made by Jaehyeok Choi

dynamic programming

Let;s save the max value in the dp array.

There is only one option for taking 1 card. Just buying one pack card. There is two option for taking 2 cards. (1+1) , (2). Save the larger value in dp array.

There is tree option for taking 3 cards. (1+1+1) , (1+2) , (3).

However, the biggest value of taking 2 cards is in the dp[2]. We already consider (1+1) (2) case in dp[2].

That means (dp[1]+dp[1]+dp[1]) case is smaller or same with (dp[2]+dp[1]) case.

So, there is two option for taking 3 cards.

3 option for taking 4 cards. (dp[1] + dp[3]) (dp[2] + dp[2]) (p[4])

3 option for taking 5 cards. (dp[1] + dp[4]) (dp[2] + dp[3]) (p[5])

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Equation founded~~~

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