Iāll have a stab at this - completely un-proof read. Please point out mistakes/ confusions 
The key point is that the sum of any break-down into cups of ranges is actually a sum of suffix-sums of a.
Assume 1-relative a with elements a_1, a_2, ā¦ a_N.
Assume weāve picked an assignment of subranges into cups. We can represent this by a sequence c_1, c_2, ā¦ c_p where c_i is the index in a of the last element in cup i. (c_{i+1} > c_i for all valid i, and c_p=N).
In other words, our assignment is:
a_1, a_2, ... a_{c_1} (cup 1)
a_{c_1 + 1}, a_{{c_1} + 2}, ... a_{c_2} (cup 2)
ā¦
a_{c_{p-2} + 1}, a_{c_{p-2} + 2}, ... , a_{c_{p-1}} (cup p - 1).
Iāll have a stab at this - completely un-proof read. Please point out mistakes/ confusions
The key point is that the sum of any break-down into cups of ranges is actually a sum of suffix-sums of a.
Assume 1-relative a with elements a_1, a_2, ā¦ a_N.
Assume weāve picked an assignment of subranges into cups. We can represent this by a sequence c_1, c_2, ā¦ c_p where c_i is the index in a of the last element in cup i. (c_{i+1} > c_i for all valid i, and c_p=N).
In other words, our assignment is:
a_1, a_2, ... a_{c_1} (cup 1)
a_{c_1 + 1}, a_{{c_1} + 2}, ... a_{c_2} (cup 2)
ā¦
a_{c_{p-2} + 1}, a_{c_{p-2} + 2}, ... , a_{c_{p-1}} (cup p - 1).
a_{c_{p-1} + 1}, a_{c_{p-1} + 2}, ... , a_{c_p} (cup p).
$a_{c_{p-1} + 1}, a_{c_{p-1} + 2}, ... , a_{c_p}$ (cup $p$)
or in other words:
\overbrace{a_1, a_2, ... ,a_{c_1}}^{\text{cup }1},\overbrace{a_{c_1 + 1}, a_{{c_1} + 2}, ... ,a_{c_2}}^{\text{cup }2},a_{{c_2}+1},\dots, \overbrace{a_{c_{p-2} +1},\dots, a_{c_{p-1}-1},a_{c_{p-1}}}^{\text{cup }(p-1)},\overbrace{a_{c_{p-1} + 1}, a_{c_{p-1} + 2}, ... , a_{c_p}}^{\text{cup }p}
Our potency then is:
1 \times \sum_{i = 1}^{c_1} a_i + 2 \times \sum_{i = c_1+1}^{c_2} a_i + ... + (p-1) \times \sum_{i = c_{p-2} + 1}^{c_{p-1}} a_i + p \times \sum_{i = c_{p-1} + 1}^{c_p} a_i
Letās examine the last two terms in more detail. Note that, since c_p=N, the last term is actually a suffix sum of a, multiplied by p.
(p-1) \times \sum_{i = c_{p-2} + 1}^{c_{p-1}} a_i + p \times \sum_{i = c_{p-1} + 1}^{c_p} a_i=
(p-1) \times \sum_{i = c_{p-2} + 1}^{c_{p-1}} a_i + (p - 1) \times \sum_{i = c_{p-1} + 1}^{c_p} a_i +\sum_{i = c_{p-1} + 1}^{c_p} a_i=
(p-1) \times \sum_{i = c_{p-2} + 1}^{c_{p}}a_i+\sum_{i = c_{p-1} + 1}^{c_p} a_i
Whatās happened here? We started off with some random term multiplied by p-1 and a suffix sum multiplied by p. Now, weāve got one suffix sum, and another suffix sum multiplied by p-1.
Letās look at the last three terms that we have now:
(p-2) \times \sum_{i = c_{p-3} + 1}^{c_{p-2}} a_i + (p-1) \times \sum_{i = c_{p-2} + 1}^{c_{p}} a_i +\sum_{i = c_{p-1} + 1}^{c_p} a_i
We can now repeat the procedure for the terms beginning with (p-2) and (p-1), ending up, I think, with:
(p-2) \times \sum_{i = c_{p-3} + 1}^{c_{p}} a_i + \sum_{i = c_{p-2} + 1}^{c_{p}} a_i +\sum_{i = c_{p-1} + 1}^{c_p} a_i
So the term beginning with (p-2) has now also become a suffix sum (multiplied by (p-2)).
In this way, we gradually, from the right hand side, break down our original sum into a sum of p (distinct) suffix sums - a formal proof via induction would probably be quite easy, but I canāt be bothered :). Finding the p suffix sums with minimum sum should then give us our minimum potency.
