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;

int n, h[111];

int func(int l, int r){
	int mi = 1e9;
	for(int i=l; i<=r; ++i) mi = min(mi, h[i]);
	int res = mi;
	for(int i=l; i<=r; ++i) h[i] -= mi;
	for(int i=l; i<=r; ++i){
		if(h[i] > 0){
			int j = i;
			while(j+1 <= r && h[j+1] > 0) ++j;
			res += func(i, j);
		}
	}
	return res;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n;
	for(int i=1; i<=n; ++i) cin >> h[i];
	cout << func(1, n) << "\n";
}