// C++ 4.3.2 program to check whether a // given number is k-hyperperfect #include    using namespace std; // function to find the sum of all // proper divisors (excluding 1 and N) int divisorSum(int N int K) {  int sum = 0;  // Iterate only until sqrt N as we are  // going to generate pairs to produce  // divisors  for (int i = 2 ; i <= ceil(sqrt(N)) ; i++)  // As divisors occur in pairs we can  // take the values i and N/i as long  // as i divides N  if (N % i == 0)  sum += ( i + N/i );  return sum; } // Function to check whether the given number // is prime bool isPrime(int n) {  //base and corner cases  if (n == 1 || n == 0)  return false;  if (n <= 3)  return true;  // Since integers can be represented as  // some 6*k + y where y >= 0 we can eliminate  // all integers that can be expressed in this  // form  if (n % 2 == 0 || n % 3 == 0)  return false;  // start from 5 as this is the next prime number  for (int i=5; i*i<=n; i=i+6)  if (n % i == 0 || n % ( i + 2 ) == 0)  return false;  return true; } // Returns true if N is a K-Hyperperfect number // Else returns false. bool isHyperPerfect(int N int K) {  int sum = divisorSum(N K);  // Condition from the definition of hyperperfect  if ((1 + K * (sum)) == N)  return true;  else  return false; } // Driver function to test for hyperperfect numbers int main() {  int N1 = 1570153 K1 = 12;  int N2 = 321 K2 = 3;  // First two statements test against the condition  // N = 1 + K*(sum(proper divisors))  if (isHyperPerfect(N1 K1))  cout << N1 << ' is ' << K1   <<'-HyperPerfect' << 'n';  else  cout << N1 << ' is not ' << K1   <<'-HyperPerfect' << 'n';  if (isHyperPerfect(N2 K2))  cout << N2 << ' is ' << K2   <<'-HyperPerfect' << 'n';  else  cout << N2 << ' is not ' << K2   <<'-HyperPerfect' << 'n';  return 0; } 

// Java program to check  // whether a given number // is k-hyperperfect import java.io.*; class GFG  {   // function to find the  // sum of all proper  // divisors (excluding  // 1 and N) static int divisorSum(int N   int K) {  int sum = 0;  // Iterate only until   // sqrt N as we are   // going to generate   // pairs to produce  // divisors  for (int i = 2 ;  i <= Math.ceil(Math.sqrt(N));   i++)  // As divisors occur in   // pairs we can take   // the values i and N/i   // as long as i divides N  if (N % i == 0)  sum += (i + N / i);  return sum; } // Function to check  // whether the given // number is prime static boolean isPrime(int n) {  // base and corner cases  if (n == 1 || n == 0)  return false;  if (n <= 3)  return true;  // Since integers can be   // represented as some   // 6*k + y where y >= 0   // we can eliminate all   // integers that can be   // expressed in this form  if (n % 2 == 0 ||   n % 3 == 0)  return false;  // start from 5 as this   // is the next prime number  for (int i = 5;   i * i <= n; i = i + 6)  if (n % i == 0 ||   n % ( i + 2 ) == 0)  return false;  return true; } // Returns true if N is  // a K-Hyperperfect number // Else returns false. static boolean isHyperPerfect(int N  int K) {  int sum = divisorSum(N K);  // Condition from the   // definition of hyperperfect  if ((1 + K * (sum)) == N)  return true;  else  return false; } // Driver Code public static void main (String[] args) {  int N1 = 1570153 K1 = 12;  int N2 = 321 K2 = 3;    // First two statements test   // against the condition   // N = 1 + K*(sum(proper divisors))  if (isHyperPerfect(N1 K1))  System.out.println (N1 + ' is ' + K1 +   '-HyperPerfect' );  else  System.out.println(N1 + ' is not ' + K1 +   '-HyperPerfect' );    if (isHyperPerfect(N2 K2))  System.out.println( N2 + ' is ' + K2 +   '-HyperPerfect' );  else  System.out.println(N2 + ' is not ' + K2 +   '-HyperPerfect'); } } // This code is contributed by ajit 

# Python3 program to check whether a  # given number is k-hyperperfect  import math # Function to find the sum of all  # proper divisors (excluding 1 and N)  def divisorSum(N K): Sum = 0 # Iterate only until sqrt N as we are  # going to generate pairs to produce  # divisors  for i in range(2 math.ceil(math.sqrt(N))): # As divisors occur in pairs we can  # take the values i and N/i as long  # as i divides N  if (N % i == 0): Sum += (i + int(N / i)) return Sum # Function to check whether the given # number is prime def isPrime(n): # Base and corner cases  if (n == 1 or n == 0): return False if (n <= 3): return True # Since integers can be represented as  # some 6*k + y where y >= 0 we can eliminate  # all integers that can be expressed in this  # form  if (n % 2 == 0 or n % 3 == 0): return False # Start from 5 as this is the next # prime number  i = 5 while (i * i <= n): if (n % i == 0 or n % (i + 2) == 0): return False i += 6 return True # Returns true if N is a K-Hyperperfect # number. Else returns false.  def isHyperPerfect(N K): Sum = divisorSum(N K) # Condition from the definition # of hyperperfect  if ((1 + K * (Sum)) == N): return True else: return False # Driver code N1 = 1570153 K1 = 12 N2 = 321 K2 = 3 # First two statements test against the condition  # N = 1 + K*(sum(proper divisors))  if (isHyperPerfect(N1 K1)): print(N1 ' is ' K1 '-HyperPerfect' sep = '') else: print(N1 ' is not ' K1 '-HyperPerfect' sep = '') if (isHyperPerfect(N2 K2)): print(N2 ' is ' K2 '-HyperPerfect' sep = '') else: print(N2 ' is not ' K2 '-HyperPerfect' sep = '') # This code is contributed by avanitrachhadiya2155 

// C# program to check  // whether a given number // is k-hyperperfect using System; class GFG {   // function to find the  // sum of all proper  // divisors (excluding  // 1 and N) static int divisorSum(int N   int K) {  int sum = 0;  // Iterate only until   // sqrt N as we are   // going to generate   // pairs to produce  // divisors  for (int i = 2 ;  i <= Math.Ceiling(Math.Sqrt(N));   i++)  // As divisors occur in   // pairs we can take   // the values i and N/i   // as long as i divides N  if (N % i == 0)  sum += (i + N / i);  return sum; } // Function to check  // whether the given // number is prime static bool isPrime(int n) {  // base and corner cases  if (n == 1 || n == 0)  return false;  if (n <= 3)  return true;  // Since integers can be   // represented as some   // 6*k + y where y >= 0   // we can eliminate all   // integers that can be   // expressed in this form  if (n % 2 == 0 ||   n % 3 == 0)  return false;  // start from 5 as this   // is the next prime number  for (int i = 5;   i * i <= n; i = i + 6)  if (n % i == 0 ||   n % ( i + 2 ) == 0)  return false;  return true; } // Returns true if N is  // a K-Hyperperfect number // Else returns false. static bool isHyperPerfect(int N  int K) {  int sum = divisorSum(N K);  // Condition from the   // definition of hyperperfect  if ((1 + K * (sum)) == N)  return true;  else  return false; } // Driver Code static public void Main () { int N1 = 1570153 K1 = 12; int N2 = 321 K2 = 3; // First two statements  // test against the  // condition N = 1 + K* // (sum(proper divisors)) if (isHyperPerfect(N1 K1))  Console.WriteLine(N1 + ' is ' + K1 +   '-HyperPerfect' ); else  Console.WriteLine(N1 + ' is not ' + K1 +   '-HyperPerfect' ); if (isHyperPerfect(N2 K2))  Console.WriteLine( N2 + ' is ' + K2 +   '-HyperPerfect' ); else  Console.WriteLine(N2 + ' is not ' + K2 +   '-HyperPerfect'); } } // This code is contributed // by akt_mit 

 // PHP 4.3.2 program to check  // whether a given number is  // k-hyperperfect // function to find the sum  // of all proper divisors  // (excluding 1 and N) function divisorSum($N $K) { $sum = 0; // Iterate only until  // sqrt N as we are // going to generate  // pairs to produce // divisors for ($i = 2 ; $i <= ceil(sqrt($N)) ; $i++) // As divisors occur in // pairs we can take the // values i and N/i as long // as i divides N if ($N % $i == 0) $sum += ( $i + $N / $i); return $sum; } // Function to check whether  // the given number is prime function isPrime($n) { // base and corner cases if ($n == 1 || $n == 0) return false; if ($n <= 3) return true; // Since integers can be  // represented as some 6*k + y  // where y >= 0 we can  // eliminate all integers that  // can be expressed in this form if ($n % 2 == 0 || $n % 3 == 0) return false; // start from 5 as this // is the next prime number for ($i = 5; $i * $i <= $n; $i = $i + 6) if ($n % $i == 0 || $n % ($i + 2) == 0) return false; return true; } // Returns true if N is a  // K-Hyperperfect number // Else returns false. function isHyperPerfect($N $K) { $sum = divisorSum($N $K); // Condition from the  // definition of hyperperfect if ((1 + $K * ($sum)) == $N) return true; else return false; } // Driver Code $N1 = 1570153; $K1 = 12; $N2 = 321; $K2 = 3; // First two statements test  // against the condition  // N = 1 + K*(sum(proper divisors)) if (isHyperPerfect($N1 $K1)) echo $N1  ' is '  $K1 '-HyperPerfect'  'n'; else echo $N1  ' is not '  $K1 '-HyperPerfect'  'n'; if (isHyperPerfect($N2 $K2)) echo $N2  ' is '  K2 '-HyperPerfect'  'n'; else echo $N2  ' is not '  $K2  '-HyperPerfect'  'n'; // This code is contributed // by akt_mit ?> 

<script> // Javascript program to check // whether a given number // is k-hyperperfect // Function to find the // sum of all proper // divisors (excluding // 1 and N) function divisorSum(N K) {  let sum = 0;  // Iterate only until sqrt N as  // we are going to generate  // pairs to produce divisors  for(let i = 2;   i <= Math.ceil(Math.sqrt(N));   i++)  // As divisors occur in  // pairs we can take  // the values i and N/i  // as long as i divides N  if (N % i == 0)  sum += (i + parseInt(N / i 10));  return sum; } // Function to check // whether the given // number is prime function isPrime(n) {    // base and corner cases  if (n == 1 || n == 0)  return false;  if (n <= 3)  return true;  // Since integers can be  // represented as some  // 6*k + y where y >= 0  // we can eliminate all  // integers that can be  // expressed in this form  if (n % 2 == 0 || n % 3 == 0)  return false;  // Start from 5 as this  // is the next prime number  for(let i = 5; i * i <= n; i = i + 6)  if (n % i == 0 || n % (i + 2) == 0)  return false;  return true; } // Returns true if N is // a K-Hyperperfect number // Else returns false. function isHyperPerfect(N K) {  let sum = divisorSum(N K);  // Condition from the  // definition of hyperperfect  if ((1 + K * (sum)) == N)  return true;  else  return false; } // Driver code let N1 = 1570153 K1 = 12; let N2 = 321 K2 = 3; // First two statements // test against the  // condition N = 1 + K* // (sum(proper divisors)) if (isHyperPerfect(N1 K1))  document.write(N1 + ' is ' + K1 +   '-HyperPerfect' + '
'); else  document.write(N1 + ' is not ' + K1 +   '-HyperPerfect' + '
'); if (isHyperPerfect(N2 K2))  document.write(N2 + ' is ' + K2 +   '-HyperPerfect' + '
'); else  document.write(N2 + ' is not ' + K2 +   '-HyperPerfect' + '
');   // This code is contributed by decode2207 </script>