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;

// 区間[i, j)に水やりをする関数
// 戻り値は水やりをした回数
int water(vector<int> &hs, int i, int j) {
  // 対象がないときにresult = 200が返っちゃう
  if (i >= j) { return 0; }

  int result = 200;
  for (int p = i; p < j; p++) {
    result = min(result, hs[p]);
  }
  for (int p = i; p < j; p++) {
    hs[p] -= result;
  }
  return result;
}

int main(void) {
  int n;
  cin >> n;
  vector<int> hs(n + 1, 0);
  for (int i = 0; i < n; i++) {
    cin >> hs[i];
  }

  int result = 0;
  int head = 0;
  int tail = 0;

  for (int i = 0; i < 100 * 1000; i++) {
    while (hs[head] == 0) { head++; }
    tail = head + 1;
    while (hs[tail] != 0) { tail++; }

    // 区間[i, j)に対して水やり
    if (head > n) { break; }
    result += water(hs, head, tail);
  }

  cout << result << endl;

  return 0;
}