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  • B. T-primes

    /*
    PROBLEMSSUBMITSTATUSSTANDINGSCUSTOM TEST
    B. T-primes
    time limit per test2 seconds
    memory limit per test256 megabytes
    inputstandard input
    outputstandard 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 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.
    
    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 <iostream>
    #include<cmath>
    #include<stdio.h>
    #include<stdlib.h>
    using namespace std;
    #define maxn 1003
    bool isprime( __int64 n)
    {
        for( __int64 i=2; i<=sqrt((float)n); i++)
            if(n%i==0)
                return 0;
        return 1;
    }
    int main()
    {
        __int64 num,n,tmp;
        scanf("%I64d",&num);
        while(num--)
    
        {
            scanf("%I64d",&n);
            if(n==1)
            {
                printf("NO
    ");
                continue;
            }
            tmp=sqrt((float)n);
            if(tmp*tmp==n)
            {
                if(isprime(tmp))
                    printf("YES
    ");
                else
                    printf("NO
    ");
            }
            else
            {
                printf("NO
    ");
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/heqinghui/p/3180681.html
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