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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<bool> vb;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define in insert
#define FOR(i,a,b) for (int i = a; i < b; i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
const int INF = 1e9 + 5;
int main(){
    int n;
    cin >> n;
    vi a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    int cnt = 0;
    bool q = true;
    while(q){
        int l = -1, r = -1;
        for(int i = 0; i < n; i++){
            if(a[i] > 0){
                l = i;
                break;
            }
        }
        if(l == -1) break;
        for(int i = l; i < n; i++){
            if(a[i] == 0){
                r = i - 1;
                break;
            }
        }
        if(r == -1) r = n-1;
        int mn = (*min_element(a.begin() + l, a.begin() + r + 1));
        for(int i = l; i <= r; i++){
            a[i] -= mn;
        }
        cnt += mn;
    }
    cout << cnt;
}