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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int counter(int count, int n) {
	for (int i =0; i < n; i++) {
		count++;
	}
	return count;
}

int main() {
	int N;
	scanf("%d",&N);
	int hight[N];
	for (int i = 0; i < N; i++) {
		scanf("%d",&hight[i]);
	}
	int count=0,a,b=0,c=0;
	for (;;) {
		b = 1;
		for (int i = 0; hight[i] == 0; i++) c = i;
			//printf("%d\n",c);
			if (c == N-1) break;
		for (int i = 0; i < N; i++) {
			if (hight[i] == 0) {
				if (i != c && i != N-1 && hight[i+1] != 0) b++;
			}
			if (hight[i] != 0) hight[i] = hight[i] - 1;
		}
		/*for (int i = 0; i < N; i++) {
			printf("%d ",hight[i]);
		}*/
		count = counter(count,b);
		//printf("%d\n",count);
	}
	printf("%d\n",count);
	return 0;
}