CF 142B T-primes

B. T-primes time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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 thas 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 ≤ 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. Output

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. Sample test(s) input

3 4 5 6

output

YES NO NO

Note

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 <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
const int MAXN = 1000000+10;
long long int a[100000+10]; bool prime[MAXN];
bool judge(long long int n){
  if (n == 1)  return false;
  long long int x = (long long)sqrt(n);
//  if (x * x != n) return false;
 // else {
  //  if (!prime[x]) return true; 
   // else return false;
//  }
//  printf("x = %I64d ", x);
  if (x * x != n || prime[x]) return false; else return true;
}
int main(void){
#ifndef ONLINE_JUDGE
  freopen("tprime.in", "r", stdin);
#endif
  int n; memset(prime, false, sizeof(prime));
  for (int i = 2; i * i <= MAXN; ++i)
    if (prime[i] == false)
      for (int j = i*2; j <= MAXN; j+=i)
        prime[j] = true;
  while (~scanf("%d", &n)){
    for (int i = 0; i < n; ++i) cin >> a[i];
    for (int i = 0; i < n; ++i)
    {
      if(judge(a[i])) printf("YES\n"); 
      else
       printf("NO\n");
    }
  }
  return 0;
}

这题卡死了……

关键就是如何判断一个数字是不是完全平方数，方法就是先开方，转化为整型，然后再平方看是不是等于原来的数字。唉，以前做过这种的，很久不做水题了，这种东西都忘了，看来水题还是要做的……