Hello Codeforces,↵
=================↵
↵
Although i did not participate in [Educational Codeforces round 190](https://mirror.codeforces.com/contest/2230/). I have tried to solve its first problem, https://mirror.codeforces.com/contest/2230/problem/A . ↵
↵
The core of the problem says that:↵
↵
_There are $n$ students whom have access to keys of cost $a$ and $b$ , which are individual keys and group keys for an online course. For which, a group can be up to three. Our task is to minimize spending for all $n$ students, getting each one a key._↵
↵
Now, if all $n$ kids need to have keys, then we can work by using the group keys for groups of 3 (max). Then as per the remainder, we can add along either $a$ or $b$ depending which is smaller.↵
↵
#### first part:↵
check whether buying individually for 3 or group wise for 3 (max) is more optimal, spend it on $floor(n/3)$ kids (as we are buying in groups of 3 or 3 individually for max).↵
mathematically : $min(3a, b)*floor(n/3)$↵
↵
#### last part:↵
after optimally buying maximum keys in less price, we check how much more is needed (remainder) and check whether buying individually for 1 or group wise for 1 (min) is better.↵
mathematically : $n \bmod 3 * min(a,b)$↵
↵
**combined:** $min(3a, b)*floor(n/3) + n \bmod 3 * min(a,b)$↵
↵
Heres the code:↵
↵
~~~~~↵
#include <bits/stdc++.h>↵
using namespace std;↵
↵
void solve(){↵
int n,a,b; cin >>n >>a >>b;↵
int price = 0;↵
if(n<3){↵
price = min(a*n, b);↵
}↵
else{↵
price = min(a*3, b)*(n/3)+(n%3)*(min(a,b));↵
}↵
cout << price << endl;↵
}↵
↵
int main() {↵
ios::sync_with_stdio(false);↵
cin.tie(NULL);↵
int t; cin >> t;↵
while(t--) solve();↵
}↵
~~~~~↵
↵
Edit: I have not talked about my $if(n<3)$ yet, sorry for that. the assumed exceptions are ${1, 2}$ as first part would become $0$ and the last part would be $n*min(a,b)$ , which causes hindrance if $b<a$ , as b is group based up to 3, we would not need more than one group key, yet still we are calculating $n*b$ . Using $min(a*n, b)$ is more reliable as it does not guarantee that the result must be individual. ↵
↵
its currently in queue (queueforces), so i would like to know if my code is correct, any edges or errors.↵
↵
**Hopefully it AC's!**
=================↵
↵
Although i did not participate in [Educational Codeforces round 190](https://mirror.codeforces.com/contest/2230/). I have tried to solve its first problem, https://mirror.codeforces.com/contest/2230/problem/A . ↵
↵
The core of the problem says that:↵
↵
_There are $n$ students whom have access to keys of cost $a$ and $b$ , which are individual keys and group keys for an online course. For which, a group can be up to three. Our task is to minimize spending for all $n$ students, getting each one a key._↵
↵
Now, if all $n$ kids need to have keys, then we can work by using the group keys for groups of 3 (max). Then as per the remainder, we can add along either $a$ or $b$ depending which is smaller.↵
↵
#### first part:↵
check whether buying individually for 3 or group wise for 3 (max) is more optimal, spend it on $floor(n/3)$ kids (as we are buying in groups of 3 or 3 individually for max).↵
mathematically : $min(3a, b)*floor(n/3)$↵
↵
#### last part:↵
after optimally buying maximum keys in less price, we check how much more is needed (remainder) and check whether buying individually for 1 or group wise for 1 (min) is better.↵
mathematically : $n \bmod 3 * min(a,b)$↵
↵
**combined:** $min(3a, b)*floor(n/3) + n \bmod 3 * min(a,b)$↵
↵
Heres the code:↵
↵
~~~~~↵
#include <bits/stdc++.h>↵
using namespace std;↵
↵
void solve(){↵
int n,a,b; cin >>n >>a >>b;↵
int price = 0;↵
if(n<3){↵
price = min(a*n, b);↵
}↵
else{↵
price = min(a*3, b)*(n/3)+(n%3)*(min(a,b));↵
}↵
cout << price << endl;↵
}↵
↵
int main() {↵
ios::sync_with_stdio(false);↵
cin.tie(NULL);↵
int t; cin >> t;↵
while(t--) solve();↵
}↵
~~~~~↵
↵
Edit: I have not talked about my $if(n<3)$ yet, sorry for that. the assumed exceptions are ${1, 2}$ as first part would become $0$ and the last part would be $n*min(a,b)$ , which causes hindrance if $b<a$ , as b is group based up to 3, we would not need more than one group key, yet still we are calculating $n*b$ . Using $min(a*n, b)$ is more reliable as it does not guarantee that the result must be individual. ↵
↵
its currently in queue (queueforces), so i would like to know if my code is correct, any edges or errors.↵
↵
**Hopefully it AC's!**
