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Codility---CountDiv

时间:2017-04-28 10:25:19      阅读:236      评论:0      收藏:0      [点我收藏+]

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Task description

Write a function:

class Solution { public int solution(int A, int B, int K); }

that, given three integers A, B and K, returns the number of integers within the range [A..B] that are divisible by K, i.e.:

{ i : A ≤ i ≤ B, i mod K = 0 }

For example, for A = 6, B = 11 and K = 2, your function should return 3, because there are three numbers divisible by 2 within the range [6..11], namely 6, 8 and 10.

Assume that:

  • A and B are integers within the range [0..2,000,000,000];
  • K is an integer within the range [1..2,000,000,000];
  • A ≤ B.

Complexity:

  • expected worst-case time complexity is O(1);
  • expected worst-case space complexity is O(1).
 
Solution
 
Programming language used: Java
Total time used: 19 minutes
Code: 01:36:49 UTC, java, final, score:  100
// 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 A, int B, int K) {
        // write your code in Java SE 8
        int n=0;
		if(A % K == 0) {
			n = A/K -1;
		}else {
			n = A/K;
		}
		return B/K - n ;
    }
}



https://codility.com/demo/results/training7BZKXP-YGT/

Codility---CountDiv

标签:pac   bug   java   height   ports   lock   dem   deb   exit   

原文地址:http://www.cnblogs.com/samo/p/6780186.html

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