完美数

如果一个数字等于其适当除数的总和，即等于其正除数的总和(不包括数字本身)，则该数字为完美数。编写一个函数来检查给定的数字是否完美。
例子：

Input: n = 15
Output: false
Divisors of 15 are 1, 3 and 5. Sum of 
divisors is 9 which is not equal to 15.

Input: n = 6
Output: true
Divisors of 6 are 1, 2 and 3. Sum of 
divisors is 6.

一个简单的解决方案是遍历从1到n-1的每个数字，并检查它是否为除数。保持所有除数之和。如果sum等于n，则返回true，否则返回false。
一个有效的解决方案是遍历数字直到n的平方根。如果数字“ i”除以n，则将“ i”和n / i相加。
以下是有效解决方案的实施。

C++

// C++ program to check if a given number is perfect
// or not
#include
using namespace std;
 
// Returns true if n is perfect
bool isPerfect(long long int n)
{
    // To store sum of divisors
    long long int sum = 1;
  
    // Find all divisors and add them
    for (long long int i=2; i*i<=n; i++)
    {
        if (n%i==0)
        {
            if(i*i!=n)
                sum = sum + i + n/i;
            else
                sum=sum+i;
        }
    }
     // If sum of divisors is equal to
     // n, then n is a perfect number
     if (sum == n && n != 1)
          return true;
  
     return false;
}
  
// Driver program
int main()
{
    cout << "Below are all perfect numbers till 10000\n";
    for (int n =2; n<10000; n++)
        if (isPerfect(n))
            cout << n << " is a perfect number\n";
  
    return 0;
}

Java

// Java program to check if a given
// number is perfect or not
class GFG
{
     
// Returns true if n is perfect
static boolean isPerfect(int n)
{
    // To store sum of divisors
    int sum = 1;
 
    // Find all divisors and add them
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i==0)
        {
            if(i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
    // If sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;
 
    return false;
}
 
// Driver program
public static void main (String[] args)
{
    System.out.println("Below are all perfect" +
                                "numbers till 10000");
    for (int n = 2; n < 10000; n++)
        if (isPerfect(n))
            System.out.println( n +
                    " is a perfect number");
}
}
 
// This code is contributed by mits

Python3

# Python3 code to check if a given
# number is perfect or not
 
# Returns true if n is perfect
def isPerfect( n ):
     
    # To store sum of divisors
    sum = 1
     
    # Find all divisors and add them
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
     
    # If sum of divisors is equal to
    # n, then n is a perfect number
     
    return (True if sum == n and n!=1 else False)
 
# Driver program
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")
         
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to check if a given
// number is perfect or not
 
class GFG
{
     
// Returns true if n is perfect
static bool isPerfect(int n)
{
    // To store sum of divisors
    int sum = 1;
 
    // Find all divisors and add them
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i==0)
        {
            if(i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
    // If sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;
 
    return false;
}
 
// Driver program
static void Main()
{
    System.Console.WriteLine("Below are all perfect" +
                                "numbers till 10000");
    for (int n = 2; n < 10000; n++)
        if (isPerfect(n))
            System.Console.WriteLine( n +
                    " is a perfect number");
}
}
 
// This code is contributed by chandan_jnu

Javascript

 
// Javascript program to check if a given number is perfect
// or not
 
// Returns true if n is perfect
function isPerfect(n)
{
    // To store sum of divisors
    sum = 1;
 
    // Find all divisors and add them
    for (let i=2; i*i<=n; i++)
    {
        if (n%i==0)
        {
            if(i*i!=n)
                sum = sum + i + n/i;
            else
                sum=sum+i;
        }
    }
    // If sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;
 
    return false;
}
 
// Driver program
  
    document.write("Below are all perfect numbers till 10000" + "<br>");
    for (let n =2; n<10000; n++)
        if (isPerfect(n))
            document.write(n + " is a perfect number" + "<br>");
 
// This code is contributed by Mayank Tyagi
 

输出：

Below are all perfect numbers till 10000
6 is a perfect number
28 is a perfect number
496 is a perfect number
8128 is a perfect number