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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef pair<ll, int> li;
typedef pair<int, ll> il;
typedef vector<ii> vii;
typedef vector<il> vil;
typedef vector<li> vli;

#define F first
#define S second
#define pb(x) push_back(x)
#define mp(a,b) make_pair(a,b)
#define sz size()
#define all(A)    A.begin(), A.end()
#define mem(a, b) memset(a, b, sizeof(a))
#define f0(i,b)   for(int i=0;i<(b);i++)
#define f1(i,b)   for(int i=1;i<=(b);i++)
#define f2(i,a,b) for(int i=(a);i<=(b);i++)
#define fr(i,b,a) for(int i=(b);i>=(a);i--)
#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))
#define mod 1000000007

#define forr(i,n)    for(i=0; i<n; i++)
#define whilee(i,t)  while(i!=t)
#define  sl(a)  scanf("%lld",&a)
#define  pl(a)  printf("%lld",a)

#define scan(x) do{while((x=getchar())<'0'); for(x-='0'; '0'<=(_=getchar()); x=(x<<3)+(x<<1)+_-'0');}while(0)
char _;

//for(auto it =A.begin(); it!=A.end(); it++)
//sort(A.begin(),A.end());
//binary_search(A.begin(),A.end(),x);
//reverse(A.begin(), A.end());
//cout << *max_element(A.begin(), A.end());
//cout << *min_element(A.begin(), A.end());
//count(A.begin(), A.end(), X); //counts the occurrences of X
//distance(A.begin(),A.end());  //distance between first to last element
//accumulate(A.begin(), A.end(), 0); //add
//next_permutation(A.begin(), A.end());
//prev_permutation(A.begin(), A.end());
//swap(S[0], T[0]);
//scanf("%d",&x);
//printf("%d\n",x);

//vector <int> A;
//deque <int> A;
//queue <int> A;
//stack <int> A;
//set <int> A;
//map <int,int> A;
//pair <int,int> A ;
//vector <pair <int,int>> A ;
//bitset<32> bset1;
//char ch;
//string S, T, U;
ll i, j, k, a, b, c, d, f, g, h, l, m, n, p, q, r, s, t, u, v, w, x, y, z, maxx=INT_MIN, minn=INT_MAX, ans=0, sum=0, cnt=0;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);//NULL
    cout.tie(nullptr);//NULL

    cin>>n;
    int A[n+1];

    A[0]=0;
    f1(i,n)
    {
        cin>>A[i];
    }

    ans=0;
    x=0;
    y=0;
    forr(i,n)
    {
        if(A[i]<=A[i+1])
        {
            x = A[i+1];
        }
        else
        {
            ans += (x-y);
            x = A[i+1];
            y = A[i+1];
        }
    }
    ans += (x-y);


    cout<<ans<<endl;

    return 0;
}