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;

/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x, long long int y) {
	return(x + y - 1) / y;

}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;

	if (number2 > number1) {
		m = number2;
		n = number1;
	}
	long long int s = -1;
	while (s != 0) {
		s = m % n;
		m = n;
		n = s;
	}
	return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;
	return m / lcm(m, n) * n;
}
int main() {
	int N;
	cin >> N;
	vector<int> s;
	int mins = 10000;
	int maxs = 0;
	for (int i = 0; i < N; i++) {
		int a;
		cin >> a;
		s.push_back(a);
		mins = min(a, mins);
		maxs = max(a, maxs);
	}
	int sum = 0;
	for (int i = 1; i <= maxs; i++) {
		int minss = -2000;
		for (int j = 0; j < N; j++) {
			if (s[j] >= i) {
				if (minss+1 != j) {
					sum++;
				}
				minss = j;
			}
		}
	}
	cout << sum;
}