package com.code;

public class Test05_1 {

     public static int solution(int A, int B, int K) {

             // handle 6,8,7 condition

             // handle 6,11,2 condition

             // handle 7,12,3 condition

             if(A%K==0){

                 return (B-A)/K + 1;

             }else{

                 int start = A+(K-A%K);

                 if(start>B){ // handle 7,8,9 condition

                     return 0;

                 }else{

                     return (B-start)/K+1;

                 }

             }

    }

    public static void main(String[] args) {

        System.out.println(solution(6, 11, 2));

        System.out.println(solution(6, 8, 7));

        System.out.println(solution(5, 5, 3));

        System.out.println(solution(6, 7, 8));

        System.out.println(solution(0, 0, 11));

    }

}

/**

1. CountDiv 数数有几个

Compute number of integers divisible by k in range [a..b].

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).

*/