)
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