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<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
using namespace std;

int main(void){
	ios::sync_with_stdio(false); cin.tie(0);
	
	int n,i,j,k;
	cin>>n;
	int h[n];
	for(i=0;i<n;i++)cin>>h[i];
	int place=0,ans=0;
	for(i=0;i<n-1;i++){
		if(h[i]<=0)continue;
		for(k=0;k<h[i];k++){
			ans++;
			j=i+1;
			while(1){
				h[j]--;
				if(h[j]<0)break;
				j++;
			}
		}
//		cout<<ans<<endl;
	}
	if(h[n-1]>0)ans+=h[n-1];
	cout<<ans<<endl;
	return 0;
}