I am having trouble in coming up with a solution for Carrot Cakes. The editorials are provided but I can’t understand how do we arrive at such a formula…

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let’s say we don’t build second one.

Then it’s just simple to calculate **TimeWithoutBuild** using formula :

* TimeWithoutBuild = ceil(n/k) t**

Now let’s build the second and calculate the **TimeWithBuild**:

as it take **d** unit time to build, we first add **d** to **TimeWithBuild** now we might have already baked some cakes while building the second one. So we need to subtract that cakes from **n**, so our remaining cakes will be **n-(d/t) k* let’s say this is

**n_rem**Now our speed of baking is doubled

So

**TimeWithBuild = d + ceil(n_rem/(2*k)) * t**

Now you will be good to code it yourself but nevertheless here is my solution…

Good Luck

[details=Click to view]

#include using namespace std; #define int long long signed main(){ int n,t,k,d; cin >> n >> t >> k >> d; int T1 = t*( (n/k) + (n%k!=0)); int T2 = d + t*(( n-(d/t)*k )/(2*k) +(( n-(d/t)*k )%(2*k)!=0)); if(T2 < T1)puts("YES"); else puts("NO"); return 0; }

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Thanks for replying… Can explain the last part i.e n-(d/t)*k in detail ? I don’t get it…