zoj1873 Let it Bead
思路:polya裸题,如果是旋转,对于旋转i格的循环节长度len=lcm(i,n)/i,个数就是n/len=gcd(i,n);如果是翻转,奇数个点对称轴就是一个点一条边,那么循环节个数即n/2+1,
偶数个点有n/2条对称轴穿过两个点,循环节个数是n/2+1,n/2条对称轴穿过两条边,循环节个数就是n/2,然后直接套polya公式即可。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std; int n,m; int gcd(int a,int b){
return b==?a:gcd(b,a%b);
} long long power(int a,int k){
if (k==) return ;
if (k==) return a;
int x=power(a,k/),ans=x*x;
if (k&) ans*=a;
return ans;
} int main(){
while (scanf("%d%d",&m,&n)!=EOF){
if (n== && m==) break;
long long ans=;
for (int i=;i<=n;i++) ans+=power(m,gcd(n,i));
if (n&) ans+=n*power(m,n/+);else ans+=n/*power(m,n/)+n/*power(m,n/+);
printf("%lld\n",ans/(n*));
}
}
zoj1873 Let it Bead的更多相关文章
- 百练_2409 Let it Bead(Polya定理)
描述 "Let it Bead" company is located upstairs at 700 Cannery Row in Monterey, CA. As you ca ...
- poj 2049 Let it Bead(polya模板)
Description Cannery Row percent of the target audience insists that the bracelets be unique. (Just ...
- poj2409 Let it Bead
Let it Bead Time Limit: 1000MS M ...
- POJ1975 Median Weight Bead floyd传递闭包
Description There are N beads which of the same shape and size, but with different weights. N is an ...
- POJ 2409 Let it Bead(polya裸题)
题目传送:http://poj.org/problem?id=2409 Description "Let it Bead" company is located upstairs ...
- 【POJ2409】Let it Bead Pólya定理
[POJ2409]Let it Bead 题意:用$m$种颜色去染$n$个点的环,如果两个环在旋转或翻转后是相同的,则称这两个环是同构的.求不同构的环的个数. $n,m$很小就是了. 题解:在旋转$i ...
- POJ-1975 Median Weight Bead(Floyed)
Median Weight Bead Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 3162 Accepted: 1630 De ...
- 珍珠 Median Weight Bead 977
描述 There are N beads which of the same shape and size, but with different weights. N is an odd numbe ...
- Median Weight Bead(最短路—floyed传递闭包)
Description There are N beads which of the same shape and size, but with different weights. N is an ...
随机推荐
- DIV遮罩层传值
今天费了很大的劲儿才搞定!下面贴出代码和总结: 1.首先是前台代码: <%@ Page Title="" Language="C#" MasterPage ...
- puppet运维配置实列
aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAABQkAAAGBCAIAAABKMKAEAAAgAElEQVR4nOydeXwU9cH/t2o9WutR+7
- 学习Visitor Pattern 有感而发!override and overload
通过阅读各位前辈写的博文,像吕震宇,idior,李建忠WebCast等,对Visitor模式有一定的了解,有感而记录下来,以备忘. Visitor Pattern 假设了这样一个场景,在一个类型层次中 ...
- linux中删除目录
在linux中删除一个目录很简单,很多人还是习惯用rmdir,不过一旦目录非空,就陷入深深的苦恼之中,现在使用rm -rf命令即可解决. 直接rm就可以了,不过要加两个参数-rf 即:rm -rf 目 ...
- 最火的Android开源项目(二)
在<直接拿来用!最火的Android开源项目(一)>中,我们详细地介绍了GitHub上最受欢迎的TOP20 Android开源项目,引起了许多读者的热议,作为开发者,你最常用的是哪些开源项 ...
- AtomicLong
Spring package com.uniubi.management.controller; import java.util.concurrent.atomic.AtomicLong; impo ...
- OR 改写union数据变少
<pre name="code" class="sql">SQL> SELECT deptno FROM emp WHERE mgr = 76 ...
- Redis学习手册(事务)
一.概述: 和众多其它数据库一样,Redis作为NoSQL数据库也同样提供了事务机制.在Redis中,MULTI/EXEC/DISCARD/WATCH这四个命令是我们实现事务的基石.相信对有关系型数据 ...
- Java基础知识强化之IO流笔记53:IO流练习之 自定义类模拟BufferedReader的readLine()功能案例
1. 用Reader模拟BufferedReader的readLine()功能: readLine():一次读取一行,根据换行符判断是否结束,只返回内容,不返回换行符 2. 代码实现和思路分析: ...
- Android(java)学习笔记211:采用httpclient提交数据(qq登录案例)
1.Apache -Httpclient HttpClient 是 Apache Jakarta Common 下的子项目,可以用来提供高效的.最新的.功能丰富的支持 HTTP 协议的客户端编程工具包 ...