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.
The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains nspace-separated integers xi (1 ≤ xi ≤ 1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64dspecifier.
Print n lines: the i-th line should contain "YES" (without the quotes), if number xi is Т-prime, and "NO" (without the quotes), if it isn't.
3 4 5 6
YES NO NO
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
#include <bits/stdc++.h> using namespace std; const int N=1000010; bool p[N],f[N]; int main(){ for(int i=2;i<N;i++) if(!f[i]) { p[i]=true; for(int j=2;j*i<N;j++) { f[i*j]=true; } } long long t; cin>>t; for(long long i=0;i<t;i++){ long long n; cin>>n; long long ans=sqrt(n); if(ans*ans==n&&p[ans]) puts("YES"); else puts("NO"); } return 0; }