##### Short Problem Definition:

Given a set *S* of *n* distinct integers, print the size of a maximal subset *S’* of *S* where the sum of any 2 numbers in *S’* are not evenly divisible by *k*.

##### Link

##### Complexity:

time complexity is `O(N)`

space complexity is `O(N)`

##### Execution:

This is by all means not an easy task and is also reflected by the high failure ratio of the participants. For a sum of two numbers to be evenly divisible by k the following condition has to hold. If the remainder of *N1%k == r* then *N2%k = k-r* for *N1+N2 % k == 0*. Let us calculate the set of all numbers with a remainder of *r* and *k-r* and pick the larger set. If the remainder is half of k such as 2 % 4 = 2 or exactly k such as 4 % 4 = 0, just one number from each of these sets can be contained in *S’.*

##### Solution:

```
def solveSubset(S, k, n):
r = [0] * k
for value in S:
r[value%k] += 1
result = 0
for a in xrange(1, (k+1)//2):
result += max(r[a], r[k-a])
if k % 2 == 0 and r[k//2]:
result += 1
if r[0]:
result += 1
return result
n, k = map(int, raw_input().split())
S = map(int, raw_input().split())
print solveSubset(S, k, n)
```

```
# RUST
use std::io;
fn get_numbers() -> Vec<u32> {
let mut line = String::new();
io::stdin().read_line(&mut line).ok().expect("Failed to read line");
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}
fn calculate_nondivisible(a: Vec<u32>, n: usize, k: usize) -> u32 {
let mut result = 0;
let mut r = vec![0; k];
for val in a {
r[(val as usize)%k] += 1;
}
for idx in 1..(k+1)/2 {
result += std::cmp::max(r[idx as usize], r[(k-idx) as usize]);
}
if k % 2 == 0 && r[k/2] != 0 {
result += 1;
}
if r[0] != 0 {
result += 1;
}
result
}
fn main() {
let line = get_numbers();
let a = get_numbers();
println!("{}", calculate_nondivisible(a, line[0] as usize, line[1] as usize) );
}
```