POJ 1286 Necklace of Beads Polya定理

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Necklace of Beads

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6253 Accepted: 2606

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

45-1

Sample Output

2139

Source

Xi'an 2002

跟这个题是一样的，只不过固定是三种颜色而已。点击打开链接

//380K0MS#include<stdio.h>#include<math.h>long long gcd(long long a,long long b){    if(!b)return a;    return gcd(b,a%b);}int main(){    long long s;    while(scanf("%I64d",&s)&&s!=-1)    {        if(s==0){printf("0\n");continue;}        long long sum=0;        for(long long i=1;i<=s;i++)        {            long long tmp=gcd(s,i);//第i次旋转的循环节数            sum+=(long long)(pow(3.0,tmp*1.0));        }        if(s&1)sum+=(long long)(s*pow(3.0,(s+1)/2.0));//s为奇数，共有s个循环节数均为(s+1)/2的置换        else//当s为偶数        {            sum+=(long long)((s/2)*pow(3.0,(s+2)/2.0));//第一种循环节数均为(s+2)/2            sum+=(long long)((s/2)*pow(3.0,s/2.0));//第二种循环节数均为s/2        }        sum/=(2*s);        printf("%I64d\n",sum);    }    return 0;}