python练习题目(二)

题目一

ABCD乘9=DCBA，A=? B=? C=? D=?

程序分析

• A和D肯定不为0，B和C取值范围为0-9
• 以A开头的四位数乘9得到四位数，由于判断A肯定为1
• A为1，那么乘以9得到的四位数，D肯定为9

  代码实现

  for A in range(1, 2):
      for B in range(0, 10):
          for C in range(0, 10):
              for D in range(9, 10):
                  if (1000*A + 100*B + 10*C + D)*9 == (D*1000 + C*100 + B*10 + A):
                      print("A={0},B={1},C={2},D={3}".format(A, B, C, D))

输出结果

题目二

九宫格
A B C
D E F
G H I
A-I代表数字，取值范围为1-9，要求横、竖、对角各方向的3个数字不重复，且3个数字之和相等，均为15

程序分析

• 数字取值范围为1-9
• A取值范围1-9，那么B取值范围为1-9同时排除A
• C取值范围1-9同时排除A和B，依次类推

  代码实现

  实现方式1

  使用copy()及remove()方法实现各数字判断

  import time
  
  start = time.clock()
  number = [x for x in range(1, 10)]
  
  for A in number:
      a = number.copy()
      a.remove(A)
      for B in a:
          b = a.copy()
          b.remove(B)
          for C in b:
              c = b.copy()
              c.remove(C)
              for D in c:
                  d = c.copy()
                  d.remove(D)
                  for E in d:
                      e = d.copy()
                      e.remove(E)
                      for F in e:
                          f = e.copy()
                          f.remove(F)
                          for G in f:
                              g = f.copy()
                              g.remove(G)
                              for H in g:
                                  h = g.copy()
                                  h.remove(H)
                                  for I in h:
                                      if A+B+C == D+E+F == G+H+I == A+D+G == B+E+H == C+F+I == A+E+I == C+E+G == 15:
                                          print('''
                                          -------------
                                          | {0} | {1} | {2} |
                                          | {3} | {4} | {5} |
                                          | {6} | {7} | {8} |
                                          -------------'''.format(A, B, C, D, E, F, G, H, I))
  
  end = time.clock()
  print('Running time: %s Seconds'%(end-start))

实现方式2

使用列表生成式实现各数字判断

import time

start = time.clock()
for A in [x for x in range(1, 10)]:
    for B in [x for x in range(1, 10) if x != A]:
        for C in [x for x in range(1, 10) if x != B and x != A]:
            for D in [x for x in range(1, 10) if x != C and x != B and x != A]:
                for E in [x for x in range(1, 10) if x != D and x != C and x != B and x != A]:
                    for F in [x for x in range(1, 10) if x != E and x != D and x != C and x != B and x != A]:
                        for G in [x for x in range(1, 10) if x != F and x != E and x != D and x != C and x != B and x != A]:
                            for H in [x for x in range(1, 10) if x != G and x != F and x != E and x != D and x != C and x != B and x != A]:
                                for I in [x for x in range(1, 10) if x != H and x != G and x != F and x != E and x != D and x != C and x != B and x != A]:
                                    if A+B+C == D+E+F == G+H+I == A+D+G == B+E+H == C+F+I == A+E+I == C+E+G == 15:
                                        print('''
                                        -------------
                                        | {0} | {1} | {2} |
                                        | {3} | {4} | {5} |
                                        | {6} | {7} | {8} |
                                        -------------'''.format(A, B, C, D, E, F, G, H, I))
end = time.clock()
print('Running time: %s Seconds'%(end-start))

输出结果

实现方式1

实现方式2

知识点

根据题目2输出结果中的运行时间可知，方式1运行0.3秒，方式2运行1.4秒，相差5倍。
根据代码逻辑，方式1在嵌套循环中直接使用remove()方式删除；方式2需要在嵌套循环中进行比较，这是导致两种算法时间差异的主要原因。
同理，题目1中通过对题目分析，确定数字A和D的取值，不需要从1-9进行循环，对程序运行时间肯定也有大幅的提升。
好代码好算法，值得关注。