Codeforces Round #599 (Div. 2) C. Tile Painting
Ujan has been lazy lately, but now has decided to bring his yard to good shape. First, he decided to paint the path from his house to the gate.
The path consists of nn consecutive tiles, numbered from 11 to nn. Ujan will paint each tile in some color. He will consider the path aesthetic if for any two different tiles with numbers ii and jj, such that |j−i||j−i| is a divisor of nn greater than 11, they have the same color. Formally, the colors of two tiles with numbers ii and jj should be the same if |i−j|>1|i−j|>1 and nmod|i−j|=0nmod|i−j|=0 (where xmodyxmody is the remainder when dividing xx by yy).
Ujan wants to brighten up space. What is the maximum number of different colors that Ujan can use, so that the path is aesthetic?
The first line of input contains a single integer nn (1≤n≤10121≤n≤1012), the length of the path.
Output a single integer, the maximum possible number of colors that the path can be painted in.
4
2
5
5
In the first sample, two colors is the maximum number. Tiles 11 and 33 should have the same color since 4mod|3−1|=04mod|3−1|=0. Also, tiles 22and 44 should have the same color since 4mod|4−2|=04mod|4−2|=0.
In the second sample, all five colors can be used.
#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;cin>>n;
int flag = ;
long long num = n;
for(long long i=;i*i<=n;i++){
if(n%i==){
num=__gcd(num,i);
num=__gcd(num,n/i);
}
}
cout<<num<<endl;
}
//求除1以外所以因子的最大公约数
/*我们枚举n的所有的因子 a[1],a[2],a[3]....a[x]。
翻译过来就是我们每a[1]个,都得相同;每a[2]个都得相同;....;每a[x]个都得相同。
那么实际上这个东西的循环节就等于他们的最小公倍数。
那么最多个颜色就是n/lcm,实际上就是gcd。因为gcd x lcm = n */
Codeforces Round #599 (Div. 2) C. Tile Painting的更多相关文章
- Codeforces Round #599 (Div. 1) A. Tile Painting 数论
C. Tile Painting Ujan has been lazy lately, but now has decided to bring his yard to good shape. Fir ...
- Codeforces Round #599 (Div. 2) D. 0-1 MST(bfs+set)
Codeforces Round #599 (Div. 2) D. 0-1 MST Description Ujan has a lot of useless stuff in his drawers ...
- Codeforces Round #599 (Div. 2)
久违的写篇博客吧 A. Maximum Square 题目链接:https://codeforces.com/contest/1243/problem/A 题意: 给定n个栅栏,对这n个栅栏进行任意排 ...
- Codeforces Round #599 (Div. 2) Tile Painting
题意:就是给你一个n,然后如果 n mod | i - j | == 0 并且 | i - j |>1 的话,那么i 和 j 就是同一种颜色,问你最大有多少种颜色? 思路: 比赛的时候,看到 ...
- Codeforces Round #353 (Div. 2) B. Restoring Painting 水题
B. Restoring Painting 题目连接: http://www.codeforces.com/contest/675/problem/B Description Vasya works ...
- Codeforces Round #461 (Div. 2) C. Cave Painting
C. Cave Painting time limit per test 1 second memory limit per test 256 megabytes Problem Descriptio ...
- Codeforces Round #599 (Div. 2)D 边很多的只有0和1的MST
题:https://codeforces.com/contest/1243/problem/D 分析:找全部可以用边权为0的点连起来的全部块 然后这些块之间相连肯定得通过边权为1的边进行连接 所以答案 ...
- Codeforces Round #599 (Div. 2) E. Sum Balance
这题写起来真的有点麻烦,按照官方题解的写法 先建图,然后求强连通分量,然后判断掉不符合条件的换 最后做dp转移即可 虽然看起来复杂度很高,但是n只有15,所以问题不大 #include <ios ...
- Codeforces Round #599 (Div. 1) C. Sum Balance 图论 dp
C. Sum Balance Ujan has a lot of numbers in his boxes. He likes order and balance, so he decided to ...
随机推荐
- Centos下安装Oracle12c
总结一次安装oracle的折腾血泪史环境准备 centos7 虚拟机VMware Workstation Pro14 IP:192.168.245.128(根据实际情况) 4G物理内存,8G虚拟内存, ...
- centos8 ftp
安装 yum install -y vsftpd 启动 systemctl start vsftpd.service 开机启动 systemctl enable vsftpd.service 查看状态 ...
- springBoot 发送邮件图片不显示
解决方案 MimeMessageHelper 的执行顺序错了,先执行 setText() 然后执行 addInline() 添加图片 <img src="cid:p03"/& ...
- windows 服务启动外部程序
服务使用Process启动外部程序没窗体 在WinXP和Win2003环境中,安装服务后,右键单击服务“属性”-“登录”选项卡-选择“本地系统帐户”并勾选“允许服务与桌面交互”即可. 在Win7及以后 ...
- Linux C/C++ 字符串逆序
/*字符串逆序*/ #include <stdio.h> #include <string.h> void nixu(char *str) { ; char tmp; for( ...
- StackExchange.Redis 之 SortedSet 类型示例
1,增加操作 RedisCacheHelper.Instance.ZSortadd(); RedisCacheHelper.Instance.ZSortadd(); RedisCacheHelper. ...
- 2018护网杯easy_tornado(SSTI tornado render模板注入)
考点:SSTI注入 原理: tornado render是python中的一个渲染函数,也就是一种模板,通过调用的参数不同,生成不同的网页,如果用户对render内容可控,不仅可以注入XSS代码,而且 ...
- 关于XXE
NJUPT CTF2019: 做题的时候,抓包看了一下,响应XML格式消息,并没有严格过滤,这道题读文件, <!DOCTYPE foo [ <!ENTITY xxe SYSTEM &quo ...
- html data-xx 及 data()注意事项
1.data-xx命名:xx可以包含“-”和“_”,但是不能有大写字母: 2.用$ele.data()获取值的时候,不需要“data-”前缀:$ele.data('xx'); 3.$ele.data( ...
- JDBC用户访问被拒绝
线程“主”java中的异常.于sq1.sQLException:用户“root”@“localhost”被拒绝访问(使用密码:YES)root密码错误