T-Prime（数学规律）

Description
We know that prime numbers are positive integers that have exactly two distinct positive divisors.

Similarly, we’ll call a positive integer t Т-prime, if t has exactly three distinct positive divisors。

You are given an array of n positive integers. For each of them determine whether it is Т-prime or not.

Input
The first line contains a single positive integer, n (1 ≤ n ≤ 10000),showing how many numbers are in the array.

The next line contains n space-separated integers xi (1 ≤ xi ≤ 10^12).

Output
Print one line: The number of T-primes number

Sample Input
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3
4 5 6
Sample Output
1
Hint
Please use long long instead of int for any Xi.

The given test has three numbers.

The first number 4 has exactly three divisors — 1, 2 and 4.

The second number 5 has two divisors (1 and 5),

The third number 6 has four divisors (1, 2, 3, 6),

hence the answer for them is 1 (only the number 4 is T-primes number).

代码实现

#include<iostream>#include<cmath>using namespace std;long long data[10000];bool isTPrime(long long n){    long long temp = sqrt(n);    if(temp * temp == n)    {        if(temp < 2)            return false;        else        {            for(long long i = 2; i <= sqrt(temp); i++){                if(temp % i == 0){                    return false;                }            }            return true;        }    }    else        return false;}int main(){    int n;    while(cin >> n){        int result = 0;        for(long long i = 0; i < n; i++){            cin >> data[i];        }         for(long long i = 0; i < n; i++){            if(isTPrime(data[i])){                result ++;            }        }        cout << result << endl;    }    return 0;}

一个数一定可以被 1 和 它本身整除。所以要考虑的是是否存在另外一个数可以整除n。
假如存在这么一个数a，
那么显然（n/a）也可以整除n。
那么也就是说 a = (n / a);
a * a = n;
所以假如 n 是一个T-Prime， 那么 n 必然是完全平方数，并且其平方根必须是素数。
由此，我们可以得到一个判定T-Prime的方法。