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
- windows namedPipe 命名管道clent and server
1.client: #include "iostream" #include "windows.h" using namespace std; void mai ...
- angular 自定义指令 link or controller
Before compilation? – Controller After compilation? – Link var app = angular.module('plunker', []); ...
- 四则运算GUI版本
要求:用户界面新增支持 Windows GUI,同时保留原有命令行下所有功能.提示: 先测试驱动开发,然后重构代码,以GUI为目标修改"核心"函数,把与GUI/Console相关的 ...
- Dreamweaver 时间轴如何打开
因为下午要辅导几个调皮捣蛋的小孩HTML, 什么能让他们感觉又简单又好玩而起又能提高他们的兴趣呢?DW中设置浮动广告吧! 想好了,动手去做.DwearmWeaver里面居然没了时间轴, 度 ...
- 安装交叉编译器arm-linux-gcc
需要交叉编译环境故安装交叉编译环境 1.在宿主机的/usr/local/arm目录存放交叉编译器 mkdir /usr/local/arm 2.解压交叉编译器包至/usr/l ...
- Headless MSBuild Support for SSDT (*.sqlproj) Projects
http://sqlproj.com/index.php/2012/03/headless-msbuild-support-for-ssdt-sqlproj-projects/ Update: bre ...
- PL/SQL编程基础
范例:编写不做任何工作的PL/SQL块 BEGIN NULL ; END ; / 范例:编写一个简单的PL/SQL程序 DECLARE v_num NUMBER ; -- 定义一个变量v_num ...
- Junit单步调试
单步调试:主要查看变量内容的变化 1.设置断点位置,设置在可能出现问题的代码 2.点击项目右键以Debug as方式运行程序 3.F5 --> step into 进入方法内部进行调试 ...
- 使用 TFDConnection 的 pooled 连接池
从开始看到这个属性,就一直认为他可以提供一个连接池管理功能, 苦于文档资料太少, 甚至在帮助中对该属性的使用都没有任何介绍,如果你搜索百度,也会发现基本没资料. 最后终于在其官方网站看到了其完整相关的 ...
- remote desktop connect btw Mac, Windows, Linux(Ubuntu) Mac,Windows,Linux之间的远程桌面连接
目录 I. 预备 II. Mac连接Windows III. Windows连接Mac IV. Windows连接Ubuntu V. Mac连接Ubuntu VI. Ubuntu连接Mac VII, ...