| 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.
-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
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 }