# Codility 1. FrogJmp

annmj·2021년 11월 15일
1

## Codility

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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:

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

1. Y =< X + N X D 이므로 배수 관계를 이용한다는 것을 생각해서 문제를 풀면 된다.

### 📝 소스코드

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
public int solution(int X, int Y, int D) {
if((Y-X) % D == 0) return (Y-X) / D;
else return ((Y-X) / D) + 1;
}
}