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<algorithm>
#include<cmath>
#include<cstdio>
#include<iostream>
#include<map>
#include<numeric>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<unordered_set>
#include<vector>

#define ll long long
#define P pair<int,int>
#define FOR(i,N) for(int i=0;i<(int)N;i++)
#define FORIN(i,a,b) for(int i=a;i<(int)b;i++)
#define ALL(x) (x).begin(),(x).end()
#define LAST(x) (x)[(x).size()-1]

using namespace std;

string to_hex(int x){
    stringstream ss;
    ss<<hex<<x;
    return ss.str();
}

inline int get_int(){
    int ret;
    scanf("%d",&ret);
    return ret;
}

inline vector<int> get_ints(int n){
    vector<int> ret(n);
    FOR(i,n){
        scanf("%d",&ret[i]);
    }
    return ret;
}

inline string get_str(){
    string ret;
    cin>>ret;
    return ret;
}

bool is_prime(int n){
    int s=sqrt(n)+1;
    for(int i=2;i<=s;++i){
        if(n%i==0){
            return 0;
        }
    }
    return 1;
}

vector<P> prime_division(int n){
    vector<P> ret;
    int s=sqrt(n)+1;
    int c=0;
    for(int i=2;i<=s;++i){
        if(n%i==0){
            c=0;
            do{
                ++c;
                n/=i;
            }while(n%i==0);
            ret.push_back(P(i,c));
        }
    }
    return ret;
}

string to_string(string s){
    return s;
}

template<class T>
string to_string(vector<T> v){
    string ret="{";
    FOR(i,v.size()-1){
        ret+=to_string(v[i])+",";
    }
    if(v.size()>0){
        ret+=to_string(LAST(v));
    }
    ret+="}";
    return ret;
}

void debug(){
    cerr<<endl;
}

template<class Head,class... Tail>
void debug(Head head,Tail... tail){
    cerr<<" "<<to_string(head);
    debug(tail...);
}

int main(){
    int N;
    cin>>N;
    int last_h;
    cin>>last_h;
    int c=0;
    FOR(i,N-1){
        int h;
        cin>>h;
        if(h<last_h){
            c+=(last_h-h);
        }
        last_h=h;
    }
    c+=last_h;
    cout<<c<<endl;
    return 0;
}