2020.6.25 练习 （python)

Posted 2020-12-12 adelinebao

1. 枚举的运用 

Your team is writing a fancy new text editor and you‘ve been tasked with implementing the line numbering.

Write a function which takes a list of strings and returns each line prepended by the correct number.

The numbering starts at 1. The format is n: string. Notice the colon and space in between.

Examples:

number([]) # => []

number(["a", "b", "c"]) # => ["1: a", "2: b", "3: c"]

1 def number(lines):
2     a=enumerate(lines,1)
3     print(a.__next__(),a.__next__(),a.__next__())  # (1, ‘a‘) (2, ‘b‘) (3, ‘c‘)
4     return [‘%d: %s‘ % v for v in enumerate(lines, 1)]  # 因为返回是是先数字，在lines 里面的元素

 

2. 菲波那契数列

The Fibonacci numbers are the numbers in the following integer sequence (Fn):

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...

such as

F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

Given a number, say prod (for product), we search two Fibonacci numbers F(n) and F(n+1) verifying

F(n) * F(n+1) = prod.

Your function productFib takes an integer (prod) and returns an array:

[F(n), F(n+1), true] or {F(n), F(n+1), 1} or (F(n), F(n+1), True)

depending on the language if F(n) * F(n+1) = prod.

If you don‘t find two consecutive F(m) verifying F(m) * F(m+1) = prodyou will return

[F(m), F(m+1), false] or {F(n), F(n+1), 0} or (F(n), F(n+1), False)

F(m) being the smallest one such as F(m) * F(m+1) > prod.

Some Examples of Return:

productFib(800) # should return (34, 55, false),

 

 

 1 def productFib(prod):
 2     x = [0,1]
 3     # print(x[-1])
 4     while True:
 5         if prod == x[-1] * x [-2]:
 6             return [x[-2],x[-1],True]
 7         if prod < x[-1] * x [-2]:
 8             return [x[-2],x[-1],False]
 9         else:
10             x.append(x[-1] + x[-2])
11             continue

def productFib(prod):
  a, b = 0, 1
  while prod > a * b:
    a, b = b, a + b
  return [a, b, prod == a * b]

def productFib(prod):
    func = lambda a, b: func(b, a+b) if a*b < prod else [a, b, a*b == prod]
    return func(0, 1)

def productFib(prod, f1=0, f2=1):
    return [f1, f2, True] if prod == f1 * f2 else [f1, f2, False] if prod < f1 * f2 else productFib(prod, f2, f1+f2)


3. 数字乘方相加


Some numbers have funny properties. For example:

89 --> 8¹ + 9² = 89 * 1

695 --> 6² + 9³ + 5?= 1390 = 695 * 2

46288 --> 4³ + 6?+ 2? + 8? + 8? = 2360688 = 46288 * 51

Given a positive integer n written as abcd... (a, b, c, d... being digits) and a positive integer p

• we want to find a positive integer k, if it exists, such as the sum of the digits of n taken to the successive powers of p is equal to k * n.

In other words:

Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k

If it is the case we will return k, if not return -1.

Note: n and p will always be given as strictly positive integers.

dig_pow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1
dig_pow(92, 1) should return -1 since there is no k such as 9¹ + 2² equals 92 * k
dig_pow(695, 2) should return 2 since 6² + 9³ + 5?= 1390 = 695 * 2
dig_pow(46288, 3) should return 51 since 4³ + 6?+ 2? + 8? + 8? = 2360688 = 46288 * 51

test.assert_equals(dig_pow(46288, 3), 51)

def dig_pow(n, p):
    a = 0
    for x in str(n):
        a += int(x) ** p
        p += 1
    if a % n == 0:
        return a//n
    else:
        return -1

def dig_pow(n, p):
  s = 0
  for i,c in enumerate(str(n)):
     s += pow(int(c),p+i)
  return s/n if s%n==0 else -1

def dig_pow(n, p):
  t = sum( int(d) ** (p+i) for i, d in enumerate(str(n)) )
  return t//n if t%n==0 else -1

4. 密码有效判断

ATM machines allow 4 or 6 digit PIN codes and PIN codes cannot contain anything but exactly 4 digits or exactly 6 digits.

If the function is passed a valid PIN string, return true, else return false.

eg:

validate_pin("1234") == True
validate_pin("12345") == False
validate_pin("a234") == False

def validate_pin(pin):
    return pin.isdecimal() and (len(pin) == 4 or len(pin) == 6)

def validate_pin(pin):
    return len(pin) in (4, 6) and pin.isdigit()

1 import re
2 def validate_pin(pin):
3     return bool(re.match(r‘^(d{4}|d{6})$‘,pin))  # ^开头，$结尾，中间要严格符合这个格式，digit 4个或6个