完全数
如果一个数等于它的适当除数之和，也就是它的正除数之和而不包括数本身，那么这个数就是一个完全数。编写一个函数来检查给定的数字是否完美。 例:
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.
一个简单的解决方法就是把从 1 到 n-1 的每一个数都过一遍，检查它是不是除数。保持所有因子的和。如果和等于 n，则返回真，否则返回假。 一个有效的解决方案是通过数直到 n 的平方根。如果一个数“I”除以 n，然后将“I”和 n/i 相加。 下面是高效解决方案的实现。
C++
// C++ program to check if a given number is perfect
// or not
#include<iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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
$sum = 1;
// Find all divisors and add them
for ($i = 2; $i * $i <= $n; $i++)
{
if ($n % $i == 0)
{
if($i * $i != $n)
$sum = $sum + $i + (int)($n / $i);
else
$sum = $sum + $i;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if ($sum == $n && $n != 1)
return true;
return false;
}
// Driver Code
echo "Below are all perfect numbers till 10000\n";
for ($n = 2; $n < 10000; $n++)
if (isPerfect($n))
echo "$n is a perfect number\n";
// This code is contributed by mits
?>
java 描述语言
<script>
// 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
</script>
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
时间复杂度: √n 下面是一些关于完全数的有趣事实: 1)每个偶数完全数都是 2T5【pT8】的形式？1 (2 p ？1)2p在哪里？1 是质数。 2)是否有奇数完全数未知。 参考文献: https://en.wikipedia.org/wiki/Perfect_number 如发现有不正确的地方，请写评论，或者想分享以上讨论话题的更多信息
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