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.
Write an efficient algorithm for the following assumptions:
- 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.
int solution(int A, int B, int K) {
// write your code in C++14 (g++ 6.2.0)
int ans = (B / K) - (A / K);
return (A % K == 0 ? ans + 1 : ans);
}
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