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1. CountDiv 数数有几个 Compute number of integers divisible by k in range [a..b].

 

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).

*/

 

posted @ 2017-02-22 20:12  stono  阅读(303)  评论(0编辑  收藏  举报