https://app.codility.com/programmers/lessons/3-time_complexity/frog_jmp/
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.
/**
개구리가 현재 x위치에 존재
y 포지션 이상으로 도달하고 싶음
언제나 고정된 거리 D만큼 이동 가능
개구리가 목표 위치로 갈 수 있는 최소한의 이동 횟수
**/
class Solution {
public int solution(int X, int Y, int D) {
int answer = 0;
if((Y-X) % D > 0) answer = ((Y-X) / D) + 1;
else answer = (Y-X) / D;
return answer;
}
}