• Codeforces Round #142 (Div. 2)B. 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 tТ-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 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 %I64dspecifier.

    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".

    猜测 题。一猜  他是平方数  发现16不行  二猜 sqrt(16)应该是素数才行,再看一看诗句范围 正好。

    /* ***********************************************
    Author        :guanjun
    Created Time  :2016/9/8 17:35:45
    File Name     :cf142b.cpp
    ************************************************ */
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <algorithm>
    #include <vector>
    #include <queue    >
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <iomanip>
    #include <list>
    #include <deque>
    #include <stack>
    #define ull unsigned long long
    #define ll long long
    #define mod 90001
    #define INF 0x3f3f3f3f
    #define maxn 100010
    #define cle(a) memset(a,0,sizeof(a))
    const ull inf = 1LL << 61;
    const double eps=1e-5;
    using namespace std;
    priority_queue<int,vector<int>,greater<int> >pq;
    struct Node{
        int x,y;
    };
    struct cmp{
        bool operator()(Node a,Node b){
            if(a.x==b.x) return a.y> b.y;
            return a.x>b.x;
        }
    };
    
    bool cmp(int a,int b){
        return a>b;
    }
    ll a[maxn];
    bool is(int x){
        for(int i=2;i*i<=x;i++){
            if(x%i==0)return false;
        }
        return true;
    }
    int main()
    {
        #ifndef ONLINE_JUDGE
        //freopen("in.txt","r",stdin);
        #endif
        //freopen("out.txt","w",stdout);
        int n;
        while(cin>>n){
            for(int i=1;i<=n;i++)cin>>a[i];
            for(int i=1;i<=n;i++){
                if(a[i]==1){
                    puts("NO");continue;
                }
                ll x=sqrt(a[i]);
                if(x*x==a[i]&&is(x)){
                    puts("YES");
                }
                else puts("NO");
            }
        }
        return 0;
    }

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  • 原文地址:https://www.cnblogs.com/pk28/p/5853932.html
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