So finally, long challenge is over which to be honest wasn't very long for me. I only did 4 problems after which I lost the interest in solving problems. I prefer the problems that are short and has a good logic or something that is worth thinking about. So here are my discussions on the problems that I solved or at-least read. (I'm talking about the Div. 2 here, btw).
This was the first problem of the Div. 2 and as expected it was a pure cakewalk. The only thing needed here was implementation and honestly I think the code is way simpler in Python than any other language.
Here is my code which is sweet and short ;)
t = int(input())
while t:
s = input()
if 'not' in s.split():
print('Real Fancy')
else:
print('regularly fancy')
t -= 1
I hope the solution is very clear if one has understood the problem. I solved it in few minutes and moved on to the next problem.
This problem was easy, but the wording was WRONG!
I mean it, here is what was wrong with the statement.
The players alternate turns. In each turn, the current player must remove a non-zero number of elements from the sequence
That a made me mad because the first time I read this, I read it like we have to remove a single non-zero element for a turn. Hence I was producing multiple WAs on submitting. As soon as I fixed the issue, the problem was solved.
There was nothing much other than the implementation for the problem, just a better statement is always nice to have.
Here is my code if you are interested:
#include <bits/stdc++.h>
using namespace std;
int main() {
int t; cin >> t;
while(t--) {
int n, a, b;
cin >> n >> a >> b;
int ca = 0, cb = 0, cc = 0;
for(int i = 0; i < n; i++) {
int x; cin >> x;
if(x % a == 0 && x % b == 0) cc++;
else if(x % a == 0) cb++;
else if(x % b == 0) ca++;
}
if(cb >= ca) {
if(cc) cout << "BOB\n";
else cout << "ALICE\n";
} else {
cout << "ALICE\n";
}
}
return 0;
}
I try to keep my codes short but readble and clean at the same time. Let me know if I can improve somewhere.
Honestly I didn't know how my solution worked. The problem was simple, basic implementation and it should produce you an AC. I think that this problem was nice and much better than the previous two compared to their difficulty. It took me some time to land upon the required observation of producing the pairs.
Hint: If you are in the middle of solving this problem then think the number
of pairs you have to print. You can get the idea of how your solution needs to be.
This is my code for the problem, I'm pretty sure the author has the same solution for this problem.
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m;
cin >> n >> m;
int a[n], b[m];
int mn = INT_MAX, mx = INT_MIN;
int mni, mxi;
for(int i = 0; i < n; i++) {
cin >> a[i];
if(a[i] < mn) {
mn = a[i];
mni = i;
}
}
for(int i = 0; i < m; i++) {
cin >> b[i];
if(b[i] > mx) {
mx = b[i];
mxi = i;
}
}
for(int i = 0; i < n; i++) {
if(i != mni) {
cout << i << " " << mxi << endl;
}
}
for(int i = 0; i < m; i++) {
cout << mni << " " << i << endl;
}
return 0;
}
Okay, this was the last and probably the most interesting problem I solved during the long challenge.
The problem was very clear and had some interesting observations. Basic required thing was the definition of modulo (%) operation. Once you split the problem into basic parts the solution gets clearer and clearer.
I liked this problem and it took me some time to arrive at the solution.
Here is my code for the same:
#include <bits/stdc++.h>
using namespace std;
int main() {
int t; cin >> t;
while(t--) {
int n, p, mx; cin >> n >> p;
if(n % 2 == 0) mx = n / 2 - 1;
else mx = n / 2;
long long ans;
if(mx == 0) {
int x = (n == 1) ? 1 : 2;
ans = 1LL * x * p * p + 1LL * (p - x) * p * x + 1LL * (p - x) * (p - x) * x + 1LL * (p - x) * (p - x) * (p - x);
} else {
ans = 1LL * (p - mx) * (p - mx) + 1LL * (p - n) * (p - mx) + 1LL * (p - n) * (p - n);
}
cout << ans << endl;
}
return 0;
}
If you are in doubt of any solution you can mail me, or leave an issue at MyCPCodes where I keep my codes updated.
So this was it, I liked the problem set and it is always fun to upsolve the long contest problems because of their sheer difficulty.
I will update this post if I solved the remaining problems.
Happy coding!