Hello Guys, The editorials for January Cook of are cooked (first two only).
Problem: SURVIVE
Problem Statement:
Given a shop opening from Monday to Saturday, you are allowed to buy N sweets everyday the shop is open and you eat K sweets everyday, determine if you can survive for next S days.
Difficulty:Cakewalk
Solution:
The constraints of problem are small enough to check for everyday.
For every ith day (1<=i<=S) not being divisible by 7, buy N sweets. Eat K sweets everyday. If at any day you have less than K sweets to eat, answer is impossible.
If you survive S days, and supposing C is the number of remaining sweets and A being number of days you bought sweets, just subtract C/N from A. (The number of sweet boxes untouched).
Link to Solution here.
Problem: MULTHREE
Problem Statement:
Given K, d0 (first digit) and d1 (second digit), we care to check that K-digit number is divisible by 3. if K> 2, Remaining digits of our Number are given by:
\begin{equation}
D_{i} = {\textstyle \sum^{j-1}{0}} D{j} (mod 10)
\end{equation}
Difficulty: Easy
Solution:
First thing to note is that if sum of digits of a number is divisible by 3, number if divisible by 3.
Another thing, (I would recommend running a brute force for all possible values of d0 and d1 to observe this pattern, do give a try before reading).
After first three digits, only the pattern “2486” repeats till end of the number. This means that we can split our number into 4 parts.
- First 3 digits, manually take sum.
- Next 0 to 3 digits (pattern may start at 4, 8 or 6, Taking care of that).
- “2486” X number of times. ** Add X*20 to sum**.
- Last 0 to 3 digits (pattern may end at 2, 4 or 8).
Finding X: Check for start point of pattern (will be from p = 3 to 6 in 0-based indexing), X = floor((K-p)/4).
Edge case: If fourth digit is 0, we take sum of first three digits and check its divisibility by 3.
Link to Solution here.
PS: I have intentionally kept the length of editorials short. If you don’t get it, feel free to post a comment. 
