Codeoforces 558 B. Duff in Love
//
2 seconds
256 megabytes
standard input
standard output
Duff is in love with lovely numbers! A positive integer x is called lovely if and only if there is no such positive integer a > 1 such that a2 is a divisor of x.

Malek has a number store! In his store, he has only divisors of positive integer n (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.
The first and only line of input contains one integer, n (1 ≤ n ≤ 1012).
Print the answer in one line.
10
10
12
6
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 22, so 12 is not lovely, while 6 is indeedlovely.
#include <stdio.h>
#define ll __int64
ll f(ll n)
{
ll i, t;
;i*i<=n;i++)
)
{
t = i;
break;
}
if(i*i>n)
return n;
else
return f(n/t);
}
int main()
{
ll n;
scanf("%I64d", &n);
printf("%I64d\n", f(n));
;
}
Codeoforces 558 B. Duff in Love的更多相关文章
- Codeforces Round #326 (Div. 2) B. Pasha and Phone C. Duff and Weight Lifting
B. Pasha and PhonePasha has recently bought a new phone jPager and started adding his friends' phone ...
- 【转】Duff's Device
在看strcpy.memcpy等的实现发现用了内存对齐,每一个word拷贝一次的办法大大提高了实现效率,参加该blog(http://totoxian.iteye.com/blog/1220273). ...
- UVA 558 判定负环,spfa模板题
1.UVA 558 Wormholes 2.总结:第一个spfa,好气的是用next[]数组判定Compilation error,改成nexte[]就过了..难道next还是特殊词吗 题意:科学家, ...
- Codeforces Round #326 (Div. 2)-Duff and Meat
题意: Duff每天要吃ai千克肉,这天肉的价格为pi(这天可以买好多好多肉),现在给你一个数值n为Duff吃肉的天数,求出用最少的钱满足Duff的条件. 思路: 只要判断相邻两天中,今天的总花费 = ...
- 【LCA】CodeForce #326 Div.2 E:Duff in the Army
C. Duff in the Army Recently Duff has been a soldier in the army. Malek is her commander. Their coun ...
- uva 558 - Wormholes(Bellman Ford判断负环)
题目链接:558 - Wormholes 题目大意:给出n和m,表示有n个点,然后给出m条边,然后判断给出的有向图中是否存在负环. 解题思路:利用Bellman Ford算法,若进行第n次松弛时,还能 ...
- Codeforces Round #326 (Div. 2) D. Duff in Beach dp
D. Duff in Beach Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/588/probl ...
- Codeforces Round #326 (Div. 2) C. Duff and Weight Lifting 水题
C. Duff and Weight Lifting Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest ...
- Codeforces Round #326 (Div. 2) B. Duff in Love 分解质因数
B. Duff in Love Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/588/proble ...
随机推荐
- struts一些实用常量配置_2015.01.04
- c#如实现将一个数字转化为其他进制字符串输出
之前在 os 上看到有人说过 一直想整理 但是一直没时间 后来 从csdn 上 知道了一份 下面内容 来自 (1)http://bbs.csdn.net/topics/60512816 tost ...
- ANT命令总结(转载)
1 Ant是什么? Apache Ant 是一个基于 Java的生成工具.生成工具在软件开发中用来将源代码和其他输入文件转换为可执行文件的形式(也有可能转换为可安装的产品映像形式).随着应用程序的生成 ...
- ipseccmd命令解析
IPSec 首先需要指出的是,IPSec和TCP/IP筛选是不同的东西,大家不要混淆了.TCP/IP筛选的功能十分有限,远不如IPSec灵活和强大.下面就说说如何在命令行下控制IPSec. XP系统用 ...
- co css规范
CSS 编码规范 1. 文件组织 (建议试试LESS) 1.1 CSS 与 HTML CSS 一律写在 CSS 文件中,原则上不写内联样式. CSS 文件命名由小写字母.下划线(_)组成. CSS 文 ...
- java运算符优先级记忆口诀
尊重原创:(口诀)转自http://lasombra.iteye.com/blog/991662 今天看到<java编程思想>中的运算符优先级助记口诀,不过"Ulcer Addi ...
- 【python cookbook】【字符串与文本】15.给字符串中的变量名做插值处理
问题:想创建一个字符串,其中嵌入的变量名称会以变量的字符串值形式替换掉 解决方法:str.format().str.format_map() >>> s = '{name} has ...
- WEB前端常用网站收集
WEB前端常用网站收集整理 w3school.w3schools 前端里.脚本之家.素材家园 17素材.frontopen NEC更好的CSS方案.一些常用的JS实例 Bootstrap 官网 h ...
- iOS抓包Charles 操作
今天就来看一下Mac上如何进行抓包,之前有一篇文章介绍了使用Fidder进行抓包 http://blog.csdn.net/jiangwei0910410003/article/details/198 ...
- Linux设置FQDN
FQDN是Fully Qualified Domain Name的缩写, 含义是完整的域名. 例如, 一台机器主机名(hostname)是www, 域后缀(domain)是example.com, 那 ...