[코딜리티] codility lesson3 Time Complexity - FrogJmp 100%
문제.
풀이.
Programming language:
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 = 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.
Complexity:
- expected worst-case time complexity is O(1);
- expected worst-case space complexity is O(1).
Copyright 2009–2017 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
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// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int X, int Y, int D) {
// write your code in Java SE 8
return (Y-X)%D == 0 ? (Y-X)/D:(Y-X)/D+1;
}
}
| cs |
설명.
함정. 없는거 같다. 문장을 수학공식으로 잘 바꾸는지 보는 것 같음.
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