Tasks Details
medium
1.
CountDiv
Compute number of integers divisible by k in range [a..b].
Task Score
100%
Correctness
100%
Performance
100%
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.
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used C++
Time spent on task 34 minutes
Notes
not defined yet
Code: 07:35:27 UTC,
java,
autosave
Code: 07:37:08 UTC,
cpp,
autosave
Code: 08:05:19 UTC,
cpp,
autosave
Code: 08:05:40 UTC,
cpp,
autosave
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(int A, int B, int K) {
// write your code in C++14 (g++ 6.2.0)
while (B % K > 0) {
B--;
}
while (A &=^)
}
Code: 08:06:04 UTC,
cpp,
verify,
result: Failed
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(int A, int B, int K) {
// write your code in C++14 (g++ 6.2.0)
while (B % K > 0) {
B--;
}
while (A % K > 0)
A++;
return (B-A) % K + 1;
}
Analysis
expand all
Example tests
1.
0.001 s
WRONG ANSWER,
got 1 expected 3
Code: 08:06:48 UTC,
cpp,
autosave
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(int A, int B, int K) {
// write your code in C++14 (g++ 6.2.0)
while (B % K > 0) {
B--;
}
while (A % K > 0)
A++;
return (B-A) K + 1;
}
Code: 08:06:51 UTC,
cpp,
verify,
result: Passed
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(int A, int B, int K) {
// write your code in C++14 (g++ 6.2.0)
while (B % K > 0) {
B--;
}
while (A % K > 0)
A++;
return (B-A) / K + 1;
}
Analysis
Code: 08:07:50 UTC,
cpp,
autosave
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(int A, int B, int K) {
// write your code in C++14 (g++ 6.2.0)
while (B % K > 0) {
B--;
}
while (A % K > 0)
A++;
return (B-A) / K + 1;
}
Code: 08:08:39 UTC,
cpp,
verify,
result: Passed
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(int A, int B, int K) {
// write your code in C++14 (g++ 6.2.0)
while (B % K > 0) {
B--;
}
while (A % K > 0)
A++;
return (B-A) / K + 1;
}
Analysis
Code: 08:08:42 UTC,
cpp,
final,
score:
100
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(int A, int B, int K) {
// write your code in C++14 (g++ 6.2.0)
while (B % K > 0) {
B--;
}
while (A % K > 0)
A++;
return (B-A) / K + 1;
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(1)
expand all
Correctness tests
1.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
4.
0.001 s
OK
5.
0.001 s
OK
6.
0.001 s
OK
expand all
Performance tests
1.
0.001 s
OK
1.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
1.
0.001 s
OK
2.
0.001 s
OK
3.
0.001 s
OK
4.
0.001 s
OK