POJ 1286 Necklace of Beads Polya定理

最新推荐文章于 2019-02-22 17:03:00 发布

ACBOY、

最新推荐文章于 2019-02-22 17:03:00 发布

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分类专栏：数学

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本文探讨了由红、蓝、绿三种颜色珠子组成的圆形项链，在忽略旋转和反射重复情况下的不同排列组合数量问题。给出了针对不同输入值计算可能的不同形式项链数量的算法实现。

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Necklace of Beads

Time Limit: 1000MS		Memory Limit: 10000K
Total 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

4
5
-1

Sample Output

21
39

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

//380K	0MS
#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;
}