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 Lua (5.3.2) (judge verdict: Accepted)

local n = io.read("*n")
local h = {}
for i = 1, n do h[i] = io.read("*n") end
local tasks = {}
local tasknum, done = 0, 0
table.insert(tasks, 1 + n * 101)
tasknum = 1

local function getmin(l, r)
  local min, minpos = h[l], l
  for i = l + 1, r do
    if(h[i] < min) then
      min, minpos = h[i], i
    end
  end
  return min, minpos
end

local tot = 0
local curtask = 0
while(done < tasknum) do
  done = done + 1
  local task = tasks[done]
  local left, right = task % 101, task // 101
  local min, minpos = getmin(left, right)
  for i =left, right do
    h[i] = h[i] - min
  end
  if(left < minpos) then
    tasknum = tasknum + 1
    table.insert(tasks, left + (minpos - 1) * 101)
  end
  if(minpos < right) then
    tasknum = tasknum + 1
    table.insert(tasks, minpos + 1 + right * 101)
  end
  tot = tot + min
end
print(tot)