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Codility lessen 3.1 FrohJmp

Codility lessen 3.1 FrohJmp

作者: 波洛的汽车电子世界 | 来源:发表于2019-08-03 17:20 被阅读0次
    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:
    
    def solution(X, Y, D)
    
    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.
    
    

    这题好简单,没揣测明白出题点

    def solution(X, Y, D):
        # write your code in Python 3.6
        if (Y-X)/D -(Y-X)//D >0:
            return (Y-X)//D+1
        else:
            return (Y-X)//D
    

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