Me.

I do, because I'm left home alone :(

And me!

the lcm of ur set of numbers is not greater than 1e9, so you will end up with collisions. you have room to make 2->4 and 3->9

How did you come up with $$$CRT$$$? I was thinking of some logic like Binary Lifting.

Technique without CRT:

ll delta = a[8], ans = mod[8];
F(i, 0, 7) {
    while (ans % a[i] != mod[i])
        ans += delta;
    delta = lcm(delta, a[i]);
}

I also have a CRT approach with divisors of $$$2, 3, 5, 7, 11, 13, 17, 19, 23$$$. But still having a wrong answer.

Will there be some $$$N$$$ that might collide with these divisors? Submission

F is Chinese Remainder Theorem. $$$4,9,5,7,11,13,17,19,23$$$ are coprime integers that satisfy the CRT conditions and product is greater than $$$10^9$$$ and sum less than $$$110$$$. From the judge, try to find $$$N$$$ mod these integers. Then by CRT, you can find $$$N$$$ mod their product, which is that number itself since $$$N$$$ is guaranteed to be less than $$$10^9$$$.

Bute Force:

void chal(){
  ll n,a,b;
  cin>>n>>a>>b;
  string st;
  cin>>st;
  ll ans,x=0;
  fo(i,(n+1)/2){
    if(st[i]!=st[n-i-1]){
      x+=b;
    }
  }
  // cout<<x<<endl;
  ans=x;
  fo(i,n){
    x=0;
    char y=st[0];
    fo(i,n-1){
      st[i]=st[i+1];
    }
    st[n-1]=y;
    x=(i+1)*a;
    fo(i,(n+1)/2){
    if(st[i]!=st[n-i-1]){
      x+=b;
    }
  }
  // cout<<st<<' '<<x<<endl;
  ans=min(ans,x);
  }
  cout<<ans<<endl;


  
}

small modification of Floyd-Warshall algorithm: https://cp-algorithms.com/graph/all-pair-shortest-path-floyd-warshall.html

How do you check the graph is drawable with one stroke?

Yes, you can do a breadth-first search from each source node to precompute all pairs of distances, but Floyd-Warshall is more suitable for this.

Well, if what I believe is correct, it is possible. Let us simply compute the entire convex hull of $$$C \cup \{ S,T \}$$$. Now, there will be two paths on that convex hull connecting $$$S$$$ and $$$T$$$, one going clockwise, one going counterclockwise (assuming both $$$S$$$ and $$$T$$$ are on the hull). The answer is the minimum of the two. If either $$$S$$$ or $$$T$$$ is not on the convex hull (at least one of the two must be on it, but one may not be on it), then the straight line segment connecting $$$S$$$ and $$$T$$$ must not intersect with $$$C$$$ (if the line intersects with $$$C$$$, then both $$$S$$$ and $$$T$$$ must be on the hull*), so in that case $$$\text{dist}(S,T)$$$ is the answer.

*UPD: Here's the proof on the intersection part. Let us assume $$$S$$$ was on the hull, and $$$T$$$ inside it. Now let's think of the hull incrementally, adding $$$S$$$ first and then $$$T$$$. For $$$T$$$ not to be added to the hull, $$$T$$$ should be inside the hull. This is either inside the polygon, or at a point where there is no intersection with the segment and the polygon. The former case is already eliminated from the task, and our proof is finished.

You can find out if two segments intersect without using anything more than some cross and dot products