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;

/* macro */
#define rep(i,a,b) for(int i=a;i<b;i++)
#define revrep(i,a,b) for(int i = a; i > b; i--)
#define int long long
#define exist(s,e) ((s).find(e)!=(s).end())
#define all(v) (v).begin(), (v).end()
#define each(s,itr) for(auto (itr) = s.begin(); (itr) != s.end(); (itr)++)
#define sum(v) accumulate(all(v), (0LL))
#define isin(tx, ty, w, h) (0 <= tx && tx <= w && 0 <= ty && ty <= h)

/* alias */
template<class T> using vec = vector<T>;
typedef vector<int> vi;
typedef pair<int, int> pi;

/* constant */
const int inf = pow(2, 62);
const int mod = 1e9 + 7;
int dx[8]={1,0,-1,0,-1,1,-1,1};
int dy[8]={0,1,0,-1,-1,-1,1,1};

/* io_method */
int input(){int tmp;cin >> tmp;return tmp;}
string raw_input(){string tmp;cin >> tmp;return tmp;}
template<class T> void printx(T n){cout << n;}
template<class T, class U> void printx(pair<T, U> p){cout << "(" << p.first << "," << p.second << ")";}
template<class T> void printx(vector<T> v){cout << "{";rep(i,0,v.size()){printx(v[i]);if(i != v.size()-1)printx(",");}cout << "}";}
template<class T> void print(T n){printx(n);cout << endl;}
template<class T> void print(set<T> s){cout << "{";each(s, e){if(e != s.begin()) printx(",");printx(*e);}cout << "}" << endl;}
template<class T, class U> void print(map<T, U> mp){cout << "{";each(mp, e){cout << "(" << e -> first << "," << e -> second << ")";}cout << "}" << endl;}

/* general_method */
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T>T cut(T &a, int l, int r){return T(a.begin()+l, a.begin()+r);}

/* main */

int n;
vi arr;

void in(){
    cin >> n;
    rep(i,0,n) arr.push_back(input());
}

void solve(){

    int ans = 0;
    while(true){
        int l = 0;
        while(l < n && arr[l] == 0) l++;

        if(l == n) break;
        int r = l;
        while(r + 1 < n && arr[r + 1] != 0) r++;

        rep(i,l,r + 1){
            arr[i]--;
        }
        ans++;
    }
    print(ans);

}

signed main(){

    cin.tie(0);
    ios::sync_with_stdio(false);
    in();
    solve();
    
}