Why it's called Panasonic contest? What is the benefit for Panasonic and participants?

Starting soon!

How to solve B and D ??

How to solve E,F?

Can someone please tell me what's wrong with my approach in E? (getting wrong answers on 15 test cases :( )

link to submission.

Hints for problems :

A. Check if H%M is zero.

B. Assuming that mangoes can provide all values between A to B inclusive, we can buy, say i mangoes only if i*A <= W <= i*B. So range of mangoes that can be bought is [ceil(W/B), floor(W/A)].

C. Count how many numbers for each digit count are there that lie between 1 to N inclusive. Say there are K numbers for D digits. Then answer gets incremented by ((D-1)/3)*(K).

D. For each query, find the respective available boxes and keep a sorted array of baggages on the bases of value. Then try to add the most expensive baggage by value to an available box that has size just more than the baggage size. Time Complexity would be O(Q*N*log(W)).

E. Let us build dynamic programming from the right of the string for the game. The states are (position, previous result modulo 7). For transition, consider if it is Takahashi, he tends to visit a new result (only 2 possible new results for a current previous result, ie, taking 0 or S[i] as the current digit) so that the next state's winner is Takahashi and the same for Aoki. Time Complexity would be O(7*N) or O(N) only.

F. Consider reading about SOS dp and meet in the middle as it is a prerequisite for the same. Since we are given, say N(<=73) numbers, so nearly half of them would be even. And among all even numbers, final set can only consist of a maximum of 1 even number. So that leaves us with nearly 36 odd numbers. So for each even number, find set of all odds that is coprime with the even number. For each such set, we need to solve the same problem. In order to solve for 36 odd numbers, we can partition the set into 2 sets of nearly 18 odd numbers. Calculate all valid subsets of both the newly built partition sets that are viable among themselves. For each element from first set, lets denote it by mask, we can obtain a supermask that corresponds to the set of allowed elements from the second set for that mask. To get the count of sets corresponding to a supermask in the 2nd set, we can build a SOS dp over the second set's mask. This gives us all possible valid masks from second set for a particular mask from first set. We can add the answer for each such element from first set to the final answer. Do not forget the case when no even number is chosen. Time Complexity would be O((N/2)*(2^(N/4))*(N/4)), where N = (B-A+1).

Is that only me or someone else also found B more difficult then A,C,D

Problem A was so easy still I did w.a(in hurry to solve it fast).by printing "no" in lower case. Ended up only solving a,b,c this time. :(