Observe that the final marks should be a non-decreasing sequence. (The observation was not true for the original problem, which was incorrect as a result.) This is because it never hurts Bob to swap marks $m_i$ and $m_j$ if $m_i>m_j$ and $i<j$, since the $v$ array is non-decreasing.

With this observation, a dynamic programming solution can be found for the first subtask. Let $dp[i][j][e]$ be the most amount of money given that only the first $i$ tests are being considered, $m_i=j$, and $e=m_1+m_2+\cdots +m_i$. Then the transition can be done in $\mathcal{O}(N)$ by considering the length of marks equal to exactly $j$. The rest of the marks will definitely be less than $m_i$ and will contribute to the money earned.

Time Complexity: $\mathcal{O}(N^{3}E)$

For the second subtask, a similar approach is used. Instead of considering the tests from $1$ to $N$, this dynamic programming solution considers the tests from $N$ to $1$. The advantage here is the dimension $[j]$ in the first subtask is no longer necessary, as $m_i$ should always be equal to $0$. So let $dp[i][e]$ be the most amount of money given that only the last $N-i+1$ tests are being considered and $e=m_i+m_{i+1}+\cdots +m_N$. If $m_i=m_j<m_{j+1}$, that is equivalent to adding $1$ to all the marks in the transition. So

$${\displaystyle dp[i][e]=\underset{j>i}{min}dp[j+1][e-(N-j)]+(j-i+1)\cdot (v_j+v_{j+1}+\cdots +v_N)}$$

Then the transition can be done in $\mathcal{O}(N)$.

Time Complexity: $\mathcal{O}(N^{2}E)$

For the final subtask, note that the convex hull optimization can be applied to the dynamic programming solution from the second subtask. This changes the transition into amortized $\mathcal{O}(1)$.

Time Complexity: $\mathcal{O}(NE)$