LintCode 728: Three Distinct Factors

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本文介绍了一种高效算法，用于检查给定的正整数（1<=n<=10^18）是否恰好有三个不同的因子。通过判断该数是否可以表示为1、其平方根和平方根的平方，且平方根为质数，来确定其因子数量。

Three Distinct Factors

Given a positive integer n (1 <= n <= 10^18). Check whether a number has exactly three distinct factors, return true if it has exactly three distinct factors, otherwise false.

Example
Example1

Input: n = 9
Output: true
Explanation:
Number 9 has exactly three factors: 1, 3, 9, so return true.
Example2

Input: n = 10
Output: false
Explanation:
Number 10 has four factors: 1, 2, 5, 10, so return false.

解法1：
注意满足3个因子的数一定是1 X sqrt X sqrt=N。其中sqrt必须是质数。

class Solution {
public:
    /**
     * @param n: the given number
     * @return:  return true if it has exactly three distinct factors, otherwise false
     */
    bool isThreeDisctFactors(long long n) {
        long long sqrtN = sqrt(n);
        if (sqrtN * sqrtN != n) return false;
        return isPrime(sqrtN);   
    }

private:
    bool isPrime(long long num) {
        long long sqrtNum = sqrt(num);
        for (int i = 2; i <= sqrtNum; ++i) {
            if (num % i == 0) return false;
        }
        return true;
    }
};