• Codefroces B. T-primes


    http://codeforces.com/problemset/problem/230/B

    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 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.

    Examples
    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".

    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;
    } 
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  • 原文地址:https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7126149.html
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