Codefroces B. T-primes
http://codeforces.com/problemset/problem/230/B
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 n space-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 %I64d specifier.
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".
只有三个因子的数,分析后即可得到1.本身,以及一个素数完全平方根
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <cmath> #include <set> #include <map> #include <algorithm> using namespace std; typedef long long ll; ll a[1000006]; void get_prime() { memset(a,0,sizeof(a)); a[1]=1; for(int i=2;i<1000006;i++) { if(a[i]==0) { for(int j=2;j*i<1000006;j++) { a[j*i]=1; } } } } int main() { ll x,n; cin>>n; get_prime(); while(n--) { cin>>x; ll ans=sqrt(x); if(ans*ans==x && !a[ans])puts("YES"); else puts("NO"); } return 0; }