Superabundant Numbers
Last Updated :
22 Apr, 2021
A number is a Superabundant Number if sigma(n)/n > sigma(m)/m for all m < n, where sigma is the sum of the divisors of n.
1, 2, 4, 6, 12, 24, 36, 48…
Check if N is a Superabundant number
Given a number N, the task is to check if N is a Superabundant Number or not. If N is a Superabundant Number then print “Yes” else print “No”.
Examples:
Input: N = 4
Output: Yes
Explanation:
sigma(4)/4 = 7/4 = 1.75
sigma(1)/1 = 1/1 = 1
sigma(2)/2 = 3/2 = 1.5
sigma(3)/3 = 4/3 = 1.333333
sigma(n)/n > sigma(m)/m for all m < n
Input: N = 10
Output: No
Approach: For all numbers i from 1 till less than N check if sigma(n)/n > sigma(m)/m then return false otherwise return true at last.
For Example:
If N = 4, sigma(4)/4 = 7/4 = 1.75
for i = 1 to 3
sigma(1)/1 = 1/1 = 1
sigma(2)/2 = 3/2 = 1.5
sigma(3)/3 = 4/3 = 1.333333
sigma(n)/n > sigma(m)/m for all m < n,
Hence return true at last
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int sigma( int n)
{
if (n == 1)
return 1;
int result = 0;
for ( int i = 2; i <= sqrt (n); i++) {
if (n % i == 0) {
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
return (result + n + 1);
}
bool isSuperabundant( int N)
{
for ( float i = 1; i < N; i++) {
float x = sigma(i) / i;
float y = sigma(N) / (N * 1.0);
if (x > y)
return false ;
}
return true ;
}
int main()
{
int N = 4;
isSuperabundant(N) ? cout << "Yes"
: cout << "No" ;
return 0;
}
|
Java
class GFG{
static int sigma( int n)
{
if (n == 1 )
return 1 ;
int result = 0 ;
for ( int i = 2 ; i <= Math.sqrt(n); i++)
{
if (n % i == 0 )
{
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
return (result + n + 1 );
}
static boolean isSuperabundant( int N)
{
for ( double i = 1 ; i < N; i++)
{
double x = sigma(( int )(i)) / i;
double y = sigma(( int )(N)) / (N * 1.0 );
if (x > y)
return false ;
}
return true ;
}
public static void main(String[] args)
{
int N = 4 ;
if (isSuperabundant(N))
System.out.print( "Yes\n" );
else
System.out.print( "No\n" );
}
}
|
Python3
def sigma(n):
if (n = = 1 ):
return 1
result = 0
for i in range ( 2 , pow (n, 1 / / 2 )):
if (n % i = = 0 ):
if (i = = (n / i)):
result + = i
else :
result + = (i + n / i)
return (result + n + 1 )
def isSuperabundant(N):
for i in range ( 1 , N):
x = sigma(( int )(i)) / i
y = sigma(( int )(N)) / (N * 1.0 )
if (x > y):
return False
return True
if __name__ = = '__main__' :
N = 4
if (isSuperabundant(N) ! = True ):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG{
static int sigma( int n)
{
if (n == 1)
return 1;
int result = 0;
for ( int i = 2; i <= Math.Sqrt(n); i++)
{
if (n % i == 0)
{
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
return (result + n + 1);
}
static bool isSuperabundant( int N)
{
for ( double i = 1; i < N; i++)
{
double x = sigma(( int )(i)) / i;
double y = sigma(( int )(N)) / (N * 1.0);
if (x > y)
return false ;
}
return true ;
}
public static void Main(String[] args)
{
int N = 4;
if (isSuperabundant(N))
Console.Write( "Yes\n" );
else
Console.Write( "No\n" );
}
}
|
Javascript
<script>
function sigma(n)
{
if (n == 1)
return 1;
var result = 0;
for ( var i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
return (result + n + 1);
}
function isSuperabundant(N)
{
for ( var i = 1; i < N; i++) {
var x = sigma(i) / i;
var y = sigma(N) / (N * 1.0);
if (x > y)
return false ;
}
return true ;
}
var N = 4;
isSuperabundant(N) ? document.write( "Yes" )
: document.write( "No" );
</script>
|
Time Complexity: O(n)
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