Codeforces Round #142 (Div. 2)B. T-primes
2 seconds
256 megabytes
standard input
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.
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.
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.
3
4 5 6
YES
NO
NO
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 << ;
const double eps=1e-;
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=;i*i<=x;i++){
if(x%i==)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=;i<=n;i++)cin>>a[i];
for(int i=;i<=n;i++){
if(a[i]==){
puts("NO");continue;
}
ll x=sqrt(a[i]);
if(x*x==a[i]&&is(x)){
puts("YES");
}
else puts("NO");
}
}
return ;
}
。
Codeforces Round #142 (Div. 2)B. T-primes的更多相关文章
- Codeforces Round #142 (Div. 1) C. Triangles
Codeforces Round #142 (Div. 1) C. Triangles 题目链接 今天校内选拔赛出了这个题,没做出来....自己思维能力还不够强吧.我题也给读错了.. 每次拆掉一条边, ...
- Codeforces Round #142 (Div. 2)
A. Dragons 按\(x\)排序. B. T-primes \(x\)是平方数,且根\(\sqrt{x}\)是个质数. C. Shifts 枚举列的位置,对于每行来说,最多只会涉及4个列. D. ...
- Codeforces Round #556 (Div. 2) - C. Prefix Sum Primes(思维)
Problem Codeforces Round #556 (Div. 2) - D. Three Religions Time Limit: 1000 mSec Problem Descripti ...
- Codeforces Round #556 (Div. 1)
Codeforces Round #556 (Div. 1) A. Prefix Sum Primes 给你一堆1,2,你可以任意排序,要求你输出的数列的前缀和中质数个数最大. 发现只有\(2\)是偶 ...
- Codeforces Round #366 (Div. 2) ABC
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate ...
- Codeforces Round #354 (Div. 2) ABCD
Codeforces Round #354 (Div. 2) Problems # Name A Nicholas and Permutation standard input/out ...
- Codeforces Round #368 (Div. 2)
直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输 ...
- cf之路,1,Codeforces Round #345 (Div. 2)
cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅..... ...
- Codeforces Round #279 (Div. 2) ABCDE
Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems # Name A Team Olympiad standard input/outpu ...
随机推荐
- nginx报错:Permission denied
vi /etc/selinux/config #SELINUX=enforcing #注释掉 #SELINUXTYPE=targeted #注释掉 SELINUX=disabled #增加 :wq! ...
- Vue2.0 —生命周期和钩子函数
vue生命周期简介 咱们从上图可以很明显的看出现在vue2.0都包括了哪些生命周期的函数了. 生命周期探究 对于执行顺序和什么时候执行,看上面两个图基本有个了解了.下面我们将结合代码去看看钩子函数的执 ...
- 判断机型是否为iphoneX
判断机型是否为iphoneX isIPhoneX() { var u = navigator.userAgent; var isIOS ...
- 安装ubuntu系统空间分配问题
以下是我安装linux系统(ubuntu)时的系统空间配置,以50G为例: 挂载点 大小 格式 分区类型 / 15G Ext4 主分区 /home 30G Ext4 逻辑分区 /boot 1G Ext ...
- buf.readUInt16BE()
buf.readUInt16BE(offset[, noAssert]) buf.readUInt16LE(offset[, noAssert]) offset {Number} 0 <= of ...
- vim基础(二)
上一篇提到了插入与删除.定位.复制与粘贴以及退出与保存几项基本操作,这篇继续整理其他常用命令. 撤销与替换 首先是我们在输入过程中不小心打错了,或者误删了,怎么恢复?在word里我们有ctrl+Z,v ...
- AtCoder Beginner Contest 131 Solution
前言 这次ABC还是有一点难度的吧. TaskA Security Solution 直接模拟就好了. Code /* mail: mleautomaton@foxmail.com author: M ...
- 负载均衡之Ocelot
Ocelot 负载均衡: 背景知识,ocelot是基于 webapi 的网关框架,要使用ocelot来做路由转发和负载均衡,需要创建一个webapi,然后以这个webapi来做gateway. ...
- IDEA建立一个可运行的struts2项目
参考博客:https://blog.csdn.net/shuai_wy/article/details/79027573 直接使用IDEA创建struts2项目,配置好tomcat后是跑不起来的 需要 ...
- HDU 1176 DP
题目大意: 在0~10这11个点上面接饼 , 每秒最多往左或往移动一格,或者保持原地不动 令dp[i][j]表示在第 i 秒在 第 j 个点上最多能得到的饼的数量 dp[i][j] = max(dp[ ...