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  • python 练习 6

    #!/usr/bin/python
    # -*- coding: utf-8 -*-
    from collections import deque
    from math import log10
    def 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 seqn
    def 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
        return
    def 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)
        return
    def 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]
        return
    def iss(n):
        s=''.join(map(str,n))
        t=set(s)
        m=len(s)
        return len(t)==m and '0' not in t
    def 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 t
    def 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 t
    def 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 t
    def 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]
        return
    def 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
        return
    def 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
                        
        return
    def 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*j
    if __name__ == '__main__':
        s=""
        for i in range(58,69):
            s+='z'+str(i)+'()
    '
        exec(s)
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  • 原文地址:https://www.cnblogs.com/kuihua/p/5521994.html
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