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

Copy sample input to clipboard

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。