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 main(void) {
	int n, h[100];
	int flower[100], l, r;
	int i, cnt = 0;
	l = r = 0;

	scanf("%d", &n);
	for (i = 0; i < n; i++){
		scanf("%d", &h[i]);
		flower[i] = 0;
	}

	while (1) {
		l = 0;
		for (i = 0; i < n; i++) {
			if (flower[l] < h[l]) {
				break;
			}
			l++;
		}

		r = l;
		for (i = 0; i <= n; i++) {
			if (flower[r] >= h[r]) {
				break;
			}
			r++;
		}

		/*
		printf("%d - %d\n", l, r);
		for (i = 0; i < n; i++) {
			printf("%d(%d) ", flower[i], h[i]);
		}
		printf("\n\n");
		*/

		if (l == n && r == n) {
			printf("%d\n", cnt);
			return 0;
		}

		for (i = l; i < r ; i++) {
			flower[i]++;
		}

		cnt++;
	}

	return 0;
}