[CODILITY] Lesson 5 Prefix Sums - CountDiv

Write a function:

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).

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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 A, int B, int K) {
        // write your code in Java SE 8
    		int rst = 0;
		int b = B/K;
		int a = A/K;
		rst = b-a;
		if(A%K == 0) rst++;		
		return rst;		
    }
}

뭐 이딴 문제를......

B가 가질 수 있는 K의 배수 - A가 가질 수 있는 K의 배수 + A가 K의 배수라면 +1