如果由sum(n)表示的数字的所有适当除数的总和大于数字n的值，则将数字n称为丰富数。这两个值之间的差异称为丰度。
从数学上讲，如果满足以下条件，则将该数字称为“充裕的数字”：
sum(n)> n
abundance = sum(n) - n
sum(n): aliquot sum - The sum of all proper divisors of n给定数字n，我们的任务是查找该数字是否为充裕数字。
前几个丰富数字是：12、18、20、24、30、36、40、42、48、54、56、60、66…..
例子 ：
Input: 21
Output: NO
Input: 12
Output: YES
Input: 17
Output: NO方法1
一个简单的解决方案是迭代从1到n-1的所有数字，并检查数字是否除以n并计算总和。检查此总和是否大于n。
这种方法的时间复杂度：O(n)
优化的解决方案：
如果我们仔细观察，数字n的除数是成对出现的。例如，如果n = 100，则所有对数除数为：(1,100)，(2,50)，(4,25)，(5,20)，(10,10)
利用这一事实，我们可以加快程序速度。在检查除数时，我们必须注意是否存在两个相等的除数，例如(10，10)。在这种情况下，我们在计算总和时将只采用其中之一。
从所有除数的总和中减去数字n，即可得到适当除数的总和。
C++
// An Optimized Solution to check Abundant Number
#include
using namespace std;
// Function to calculate sum of divisors
int getSum(int n)
{
int sum = 0;
// Note that this loop runs till square root
// of n
for (int i=1; i<=sqrt(n); i++)
{
if (n%i==0)
{
// If divisors are equal,take only one
// of them
if (n/i == i)
sum = sum + i;
else // Otherwise take both
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
// calculate sum of all proper divisors only
sum = sum - n;
return sum;
}
// Function to check Abundant Number
bool checkAbundant(int n)
{
// Return true if sum of divisors is greater
// than n.
return (getSum(n) > n);
}
/* Driver program to test above function */
int main()
{
checkAbundant(12)? cout << "YES\n" : cout << "NO\n";
checkAbundant(15)? cout << "YES\n" : cout << "NO\n";
return 0;
} Java
// An Optimized Solution to check Abundant Number
// in JAVA
import java.io.*;
import java.math.*;
// Function to calculate sum of divisors
class GFG{
static int getSum(int n)
{
int sum = 0;
// Note that this loop runs till square
// root of n
for (int i=1; i<=(Math.sqrt(n)); i++)
{
if (n%i==0)
{
// If divisors are equal,take only
// one of them
if (n/i == i)
sum = sum + i;
else // Otherwise take both
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
// calculate sum of all proper divisors
// only
sum = sum - n;
return sum;
}
// Function to check Abundant Number
static boolean checkAbundant(int n)
{
// Return true if sum of divisors is
// greater than n.
return (getSum(n) > n);
}
/* Driver program to test above function */
public static void main(String args[])throws
IOException
{
if(checkAbundant(12))
System.out.println("YES");
else
System.out.println("NO");
if(checkAbundant(15))
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed by Nikita Tiwari.Python
# An Optimized Solution to check Abundant Number
# in PYTHON
import math
# Function to calculate sum of divisors
def getSum(n) :
sum = 0
# Note that this loop runs till square root
# of n
i = 1
while i <= (math.sqrt(n)) :
if n%i == 0 :
# If divisors are equal,take only one
# of them
if n/i == i :
sum = sum + i
else : # Otherwise take both
sum = sum + i
sum = sum + (n / i )
i = i + 1
# calculate sum of all proper divisors only
sum = sum - n
return sum
# Function to check Abundant Number
def checkAbundant(n) :
# Return true if sum of divisors is greater
# than n.
if (getSum(n) > n) :
return 1
else :
return 0
# Driver program to test above function */
if(checkAbundant(12) == 1) :
print "YES"
else :
print "NO"
if(checkAbundant(15) == 1) :
print "YES"
else :
print "NO"
# This code is contributed by Nikita Tiwari.C#
// An Optimized Solution to check Abundant Number
// in C#
// Function to calculate sum of divisors
using System;
class GFG {
// Function to calculate sum of divisors
static int getSum(int n)
{
int sum = 0;
// Note that this loop runs till square
// root of n
for (int i = 1; i <= (Math.Sqrt(n)); i++) {
if (n % i == 0) {
// If divisors are equal, take only
// one of them
if (n / i == i)
sum = sum + i;
else // Otherwise take both
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
// calculate sum of all proper divisors
// only
sum = sum - n;
return sum;
}
// Function to check Abundant Number
static bool checkAbundant(int n)
{
// Return true if sum of divisors is
// greater than n.
return (getSum(n) > n);
}
/* Driver program to test above function */
public static void Main()
{
if (checkAbundant(12))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
if (checkAbundant(15))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code is contributed by vt_m.PHP
$n);
}
// Driver Code
$k = checkAbundant(12) ? "YES\n" : "NO\n";
echo($k);
$k = checkAbundant(15) ? "YES\n" : "NO\n";
echo($k);
// This code is contributed by Ajit.
?>Javascript
输出 ：
YES
NO