【数组】【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).

思路：

这题有一点Prefix Sums的思想，但不需要存一张表，实时计算N/K就好。需要注意的是区间的边界是A ≤ i ≤ B，分别计算0 < x ≤ A 和0 < x ≤ B间的数量之后，还要考虑A是否也算一个。

代码：

01.

int solution(int A, int B, int K) {

02.int c = (A%K == 0)?1:0;

03.return B/K - A/K + c;

}