codeforces-230B-T-primes

Codeforces B题解析

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本文解析了Codeforces平台上的B.T-primes问题，介绍了如何判断一个数是否为Т-prime，即该数是否恰好拥有三个不同的正除数，并提供了一个有效的解决方案。

题目链接：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.

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 nspace-separated integers x_i (1 ≤ x_i ≤ 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.

Output

Print n lines: the i-th line should contain "YES" (without the quotes), if number x_i 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".

解析：

对一个不等 1 的正整数 x 而言，1 和它本身是它的两个约数。

若它有且仅有三个不同约数，则另一个是根号x，设根号x为看，则k*k==x，且为质数。

参考代码如下：

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int ans[1000010];

int main()
{
    int n, i;
    long long x;
    for(i=0; i<1000010; i++);
    {    ans[i] = 0;   }
    ans[1] = 1;
    for(i=2; i<1000010; i++)
    {
        if(ans[i]) continue;
            for(int j=2*i; j<1000010; j+=i)
            {    ans[j] = 1;    }
    }
    while(scanf("%d", &n) != EOF)
    {
        while(n--)
        {
            scanf("%I64d", &x);
            long long tmp=sqrt(x);
            if(!ans[tmp] && tmp*tmp==x)
                printf("YES\n");
            else 
                printf("NO\n");         
        }                 
    }
    return 0;    
}