일 | 월 | 화 | 수 | 목 | 금 | 토 |
---|---|---|---|---|---|---|
1 | 2 | 3 | 4 | 5 | 6 | 7 |
8 | 9 | 10 | 11 | 12 | 13 | 14 |
15 | 16 | 17 | 18 | 19 | 20 | 21 |
22 | 23 | 24 | 25 | 26 | 27 | 28 |
29 | 30 | 31 |
Tags
- Mac
- ELASTIC
- MySQL
- aggs
- Elasticsearch
- query
- springboot
- licence delete curl
- 900gle
- Python
- docker
- API
- plugin
- aggregation
- Test
- zip 파일 암호화
- matplotlib
- 차트
- License
- token filter test
- 파이썬
- analyzer test
- TensorFlow
- zip 암호화
- flask
- license delete
- high level client
- sort
- Java
- Kafka
Archives
- Today
- Total
개발잡부
코딜리티 11 본문
반응형
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.
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.
public static int solution(int A, int B, int K) {
// write your code in Java SE 8
int ret = 0;
ret = (B - A) / K;
int r_a = A / K;
int r_b = B / K;
if (A % K == 0) {
r_a = r_a - 1;
ret = ret + 1;
}
ret = r_b - r_a;
return ret;
}
반응형
Comments