# Check if the sum of perfect squares in an array is divisible by x

Given an array arr[] and an integer x, the task is to check whether the sum of all the perfect squares from the array is divisible by x or not. If divisible then print Yes else print No.

Examples:

Input: arr[] = {2, 3, 4, 6, 9, 10}, x = 13
Output: Yes
4 and 9 are the only perfect squares from the array
sum = 4 + 9 = 13 (which is divisible by 13)

Input: arr[] = {2, 4, 25, 49, 3, 8}, x = 9
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Run a loop from i to n – 1 and check whether arr[i] is a perfect square or not. If arr[i] is a perfect square then update sum = sum + arr[i]. If in the end sum % x = 0 then print Yes else print No. To check whether an element is a perfect square or not, follow the following steps:

Let num be an integer element
float sq = sqrt(x)
if floor(sq) = ceil(sq) then num is a perfect square else not.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if the sum of all the ` `// perfect squares of the given array are divisible by x ` `bool` `check(``int` `arr[], ``int` `x, ``int` `n) ` `{ ` `    ``long` `long` `sum = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``double` `x = ``sqrt``(arr[i]); ` ` `  `        ``// If arr[i] is a perfect square ` `        ``if` `(``floor``(x) == ``ceil``(x)) { ` `            ``sum += arr[i]; ` `        ``} ` `    ``} ` ` `  `    ``if` `(sum % x == 0) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 3, 4, 9, 10 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `x = 13; ` ` `  `    ``if` `(check(arr, x, n)) { ` `        ``cout << ``"Yes"``; ` `    ``} ` `    ``else` `{ ` `        ``cout << ``"No"``; ` `    ``} ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `public` `class` `GFG{ ` ` `  `    ``// Function that returns true if the the sum of all the  ` `    ``// perfect squares of the given array is divisible by x  ` `    ``static` `boolean` `check(``int` `arr[], ``int` `x, ``int` `n)  ` `    ``{  ` `        ``long` `sum = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < n; i++) {  ` `            ``double` `y = Math.sqrt(arr[i]);  ` `     `  `            ``// If arr[i] is a perfect square  ` `            ``if` `(Math.floor(y) == Math.ceil(y)) {  ` `                ``sum += arr[i];  ` `            ``}  ` `        ``}  ` `     `  `        ``if` `(sum % x == ``0``)  ` `            ``return` `true``;  ` `        ``else` `            ``return` `false``;  ` `    ``}  ` ` `  ` `  ` `  `    ``// Driver Code  ` `    ``public` `static` `void` `main(String []args){ ` `        ``int` `arr[] = { ``2``, ``3``, ``4``, ``9``, ``10` `};  ` `        ``int` `n = arr.length ; ` `        ``int` `x = ``13``;  ` ` `  `        ``if` `(check(arr, x, n)) {  ` `            ``System.out.println(``"Yes"``);  ` `        ``}  ` `        ``else` `{  ` `           ``System.out.println(``"No"``);  ` `        ``}  ` `    ``} ` `    ``// This code is contributed by Ryuga ` `} `

## Python3

 `# Python3 implementation of the approach ` `import` `math ` ` `  `# Function that returns true if the the sum of all the  ` `# perfect squares of the given array is divisible by x ` `def` `check (a, y): ` `    ``sum` `=` `0` `    ``for` `i ``in` `range``(``len``(a)): ` `         `  `        ``x ``=` `math.sqrt(a[i]) ` ` `  `        ``# If a[i] is a perfect square ` `        ``if` `(math.floor(x) ``=``=` `math.ceil(x)): ` `            ``sum` `=` `sum` `+` `a[i] ` `     `  `    ``if` `(``sum` `%` `y ``=``=` `0``): ` `        ``return` `True` `    ``else``: ` `        ``return` `False` `         `  ` `  `# Driver code ` `a ``=` `[``2``, ``3``, ``4``, ``9``, ``10``] ` `x ``=` `13` ` `  `if` `check(a, x) : ` `    ``print``(``"Yes"``) ` `else``: ` `    ``print``(``"No"``) `

## C#

 `// C# implementation of the approach  ` ` `  `using` `System; ` `public` `class` `GFG{ ` `  `  `    ``// Function that returns true if the the sum of all the  ` `    ``// perfect squares of the given array is divisible by x  ` `    ``static` `bool` `check(``int``[] arr, ``int` `x, ``int` `n)  ` `    ``{  ` `        ``long` `sum = 0;  ` `        ``for` `(``int` `i = 0; i < n; i++) {  ` `            ``double` `y = Math.Sqrt(arr[i]);  ` `      `  `            ``// If arr[i] is a perfect square  ` `            ``if` `(Math.Floor(y) == Math.Ceiling(y)) {  ` `                ``sum += arr[i];  ` `            ``}  ` `        ``}  ` `      `  `        ``if` `(sum % x == 0)  ` `            ``return` `true``;  ` `        ``else` `            ``return` `false``;  ` `    ``}  ` `  `  `  `  `  `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main(){ ` `        ``int``[] arr = { 2, 3, 4, 9, 10 };  ` `        ``int` `n = arr.Length ; ` `        ``int` `x = 13;  ` `  `  `        ``if` `(check(arr, x, n)) {  ` `            ``Console.Write(``"Yes"``);  ` `        ``}  ` `        ``else` `{  ` `           ``Console.Write(``"No"``);  ` `        ``}  ` `    ``}     ` `} `

## PHP

 ` `

Output:

```Yes
```

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