For the first and second subtask, a simple greedy method can be used to search for which pumpkins Penelope should take home which would give the minimum cost.

For the third subtask, a brute-force method can be used to search for which pumpkins Penelope can take home and finding the minimum cost out of all combinations.

For the full solution, if Penelope places a pumpkin of type ~T_i~ at indexes ~j~, ~j + (X + 1)~, ~j + 2 \times (X + 1)~ and so on, assuming ~N~ is divisible by ~X~, she can place each type of pumpkin up to a maximum of ~\lfloor \frac{K}{X+1} \rfloor~ times. If ~N~ is not divisible by ~X~, she can place ~K % (X+1)~ pumpkins 1 more time. Greedily search for the smallest values, without taking ~K % (X+1)~ pumpkins of a single type more than ~\lfloor \frac{K}{X+1} \rfloor + 1~ times and the remaining pumpkins of a single type more than ~\lfloor \frac{K}{X+1} \rfloor~ times.