import time
import matplotlib.pyplot as plt
import numpy as np
import time
time_start=time.time()#计算时间开始

counter=0;
def draw_nklc(nk,n):
    plt.close()
    fig = plt.figure()
    ax1=fig.add_subplot(111)
    ang = np.arange(0,2*np.pi,2*np.pi/7)
    plc= [(12*np.cos(theta),16*np.sin(theta)) for theta in ang]
    for i in range(7):
        bll=plt.Circle(plc[i],3,color = nk[i])
        ax1.add_patch(bll)
    plt.xlim((-30,30))
    plt.ylim((-30,30))
    plt.savefig('三色珠'+str(n)+'.png')

bc_st=('red','green','blue')                #总共三种颜色
nk_lc=[(x1,x2,x3,x4,x5,x6,x7)\
       for x1 in bc_st\
       for x2 in bc_st\
       for x3 in bc_st\
       for x4 in bc_st\
       for x5 in bc_st\
       for x6 in bc_st\
       for x7 in bc_st]                     #列出所有可能的序列，共计2184种序列


ns = list()                                 #建立新的序列集合

for item in nk_lc:                          #将原序列集合的元素添加至新序列集合
    fg = 1
    a = [item[i:]+item[:i]\
         for i in range(len(item))]         #添加前检查元素是否已在新序列中
    for it in a:                            #检查时先将序列循环旋转一遍
        if it in ns:
            fg = 0
            break
    if fg == 1:
        c = tuple([x for x in\
                   reversed(item)])         #然后将元素倒序再次循环旋转一遍依次检查
        b = [c[i:]+c[:i]\
             for i in range(len(c))]
        for it in b:
            if tuple(it) in ns:
                fg = 0
                break
    if fg ==1:
        ns.append(item)
        counter += 1 # 计算个数.

        

#for i in range(len(ns)):
    #draw_nklc(ns[i],i)

print('-'*60)

print('珠宝商一共有%d种独一无二的彩珠手链出售。'%(counter))
print('-'*60)
time_end=time.time()#计算时间结束
print('python3.6程序运行',time_end-time_start,'秒。')
print('-'*60)