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题目地址：

We define the Perfect Number is a positive integer that is equal to the sum of all its positive divisors except itself.

Now, given an integer n, write a function that returns true when it is a perfect number and false when it is not.

Example:

Input: 28

Output: True

Explanation: 28 = 1 + 2 + 4 + 7 + 14

Note: The input number n will not exceed 100,000,000. (1e8)

找出所有的质数，然后累加，累加结果等于原来的数字返回true，否则返回false。

计算所有的质数不必从1到num遍历一次，只遍历到$\sqrt{num}$即可。

public class PerfectNumber {
       public static boolean checkPerfectNumber(int num) {        if (num <= 1)            return false;        int sum = 1;        int s = (int)Math.sqrt(num);        //以免加两次平方根，例如Math.sqrt(9) = 3，但是3只应该出现一次。        boolean bol = (num / s == 0) && (s == (num / s));        for (int i = 2; i <= s; i++) {            if (num % i == 0) {                if (i == s && bol)                    sum = sum + i;                else                    sum = sum + i + num / i;            }        }        if (sum == num)            return true;        return false;    }    public static void main(String[] args) {        for (int i = 1; i < 10000; i++) {            if (checkPerfectNumber(i)) {                System.out.println(i);            }        }    }}

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