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
- OpenTSDB监控
OpenTSDB监控
- cstring 转string
(1)CString转换为string CString cs(_T("cs")); string s; s = (LPCSTR)(CStringA)(cs); (2)string转 ...
- ThinkPHP---thinkphp模型(M)
(1)配置数据库连接 数据库的连接配置可以在系统配置文件ThinkPHP/Conf/convention.php中找到 /* 数据库设置 */ 'DB_TYPE' => '', // 数据库类型 ...
- 08Oracle Database 完整性约束
Oracle Database 完整性约束 非空约束 创建表时 Create table table_name( Column_name datatype NOT NULL,… ); 修改表时 Alt ...
- c++通过CMake实现debug开关
刚学cmake,很多东西还不是很懂,不过今天刚刚实现了通过CMake控制debug的开关,兴奋之余记录一下. 背景介绍: 最近参与到了一个大的C++项目,很多代码已经非常成熟,我来添加一些辅助功能,但 ...
- ruby学习之路(一)
学习ruby最好的方法就是下载源码包,里面带有sample和test,是入门学习的最好实例. 我下载的是2.1.0版本,首先./configure,然后make,sudo make install.从 ...
- Luogu P1297 [国家集训队]单选错位
P1297 [国家集训队]单选错位 题目背景 原 <网线切割>请前往P1577 题目描述 gx和lc去参加noip初赛,其中有一种题型叫单项选择题,顾名思义,只有一个选项是正确答案.试卷上 ...
- Jmeter使用基础笔记-写一个http请求
前言 本篇文章主要讲述2个部分: 搭建一个简单的测试环境 用Jmeter发送一个简单的http请求 搭建测试环境 编写flask代码(我参考了开源项目HttpRunner的测试服务器),将如下的代码保 ...
- SQL学习笔记:库和表的创建
目录 创建和删除数据库 创建和删除表 添加.修改和删除字段 创建和删除数据库 CREATE DATABASE justForLearn; DROP DATABASE justForLearn; 创建和 ...
- 08.C语言:特殊函数
C语言:特殊函数 1.递归函数: 与普通函数比较,执行过程不同,该函数内部调用它自己,它的执行必须要经过两个阶段:递推阶段,回归阶段: 当不满足回归条件,不再递推: #include <stdi ...