# -*- coding: cp936 -*-#第八章----数字0~9的奇妙变换#《c趣味编程》58-68题#21:48 2006-11-4from collections import dequefrom math import log10def permute(seq, index): seqc = seq[:] seqn = [seqc.pop()] divider = 2 while seqc: index, new_index = divmod(index,divider) seqn.insert(new_index, seqc.pop()) divider += 1 return seqndef z62(): #0~7八个数放到正方体的八个顶点上,每个面的四个数之和都相等 a=[0, 1, 2, 3, 4, 5, 6, 7] n=sum(a) if n%2==1: return else: n/=2 for i in range(0,40320): v1=permute(a,i) if (v1[0] + v1[1] + v1[2] + v1[3] == n and v1[0] + v1[1] + v1[4] + v1[5]== n and v1[0] + v1[2] + v1[4] + v1[6] == n and v1[0] + v1[4] + v1[4]+ v1[7] == n and v1[0] + v1[1] > n / 2 and v1[5] > v1[7]): print v1 returndef z58(): #构造拉丁方：拉丁方是指n*n方阵其中每行每列1到n这些数字,只出现一次 ss=6 d = deque(range(1,ss+1)) for i in range(ss): d.rotate(1) print list(d) returndef z59(): '''答案如 1 2 3 4 5 6 其中每一列右大于左 每一行 下大于上 填入1~6 6个数字 ''' t=range(2,6) for i in t: for j in t: for m in t: n=14-i-j-m if i!=j!=m!=i and j>i and n>m and n>i: print [1,i,j] print [m,n,6] returndef iss(n): s=''.join(map(str,n)) t=set(s) m=len(s) return len(t)==m and '0' not in tdef z60(): '''如192，384，576 384数第二数是192的二倍 576数第二数是192的三倍 同时各位数是1~9''' for i in range(111,333): t=[i,2*i,3*i] if iss(t):print tdef z61(): #如361，529，784,他们三数是完全平方,如361=19*19同时各位数是1~9 ge=[x*x for x in range(11,31)] for i in ge: for j in ge: for m in ge: t=[i,j,m] if i<j<m and iss(t):print tdef zz68(n): if n==2 :return filter(lambda x: x%2==0,range(11,99)) t=[] p=zz68(n-1) for j in p: for i in range(10): nn=j*10+i if iss([nn]) and nn%n==0: t.append(nn) return tdef z66(): '''除式还原 ~ x7x ~ xx /-------- ~ / xxxxx ~ x77 ~ ------ ~ x7x ~ x7x ~ ---- ~ xx ~ xx ~ --- ~ 0 ''' for i in range(1,10): for j in range(11,100): m2 = i * j m3 = 7 * j if ((m2 % 100 == 77 and m2 >100) and (m3 / 10 % 10 == 7 and m3 > 100)): for k in range(1,10): if k*j<100: print [i*100+70+k,j] returndef z67(): '''除式还原2 ~ x7xxx ~ xxx /----------- ~ / xxxxxxxx ~ xxxx ~ ------ ~ xxx ~ xxx ~ ------- ~ xxxx ~ xxx ~ --------- ~ xxxx ~ xxxx ~ --------- ~ 0 ''' g=lambda x,y:map(lambda a,b=x:int(log10(a*b))+1,y)==[4,3,3,4] t=[8,9] for i in range(100,142): for j in t: for m in t: for n in t: n1=j * 10000+7000+m * 100+n e=[j,7,m,n] if g(i,e) and (m * 100+n)*i/10000==10: print n1,i,n1*i returndef z68(): #个数由123456789组成,其中前两位如12 可以被2 整除,123可以被3 整除, #前n位可以被n 整除,问他是多少 print zz68(9)def z63(): '''减式还原 ~ PEAR - ARA ----------------- ~ PEA 其中不同的字母代表不同的数字''' t=range(0,10) for i in t: for j in t: for k in t: a,b,c= 100+k * 10+i,1000+k * 100+i * 10+j,i * 100+j * 10+i if a==b-c: print [a,b,c]def z64(): '''乘式还原 其中 a代表0~9前五个数字 z代表0~9后五个数字 ~ AZA * AAZ ----------------- ~ AAAA ~ AAZZ ~ ZAA ----------------- ~ ZAZAA''' g=lambda y:''.join(map(lambda x:x> '4' and '0' or '1' ,str(y))) t=range(1,5) m=range(5,10) t1=filter(lambda y:g(y)=='101',range(100,500)) for i in t1: for j in filter(lambda y,x=i:g(y*x)=='011',t): for k in filter(lambda y,x=i:g(y*x)=="1100",t): for mm in filter(lambda y,x=i:g(y*x)=="1111",m): jj =j *100+k *10+mm if g(jj*i)=='01011': print i,jj returndef z65(): ''' 乘式还原2 ~ PPP * PP ----------- ~ PPPP ~ PPPP ------------- ~ PPPPP 其中18个p的位置上全是素数2，3，5，7''' def g(n): t=str(n) return len(filter(lambda x: int(x) in (2,3,5,7),t))==len(t) for i in range(222,778): for j in range(11,78): if g(i) and g(j) and g(i*j) and g(j%10*i)and g(int(j/10)*i): print i,j,i*jif __name__ == '__main__': s="" for i in range(58,69): s+='z'+str(i)+'()\n' exec(s)
