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 <stdio.h>
#include <iostream>
#include <vector>
#include <cassert>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <math.h>
#include <map>
using namespace std;

int n;
vector<int> h;

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        int tmp;
        cin >> tmp;
        h.push_back(tmp);
    }

    int res = 0;
    int ct = 0;
    while (true)
    {
        ct++;
        //水やり終わってるかチェック
        bool flag = true;
        for (int i = 0; i < n; i++)
        {
            if (h[i] != 0)
            {
                flag = false;
            }
        }
        if (flag)
        {
            cout << res;
            return 0;
        }

        //rとl決定
        int l = -1, r = -1;
        for (int i = 0; i < n; i++)
        {
            if (h[i] != 0 && l == -1)
            {
                //cout << "L!" << endl;
                l = i;
                continue;
            }

            if (l != -1 && h[i] == 0)
            {
                //cout << "R!" << endl;
                r = i - 1;
                break;
            }
        }

        if (r == -1)
        {
            r = n - 1;
        }

        //cout << l << " " << r << endl;

        //水やり実行
        res++;
        for (int i = l; i <= r; i++)
        {
            h[i]--;
        }
    }
    return 0;
}