[Codility] Lesson3. FrogJmp (C++)

🔮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:

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

(Y-X)/D 만큼 점프하고 그래도 Y에 도달하지 못하면 (Y-X)/D + 1 만큼 점프한다.

단순하게 생각하기..!!!

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🔮코드

#include <iostream>

int solution(int X, int Y, int D) {
    int answer = 0;
    if(X == Y){
        return 0;
    }
    else if(X + D >= Y){
        return 1;
    }
    else{

        int diff = Y-X;
        answer = diff/D;
        if(diff%D != 0)
        {
            answer += 1;
        }
    
        return answer;
    }
}