Problem 18

// approach:
// if remainder of any two no’s is k, then sum of those 2 divisible by k
// sum of elements having remainders i and k-i also divisible by k as i+k-i=k
// if k is even then i and k-i will be same i.e k/2 so sum of them will be k i.e divisible again so we should add only min(1, a[k/2]);
// if any no. having remainder zero then sum of such no. will also have remainder zero i.e will become divisible by k so we should add
// only min(1, a[0]);

int main(){
int, n, k, a[], retur_n; //a[] for storing remainders
cin>>n>>k;
for(int i=0; i>ai;
++a[ai];
}

if(k%2==0){

a[k/2]=min(1, a[k/2]);
retur_n=0;

for(int i=1; i<k/2; i++){
retur_n+=max(a[i], a[k-i]);
}
retur_n+=min(1, a[0]);
cout<<"the size of maximal subset of S is"<<retur_n<<endl;
}
}