[Codility][Kotlin] FrgJmp

문제

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:

class Solution { public int solution(int X, int Y, int 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.

 

풀이

// 개구리의 위치 X, 목적지 Y, 점프 거리 D
private fun solution(X: Int, Y: Int, D: Int): Int {
    if ((Y-X)%D > 0)
        return (Y-X)/D+1
    else
        return (Y-X)/D
}

private fun solution2(X: Int, Y: Int, D: Int): Int {
    return ceil((Y-X)/D.toDouble()).toInt()
}

 

ceil()함수를 간과하고 풀었다..허허....ceil()은 올림함수! floor()는 내림함수! round()는 반올림함수!