FrogJmp

🔗 문제 링크

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

❔ 문제 설명

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

⚠️ 제한사항

• X, Y and D are integers within the range [1..1,000,000,000];

• X ≤ Y.

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💡 풀이 (언어 : Java & Python)

X, Y, D 간의 관계를 X에 대한 방정식으로 정리하고 이상이므로 소수점을 올림수로 하면 정답이다.

Java

class Solution {
    public int solution(int X, int Y, int D) {
        double result = (Y-X) / (double) D;
        return (int) Math.ceil(result);
    }
}

Python

import math

def solution(X, Y, D):
    return math.ceil((Y - X) / D)