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).
思路：题就是判断一个数是否仅含有3个因子，其实也就是除了1和它自身外，还存在另一个因子。如果找到这个另外的因子就输出yes，否则为no。数据范围1e15，不小了 ，
一个数n是T素数，一定是1，n,sqrt(n)，且sqrt(n)为素数
也就是如果一个数是素数的平方，那么这个数有且仅有三个因子。
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,i,j;
long long a,b;
cin>>n;
for(i=1;i<=n;i++)
{
cin>>a;
b = sqrt(a);
for(j=2; j*j <= b; j++) {
if(b%j == 0){
break; //判断根号后的a即b是不是素数
}
}
if(j*j > b && b*b == a && a>1){
cout<<"YES"<<endl;
}
else{
cout<<"NO"<<endl;
}
}
return 0;
}