Problem:
Score : 300 points

Problem Statement

In a flower bed, there are N flowers, numbered 1,2,......,N. Initially, the heights of all flowers are 0.
You are given a sequence h=\{h_1,h_2,h_3,......\} as input. You would like to change the height of Flower k to h_k for all k (1 \leq k \leq N), by repeating the following "watering" operation:

Specify integers l and r. Increase the height of Flower x by 1 for all x such that l \leq x \leq r.

Find the minimum number of watering operations required to satisfy the condition.

Constraints

1 \leq N \leq 100

0 \leq h_i \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
h_1 h_2 h_3 ...... h_N

Output

Print the minimum number of watering operations required to satisfy the condition.

Sample Input 1

4
1 2 2 1

Sample Output 1

2

The minimum number of watering operations required is 2.
One way to achieve it is:

Perform the operation with (l,r)=(1,3).

Perform the operation with (l,r)=(2,4).

Sample Input 2

5
3 1 2 3 1

Sample Output 2

5

Sample Input 3

8
4 23 75 0 23 96 50 100

Sample Output 3

221

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let sub hs b = List.map (fun a -> a - b) hs 

let divide hs = 
  (*
  Printf.printf "hs@divide: %s\n" (List.fold_left (fun acc h -> (string_of_int h) ^ "; " ^ acc) "" hs);
   *)
  let rec inner left temp min_h = function
  | [] -> temp::left, min_h
  | h::hs when h = 0 -> inner (temp::left) [] 101 hs
  | h::hs -> inner left (h::temp) (min min_h h) hs
  in inner [] [] 101 hs

let rec solve = function
  | [] -> 0
  | hs -> 
     match divide hs with
     | [], min_h -> 0
     | hss, min_h when List.length hss = 1 -> min_h + (solve @@ sub hs min_h)
     | hss, _ -> List.fold_left (fun acc hs -> acc + solve hs) 0 hss

let _ =
  let n = Scanf.scanf "%d\n" (fun n -> n) in
  let hs = ref [] in
  for i = 0 to n - 1 do
    Scanf.scanf (if i <> n - 1 then "%d " else "%d\n") (fun h -> hs := h::!hs)
  done;
  solve !hs |> print_int