How I think

f(x) has a period of k

proof:
f(x)=(x%k)*((n-x)%k)
f(x+k)=((x+k)%k)*((n-x-k)%k)
f(x+k)=(x%k)*((n-x)%k)=f(x)

that means we need to find value of f(x) for 0..k only to get the answer

now take n%k
let q=n%k
q, . . ,0,k-1,k-2,...
0,1,2,3,4,5,6,7,8,9
q=(n-x)%k
.
.
(0+x)*(q-x)=qx-x2
from 0 to q
f(x) will be max at x=q/2

now from k-1 position
(q+1+x)*(k-1-x)
c=q+1
d=k-1
(c+x)(d-x)=cd+(d-c)x-x2
so f(x) will be max at x=(d-c)/2
where d-c=(k-1)-(q+1)
so q+1+x=q+1+(k-q-2)/2

```
 ll n,k; cin>>n>>k;
    
    ;
    ll oans=0;
    if(k>=n){
        cout<<n/2<<endl; return;
    }
    ll rem=(n%k);
    ll tp=(rem)/2;
    //1 0
    //(1-x)*x=1-x2 x=0;
    //1*0
    ll ans=(rem-tp)*tp;
    if(rem+1>=k){
        verify(0);
        cout<<tp<<endl; return;
        
    }
    
    ll atp=(k-1)-(rem+1);
    ll r=(atp)/2;
    ll sans=(r+rem+1)*(k-1-r);
    if(max(ans,sans)==ans){
        verify(0);
        cout<<tp<<endl; return;
        
    }else{
        verify(0); 
        cout<<rem+1+r<<endl;  
        
    }
```

Thank you for solving the issue :)