FrogJmp_codility

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:

function 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

 

Assume that:

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

 

풀이

 

 X와Y 사이의 거리를 D로 나눈만큼 점프한 후 Y에 도착하지 못했다면 한번 더 점프하도록 구현했다.

function solution(X, Y, D) {
    let result = Math.floor((Y-X)/D);
    
    if ((Y-X) % D !== 0) {
        result++;
    }
 
    return result;
}

다른 풀이

function solution(X, Y, D) {
  return Math.ceil((Y - X) / D);
}

 

 

참고 https://developer.mozilla.org/ko/docs/Web/JavaScript/Reference/Global_Objects/Math/ceil