[Swift]コード性-ProgJmpプール


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A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.
Count the minimal number of jumps that the small frog must perform to reach its target.
Write a function:
public func solution(_ X : Int, _ Y : Int, _ D : Int) -> Int
that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.
For example, given:
  X = 10  Y = 85  D = 30
the function should return 3, because the frog will be positioned as follows:
  • after the first jump, at position 10 + 30 = 40
  • after the second jump, at position 10 + 30 + 30 = 70
  • after the third jump, at position 10 + 30 + 30 + 30 = 100
  • Write an **efficient** algorithm for the following assumptions:
  • X, Y and D are integers within the range [1..1,000,000,000];
  • X ≤ Y.
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    答案用紙

    public func solution(_ X : Int, _ Y : Int, _ D : Int) -> Int {
        let remainder = (Y - X) % D
        return ((Y - X) / D) + (0 < remainder ? 1 : 0)
    }
  • の総移動距離(Y-X)において、カエルは(D)まで跳ぶことができる.
  • の総移動距離(Y-X)において、カエルが跳ぶことができる距離(D)を残りで割った.なしで0処理します.
  • 1と2を追加します.
  • Tasks Details



    問題のソース

  • https://app.codility.com/programmers/lessons/3-time_complexity/frog_jmp/