SGU 294 He's Circles

He's Circles He wrote n letters "X" and "E" in a circle. He thought that there were 2n possibilities to do it, because each letter may be either "X" or "E". But Qc noticed that some different sequences of letters ca

对Polya定理的个人认识     我们先来看一道经典题目:     He's Circles(SGU 294)         有一个长度为N的环,上面写着“X”和“E”,问本质不同的环有多少个(不能旋转重复就称之为本质不同) 输入样例:4 输出样例:6 那么要怎么办呢?暴力显然暴不出来…… 我们可以考虑使用置换群. 我们有两种算法: ①Burnside引理: 答案直接为1/|G|*(D(a1)+D(a2)+D(a3)+……+D(as)) 其中D(ak)为在进行置换群置换操作ak下不变的元素的

原题链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1412 题目要求判断是否有一条直线可以穿过所有的圆. 做法:把所有圆心做一次凸包,然后判断这个凸包是否能通过一个宽度为2*R的通道. 做法和求凸包直径差不多,只是判断的时候把点到两个端点的距离换成点到直线的距离. #include <stdio.h> #include <string.h> #include <math.h> #include <stdli

There are n circles on a infinitely large table.With every two circle, either one contains another or isolates from the other.They are never crossed nor tangent.Alice and Bob are playing a game concerning these circles.They take turn to play,Alice go

Tickets Description Conductor is quite a boring profession, as all you have to do is just to sell tickets to the passengers. So no wonder that once upon a time in a faraway galaxy one conductor decided to diversify this occupation. Now this conductor

sgu 101 - Domino Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u Description Dominoes – game played with small, rectangular blocks of wood or other material, each identified by a number of dots, or pips, on its face. The bl

题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=116 题意好晦涩 给你一个不超过一万的数 问它最少可以用多少个“超级素数”来表示 使“超级素数”之和等于它 如果无法这样表示 输出0 否则 按非降序形式输出方案 数论部分就模板的问题 没什么说的 完全背包方面也很常规 说说[输出方案] 背包九讲的伪码给的是二维dp[]的方法 实际上稍加改动就可以用在一维数组上 用一个rec[]记录dp[]的当前状态是从哪个状态转移而来(即上一个状态) 通过

SGU 100 A+B :太神不会 SGU 101 Domino: 题目大意:有N张骨牌,两张骨牌有两面有0到6的数字,能相连当且仅当前后数字相同,问能否有将N张骨牌连接的方案?思路:裸的欧拉回路,注意自环,连通 1 //sgu101 2 #include<iostream> 3 #include<cstdio> 4 #include <math.h> 5 #include<algorithm> 6 #include<string.h> 7 #i

题目地址:SGU 275 首先,贪心的思想,每一二进制位上要尽量是1,而能不能是1用高斯消元来解决.当该位有一个可以使之为1的变元时,就说明这位可以为1,而且令该变元控制该位,然后向低位消元. 代码如下: #include <iostream> #include <string.h> #include <math.h> #include <queue> #include <algorithm> #include <stdlib.h>