codeforces-230B T-primes

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codeforces-230B T-primes

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                    time limit per test2 seconds   memory limit per test256 megabytes

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 ≤ 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.

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”.

题目大意：找出一个数字，存在3个约数包括1和他本身。

题目思路：这个数只能是另一个数的平方，并且不存在其他约数。

题目链接：230B T-primes

以下是代码：

#include <iostream>#include <cmath>using namespace std;int a[1000100];void isans(){    for (int i = 1; i <= 1000000; i++)  a[i] = 1;    a[1] = 0;    for (int i = 2; i <= 1000000; i++)    {         if ((long long)i * i <= 1000000 && a[i])        {            for (int j = i * i; j <= 1000000; j += i)            {                    a[j] = 0;            }               }    } }int main(){    int n;    cin >> n;    isans();    while(n--)    {        long long x;        cin >> x;        long long num = sqrt(x);        if (num * num == x && a[num])            cout << "YES\n";        else            cout << "NO\n";    }     return 0;}

附上超时的代码：

#include <iostream>#include <cmath>using namespace std;int main(){    int n;    cin >> n;    while(n--)    {        long long x;        cin >> x;        long long  num = sqrt(x);        int i;        for (i = 2;i * i<= num; i++)        {            if (num % i == 0)            {                break;            }        }        if (num * num == x && i * i > num && x > 1)            cout << "YES\n";        else            cout << "NO\n";    }     return 0;}