[Codility] Lesson3 - Time Complexity : FrogJmp

app.codility.com/programmers/lessons/3-time_complexity/frog_jmp/

Problem

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.

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Solution(c++)

// you can use includes, for example:
#include <bits/stdc++.h>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(int X, int Y, int D) {
    // write your code in C++14 (g++ 6.2.0)
    long long int minJump;
    long double tmp;
    tmp = double((Y-X))/double(D);
    // cout << tmp;

    minJump = ceil(tmp);
    return minJump;
}

 

Test Result

** 이 문제를 풀 때 주의 사항

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

위와 같이 범위가 설정되어있기 때문에 

정수 -> long long int로

실수 -> long double 로 푸는 것이 안전하다

 

알고리즘 문제를 풀 때는 범위에 항상 주의해야한다!!

 

이 문제를 풀 때 범위를 고려하지 않아 처음에 44%가 나왔었다 ㅠㅠ

app.codility.com/demo/results/trainingV9JCAR-XNU/

Test results - Codility