POJ1286 Necklace of Beads

Posted 2020-09-09 SilverNebula

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8263		Accepted: 3452

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 are all neglected, how many different forms of the necklace are there? 

Input

The input has several lines, and each line contains the input data n. 
-1 denotes the end of the input file. 

Output

The output should contain the output data: Number of different forms, in each line correspondent to the input data.

Sample Input

4
5
-1

Sample Output

21
39

Source

Xi‘an 2002

 

数学问题 统计 polya原理

和POJ2409一样的套路

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #include<vector>
 8 #define LL long long
 9 using namespace std;
10 int read(){
11     int x=0,f=1;char ch=getchar();
12     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
13     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
14     return x*f;
15 }
16 LL phi(int x){
17     int m=sqrt(x+0.5);
18     LL res=x;
19     for(int i=2;i<=m;i++)
20         if(x%i==0){
21             res=res/i*(i-1);
22             while(x%i==0)x/=i;
23         }
24     if(x>1)res=res/x*(x-1);
25     return res;
26 }
27 int n;
28 int gcd(int a,int b){
29     return (!b)?a:gcd(b,a%b);
30 }
31 LL ksm(LL c,LL k){
32     LL res=1;
33     while(k){
34         if(k&1)res=res*c;
35         c*=c;
36         k>>=1;
37     }
38     return res;
39 }
40 int main(){
41     int i,j;
42     while(1){
43         n=read();
44         if(n==-1)break;
45         if(!n){
46             cout<<0<<endl;
47             continue;
48         }
49         LL ans=0;
50         for(i=1;i<=n;i++){
51             if(n%i==0)ans+=ksm(3,i)*phi(n/i);
52         }
53         if(!(n&1)){
54             ans+=ksm(3,n/2)*n/2;
55             ans+=ksm(3,n/2+1)*n/2;
56         }
57         else ans+=ksm(3,(n+1)/2)*n;
58         ans/=2*n;
59         cout<<ans<<endl;
60     }
61     return 0;
62 }