Description Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < ). If the repetitions that are produced by rotation around the center of the circular necklace or reflection to the axis of symmetry are…

http://poj.org/problem?id=1286 题意:有红.绿.蓝三种颜色的n个珠子.要把它们构成一个项链,问有多少种不同的方法.旋转和翻转后同样的属于同一种方法. polya计数. 搜了一篇论文Pólya原理及其应用看了看polya究竟是什么东东.它主要计算所有互异的组合的个数.对置换群还是似懂略懂.用polya定理解决这个问题的关键是找出置换群的个数及哪些置换群,每种置换的循环节数.像这样的不同颜色的珠子构成项链的问题能够把N个珠子看成正N边形. Polya定理:(1)设G是p…

点我看题目 题意 :给你3个颜色的n个珠子,能组成多少不同形式的项链. 思路 :这个题分类就是polya定理,这个定理看起来真的是很麻烦啊T_T.......看了有个人写的不错: Polya定理: (1)设G是p个对象的一个置换群,用k种颜色突然这p个对象,若一种染色方案在群G的作用下变为另一种方案,则这 两个方案当作是同一种方案,这样的不同染色方案数为: : (2)置换及循环节数的计算方法:对于有n个位置的手镯,有n种旋转置换和n种翻转置换.对于旋转置换: c(fi) = gcd(n,i) …

Necklace of Beads 大意:3种颜色的珠子,n个串在一起,旋转变换跟反转变换假设同样就算是同一种,问会有多少种不同的组合. 思路:正规学Polya的第一道题,在楠神的带领下,理解的还算挺快的.代码没什么好说的,裸的Polya.也不须要优化. /************************************************************************* > File Name: POJ1286.cpp > Author: GLSilence &…

题目:http://poj.org/problem?id=1286 真·Polya定理模板题: 写完以后感觉理解更深刻了呢. 代码如下: #include<iostream> #include<cstdio> #include<cstring> using namespace std; typedef long long ll; int n; ll ans; ll pw(ll a,int b) { ll ret=; ,a*=a) )ret*=a; return ret;…

http://poj.org/problem?id=1286 题意:求用3种颜色给n个珠子涂色的方案数.polya定理模板题. #include <stdio.h> #include <math.h> long long gcd(long long a,long long b) { return b?gcd(b,a%b):a; } int main() { long long n; while(~scanf("%lld",&n)) { ) break;…

这是做的第一道群论题,自然要很水又很裸.注意用long long. 就是用到了两个定理 burnside :不等价方案数=每个置换的不动置换方案数的和 / 置换个数 polya: 一个置换的不动置换方案数=k^(这个置换的循环个数) 先看第一个博客再看第二个 http://cxjyxx.me/?p=198 http://endlesscount.blog.163.com/blog/static/82119787201221324524202/ 这两个蛮好的,上代码: #include <cstd…

Necklace of Beads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7874   Accepted: 3290 Description Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are pro…

链接:http://poj.org/problem?id=1286 http://poj.org/problem?id=2409 #include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long LL; LL P_M( LL a, LL b ) {…

Necklace of Beads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7763   Accepted: 3247 Description Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are pro…

两种置换 旋转:有n种,分别是旋转1个2个--n个,旋转i的循环节数位gcd(i,n) 翻转:分奇偶,对于奇数个,只有一个珠子对一条边的中点,循环节数为n/2+1:对于偶数个,有珠子对珠子和边对边,循环节个数为n/2+1个和n/2个 然后用polya定理即可 #include<iostream> #include<cstdio> using namespace std; long long n,k,ans; long long ksm(long long a,long long b…

参考:刘汝佳<算法竞赛入门经典训练指南> 感觉是非常远古的东西了,几乎从来没有看到过需要用这个的题,还是学一发以防翻车. 置换:排列的一一映射.置换乘法相当于函数复合.满足结合律,不满足交换律. 置换的循环分解:即将置换看成一张有向图,分解成若干循环.循环的数量称为循环节. 以置换集合来描述等价关系.如果存在一个置换将一个方案映射到另一个方案,则这两个方案等价.置换集合应当构成置换群. 不动点:方案s经过置换f不变,则s为f的不动点. Burnside引理:等价类数量=所有置换的不动点数量的平…

和poj 2409差不多,就是k变成3了,详见 还有不一样的地方是记得特判n==0的情况不然会RE #include<iostream> #include<cstdio> using namespace std; long long n,ans; long long ksm(long long a,long long b) { long long r=1; while(b) { if(b&1) r=r*a; a=a*a; b>>=1; } return r; }…

题目:http://poj.org/problem?id=2409 题意:用k种不同的颜色给长度为n的项链染色 网上大神的题解: 1.旋转置换:一个有n个旋转置换,依次为旋转0,1,2,```n-1.对每一个旋转置换,它循环分解之后得到的循环因子个数为gcd(n,i). 2.翻转置换:分奇偶讨论. 奇数的时候 翻转轴 = (顶点+对边终点的连线),一共有n个顶点,故有n个置换,且每个置换分解之后的因子个数为n/2+1; 偶数的时候 翻转轴 = (顶点+顶点的连线),一共有n个顶点,故有n/2个置…

Description Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are produced by rotation around the center of the circular necklace or reflection to the axis of symmetry ar…

[题目分析] 题目大意:一个环有n个点,共染三种颜色.问 在旋转和对称的情况下有多少种本质不同的方案数. Burnside直接做. [代码] #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define ll long long #define F(i,j,k) for (int i=j;i<=k;++…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9359   Accepted: 3862 Description Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are produced by rotation…

Necklace of Beads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7451   Accepted: 3102 Description Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are pro…

题意 用k种颜色对n个珠子构成的环上色,旋转翻转后相同的只算一种,求不等价的着色方案数. 思路 Polya定理 X是对象集合{1, 2, --, n}, 设G是X上的置换群,用M种颜色染N种对象,则不同的染色方案数为: λ(g)表示置换g的轮换个数,且λ(g) = λ1(g) + λn(g) + -- + λn(g),其中λi(g)表示g中长度为i的轮换(循环)个数. 本题是对一个n个珠子的圆珠的颜色,而圆珠的置换群有: Ⅰ翻转:1.当n为奇数时,有n种翻转,每种翻转的轴都是一个顶点和该顶点对边…

Necklace of Beads Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1049    Accepted Submission(s): 378 Problem Description Beads of red, blue or green colors are connected together into a circula…

Color Description Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the necklace might not use up all…

题目链接:http://vjudge.net/problem/viewProblem.action?id=11117 就是利用每种等价情形算出置换节之后算组合数 #include <stdio.h> #include <cstring> #include <cstdlib> #include <algorithm> #include <cmath> using namespace std; #define lson o<<1 #def…

题解 群论,我们只要找出所有的置换群的所有循环节 具体可参照算法艺术与信息学竞赛 旋转的置换有n个,每一个的循环节个数是gcd(N,i),i的范围是0到N - 1 翻转,对于奇数来说固定一个点,然后剩下的交换,循环节个数是(N - 1)/2 +1 对于偶数来说,不经过球的有N/2个,循环节个数是(N / 2) 经过球也也有N/2,循环节个数是(N / 2) + 1 代码 #include <iostream> #include <cstdio> #include <vecto…

题目描述 用 $c$ 种颜色去染 $r$ 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. $r·c\le 32$ . 题解 Polya定理 Burnside引理:一个置换群的等价类数目等于这个置换群中所有置换的不动点数目的平均值:Polya定理:设有限群G有 $m$ 个置换,第 $i$ 个置换有 $a_i$ 个循环,现在要将所有的点染成 $c$ 种颜色,那么染色后群G的等价类数目为:$L=\frac{c^{a_1}+c^{a_2}+…+c^{a_m}}m$…

Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are produced by rotation around the center of the circular necklace or reflection to the axis of symmetry are all neglec…

Necklace of Beads Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 630    Accepted Submission(s): 232 Problem Description Beads of red, blue or green colors are connected together into a circular…

Necklace of Beads Description Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are produced by rotation around the center of the circular necklace or reflection to the a…

Polya定理:设G={π1,π2,π3........πn}是X={a1,a2,a3.......an}上一个置换群,用m中颜色对X中的元素进行涂色,那么不同的涂色方案数为:1/|G|*(mC(π1)+mC(π2)+mC(π3)+...+mC(πk)). 其中C(πk)为置换πk的循环节的个数. Polya定理的基础应用. 你得算出旋转和翻转时,每种置换的循环节数. 旋转时,每种置换的循环节数为gcd(n,i): 翻转时,若n为奇数,共有n个循环节数为n+1>>1的置换, 若n为偶数,共有n…

点我看题目 题意 :给你c种颜色的n个珠子,问你可以组成多少种形式. 思路 :polya定理的应用,与1286差不多一样,代码一改就可以交....POJ 1286题解 #include <stdio.h> #include <iostream> #include <string.h> #include <math.h> #include <algorithm> using namespace std; int gcd(int a,int b) {…

Necklace of Beads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9162   Accepted: 3786 Description Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are pro…