Lesson 3. Time Complexity - FrogJmp

1. 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:
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 = 40after the second jump, at position 10 + 30 + 30 = 70after 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.

현재 위치(X)

목표 지점(Y)

한번에 이동 가능한 거리(D)

몇 번 이동해야 X에서 Y까지 갈 수 있는지 계산하는 문제

 

2. 풀이 전략

단순 계산이므로, 그냥 계산하면 된다. 수식으로 표현하자면 다음과 같다. 

\( X + DN >= Y \) , (N은 점프 횟수)

\( DN >= Y - X \)

\(  N >=  \frac{Y - X}{D} \)

 

3. 코드 (C++)

int solution(int X, int Y, int D) {
    if( ((Y - X) % D) == 0 ) return (Y - X) / D; //나머지가 없다면, 몫 만큼만 점프하면된다.
    else return ((Y - X) / D) + 1; //나머지가 있다면, 한번 더 점프해야 한다.
}