Given an integer N, the task is to check if N is an Admirable Number.
An admirable number is a number, if there exists a proper divisor D’ of N such that sigma(N)-2D’ = 2N, where sigma(N) is the sum of all divisors of N
Examples:
Input: N = 12
Output: Yes
Explanation:
12’s proper divisors are 1, 2, 3, 4, 6, and 12
sigma(N) = 1 + 2 + 3 + 4 + 6 + 12 = 28
sigma(N) – 2D’ = 2N
28 – 2*2 = 2*12
24 == 24Input: N = 28
Output: No
Approach: The idea is to find the sum of all factors of a number that is sigma(N). And then we will find every proper divisor of a number D’ and check if there exists a proper divisor D’ of N such that
Below is the implementation of the above approach:
// C++ implementation to check if // N is an admirable number #include <bits/stdc++.h> using namespace std;
// Function to calculate the sum of // all divisors of a given number int divSum( int n)
{ // Sum of divisors
int result = 0;
// Find all divisors
// which divides 'num'
for ( int i = 2; i <= sqrt (n); i++) {
// if 'i' is divisor of 'n'
if (n % i == 0) {
// if both divisors are same
// then add it once else add
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above loop
// considers proper divisors greater
return (result + n + 1);
} // Function to check if there // exists a proper divisor // D' of N such that sigma(n)-2D' = 2N bool check( int num)
{ int sigmaN = divSum(num);
// Find all divisors which divides 'num'
for ( int i = 2; i <= sqrt (num); i++) {
// if 'i' is divisor of 'num'
if (num % i == 0) {
// if both divisors are same then add
// it only once else add both
if (i == (num / i)) {
if (sigmaN - 2 * i == 2 * num)
return true ;
}
else {
if (sigmaN - 2 * i == 2 * num)
return true ;
if (sigmaN - 2 * (num / i) == 2 * num)
return true ;
}
}
}
// Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num)
return true ;
return false ;
} // Function to check if N // is an admirable number bool isAdmirableNum( int N)
{ return check(N);
} // Driver code int main()
{ int n = 12;
if (isAdmirableNum(n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
// Java implementation to check if N // is a admirable number class GFG{
// Function to calculate the sum of // all divisors of a given number static int divSum( int n)
{ // Sum of divisors
int result = 0 ;
// Find all divisors
// which divides 'num'
for ( int i = 2 ; i <= Math.sqrt(n); i++)
{
// if 'i' is divisor of 'n'
if (n % i == 0 )
{
// if both divisors are same
// then add it once else add
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above loop
// considers proper divisors greater
return (result + n + 1 );
} // Function to check if there // exists a proper divisor // D' of N such that sigma(n)-2D' = 2N static boolean check( int num)
{ int sigmaN = divSum(num);
// Find all divisors which divides 'num'
for ( int i = 2 ; i <= Math.sqrt(num); i++)
{
// if 'i' is divisor of 'num'
if (num % i == 0 )
{
// if both divisors are same then add
// it only once else add both
if (i == (num / i))
{
if (sigmaN - 2 * i == 2 * num)
return true ;
}
else
{
if (sigmaN - 2 * i == 2 * num)
return true ;
if (sigmaN - 2 * (num / i) == 2 * num)
return true ;
}
}
}
// Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num)
return true ;
return false ;
} // Function to check if N // is an admirable number static boolean isAdmirableNum( int N)
{ return check(N);
} // Driver code public static void main(String[] args)
{ int n = 12 ;
if (isAdmirableNum(n))
System.out.println( "Yes" );
else
System.out.println( "No" );
} } // This code is contributed by shubham |
# Python3 implementation to check if # N is an admirable number import math
# Function to calculate the sum of # all divisors of a given number def divSum(n):
# Sum of divisors
result = 0
# Find all divisors
# which divides 'num'
for i in range ( 2 , int (math.sqrt(n)) + 1 ):
# If 'i' is divisor of 'n'
if (n % i = = 0 ):
# If both divisors are same
# then add it once else add
if (i = = (n / / i)):
result + = i
else :
result + = (i + n / / i)
# Add 1 and n to result as above loop
# considers proper divisors greater
return (result + n + 1 )
# Function to check if there # exists a proper divisor # D' of N such that sigma(n)-2D' = 2N def check(num):
sigmaN = divSum(num)
# Find all divisors which divides 'num'
for i in range ( 2 , int (math.sqrt(num)) + 1 ):
# If 'i' is divisor of 'num'
if (num % i = = 0 ):
# If both divisors are same then add
# it only once else add both
if (i = = (num / / i)):
if (sigmaN - 2 * i = = 2 * num):
return True
else :
if (sigmaN - 2 * i = = 2 * num):
return True
if (sigmaN - 2 * (num / / i) = = 2 * num):
return True
# Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 = = 2 * num):
return True
return False
# Function to check if N # is an admirable number def isAdmirableNum(N):
return check(N)
# Driver code n = 12
if (isAdmirableNum(n)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by divyeshrabadiya07 |
// C# implementation to check if N // is a admirable number using System;
class GFG{
// Function to calculate the sum of // all divisors of a given number static int divSum( int n)
{ // Sum of divisors
int result = 0;
// Find all divisors
// which divides 'num'
for ( int i = 2; i <= Math.Sqrt(n); i++)
{
// If 'i' is divisor of 'n'
if (n % i == 0)
{
// If both divisors are same
// then add it once else add
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above loop
// considers proper divisors greater
return (result + n + 1);
} // Function to check if there // exists a proper divisor // D' of N such that sigma(n)-2D' = 2N static bool check( int num)
{ int sigmaN = divSum(num);
// Find all divisors which divides 'num'
for ( int i = 2; i <= Math.Sqrt(num); i++)
{
// If 'i' is divisor of 'num'
if (num % i == 0)
{
// If both divisors are same then add
// it only once else add both
if (i == (num / i))
{
if (sigmaN - 2 * i == 2 * num)
return true ;
}
else
{
if (sigmaN - 2 * i == 2 * num)
return true ;
if (sigmaN - 2 * (num / i) == 2 * num)
return true ;
}
}
}
// Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num)
return true ;
return false ;
} // Function to check if N // is an admirable number static bool isAdmirableNum( int N)
{ return check(N);
} // Driver code public static void Main()
{ int n = 12;
if (isAdmirableNum(n))
Console.Write( "Yes" );
else
Console.Write( "No" );
} } // This code is contributed by Code_Mech |
<script> // Javascript implementation to check if N // is a admirable number // Function to calculate the sum of
// all divisors of a given number
function divSum( n)
{
// Sum of divisors
let result = 0;
// Find all divisors
// which divides 'num'
for ( let i = 2; i <= Math.sqrt(n); i++)
{
// if 'i' is divisor of 'n'
if (n % i == 0)
{
// if both divisors are same
// then add it once else add
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above loop
// considers proper divisors greater
return (result + n + 1);
}
// Function to check if there
// exists a proper divisor
// D' of N such that sigma(n)-2D' = 2N
function check( num) {
let sigmaN = divSum(num);
// Find all divisors which divides 'num'
for (let i = 2; i <= Math.sqrt(num); i++) {
// if 'i' is divisor of 'num'
if (num % i == 0) {
// if both divisors are same then add
// it only once else add both
if (i == (num / i)) {
if (sigmaN - 2 * i == 2 * num)
return true ;
} else {
if (sigmaN - 2 * i == 2 * num)
return true ;
if (sigmaN - 2 * (num / i) == 2 * num)
return true ;
}
}
}
// Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num)
return true ;
return false ;
}
// Function to check if N
// is an admirable number
function isAdmirableNum( N) {
return check(N);
}
// Driver code
let n = 12;
if (isAdmirableNum(n))
document.write( "Yes" );
else
document.write( "No" );
// This code is contributed by todaysgaurav </script> |
Output:
Yes
Time Complexity: O(N1/2)
References: OEIS