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>

#define debug(x) cerr << #x << ": " << x << endl
#define debugArray(x, n)                                                       \
    for(long long hoge = 0; (hoge) < (n); ++(hoge))                            \
    cerr << #x << "[" << hoge << "]: " << x[hoge] << endl
using namespace std;

using ll = long long;
using ull = unsigned long long;
const ll INF = LLONG_MAX / 2;
const ll MOD = 1e9 + 7;

signed main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    int N;
    cin >> N;
    int h[N];
    for(int i = 0; i < N; i++) {
        cin >> h[i];
    }
    int ans = 0;
    bool update = true;
    while(update) {
        update = false;
        for(int l = 0; l < N; l++)
            if(h[l]) {
                update = true;
                int hmin = h[l];
                int r;
                for(r = l + 1; r < N && h[r]; r++) {
                    hmin = min(hmin, h[r]);
                }
                ans += hmin;
                for(int i = l; i < r; i++) {
                    h[i] -= hmin;
                }
                break;
            }
    }
    cout << ans << endl;
    return 0;
}