D[i][i + 1] = max(V[i] - V[j], V[j] - V[i], V[i] + V[j]), 1 ≤ i ≤ N - 1
D[i][j] = max(D[i][j], V[i] + V[j] - D[i + 1][j - 1]), 1 ≤ i ≤ j ≤ N, j - i ≥ 2
Solutia se va afla in D[1][N].

 
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Pentru a obtine insa 100 de puncte nu trebuie retinuta intreaga matrice, ci doar ultimele doua-trei diagonale. Complexitate finala O(N2).
This is a classic dynamic programming problem and, like most DP problems, is simple if and only if you "get" the induction piece. Additionally, getting full points required figuring out (or knowing) a nice trick for when you need to calculate the sums of lots of different ranges.