Example problem: (I dont have the link as this problem only exists in my school's private judge)

We have an array of n positive intergers and we need to group them into k segments such that each element of the array is in a segment and that all elements in a segment are contigous. The cost of a segment is the sum of the numbers in the segment squared.

We need to minimize the total cost of all the segments.

Example 1, we have n=5 and k=3. Where the array is [1,2,3,4,5]. The optimal answer is by grouping (1+2+3) , (4) and (5) into different segments. The total cost is 6^2 + 4^2 + 5^2 =77.

Example 2, we have n=4, k=2. Where the array is [1,1000000,3,4]. The optimal answer is by grouping (1,1000000) and (3,4) into different segments. The total cost is 1000001^2+7^2=10000200050

Now I asked some people and they said this can be solved by doing the lagrange speedup trick. We shall define a cost C. Then do a convex hull trick where we try to minimize the value of all our segments but we also add C to our cost whenever we make a new segment. Now, if C is INF, then we will only have 1 segment, while if C is 0 we will have n segments. So people claim that we can binary search on C to find some C where there will be K segments existing.

This allowed me to AC the problem, but my solution was not always correct as the test cases were weak. I thought of this test case: n=4, k=3. Where is array is [1,1,1,1] Now, when C is 1, we get 4 segments. But when C is 2, we get 2 segments only.