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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = unsigned long long;

int main() {
  int N;
  cin >> N;
  vector<int> h(N);
  for (int i = 0; i < N; ++i) {
    cin >> h[i];
  }
  
  int result = 0;
  while (true) {
    int l = N, r;
    int cnt = 0;
    bool flg = false;
    for (int i = 0; i < N; ++i) {
      int hi = h[i];
      if (0 < hi) {
        l = min(l, i);
        flg = true;
      }
      if (hi == 0 && flg) {
        r = i - 1;
        for (int j = l; j <= r; ++j) {
          --h[j];
        }
        ++result;
        break;
      } else if (i == N - 1 && flg) {
        r = i;
        for (int j = l; j <= r; ++j) {
          --h[j];
        }
        ++result;
      } else if (hi == 0) {
        ++cnt;
      }
    }
    if (cnt == N) {
      break;
    }
  }
  cout << result << endl;
}