iT邦幫忙

第 12 屆 iT 邦幫忙鐵人賽

DAY 10
1
自我挑戰組

見習村-30 Day CodeWars Challenge系列 第 10

見習村10 - Playing with digits

10 - Playing with digits

Don't say so much, just coding...

Instruction

Some numbers have funny properties. For example:

89 --> 8¹ + 9² = 89 * 1

695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2

46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 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 + 5⁴= 1390 = 695 * 2
  dig_pow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51

Ruby

Init

  def dig_pow(n, p)
    # your code
  end

Sample Testing

  Test.assert_equals(dig_pow(89, 1), 1)
  Test.assert_equals(dig_pow(92, 1), -1)
  Test.assert_equals(dig_pow(46288, 3), 51)

Javascript

Init

  function digPow(n, p){
    // ...
  }

Sample Testing

  Test.assertEquals(digPow(89, 1), 1)
  Test.assertEquals(digPow(92, 1), -1)
  Test.assertEquals(digPow(46288, 3), 51)

Thinking

想法(1): 與 見習村07 - Sum of Digits / Digital Root 中運用的 recursive 類似
想法(2): 要多判斷今天傳進來的值有可能不是從 1 開始,然後再給他平方後加總,再取餘數來判斷回傳值

https://ithelp.ithome.com.tw/upload/images/20200925/20120826vy9gRts9zq.jpg
圖片來源:Unsplash Jason Strull

Hint & Reference

Solution

Ruby

  # Solution 1
  def dig_pow(n, p)
    number = n.to_s.split('').map(&:to_i)
    sum = number.map.with_index(p) { |num, idx| num ** idx }.reduce(:+)
    sum % n == 0 ? sum/n : -1
  end
  
  def dig_pow(n, p)
    sum = n.to_s.chars.map.with_index(p) { |num, idx| num.to_i ** idx }.reduce(:+)
    sum % n == 0 ? sum / n : -1
  end

Javascript

  // Solution 1
  function digPow(n, p){
    var sum = n.toString().split('')
               .map((num, idx) => Math.pow(parseInt(num), idx+p))
               .reduce((a,b) => a + b) / n;
    return sum % 1 == 0 ? sum : -1;
  }

上一篇
見習村09 - Sums of Parts
下一篇
見習村11 - Extract the domain name from a URL
系列文
見習村-30 Day CodeWars Challenge30

尚未有邦友留言

立即登入留言