Tech It Yourself
tháng 1 19, 2020
0
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);
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++11 (g++ 4.8.2) if (A % K != 0) A = (A / K + 1) * K; if (B < A) return 0; return (B - A) / K + 1; }
