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: Accepted)

#include <stdio.h>

int main(void)
{
	int N;
	int h[100];
	int i,f,sum,color,max=0;
	
	scanf("%d",&N);
	
	for(i=0;i<N;i++){
		scanf("%d",&h[i]);
		if(h[i]>=max)max=h[i];
	}
	
	int H[100];
	
	for(i=0;i<max;i++){
		//Hにバーコードを作成
		for(f=0;f<N;f++){
			if(h[f]==0){
			H[f]=0;
			//printf("0 ");
			}else{
			H[f]=1;
			h[f]-=1;;
			//printf("1 ");
			}
		}
		//バーコード読み取り
		if(H[0]==1){sum+=1;}
		
		for(f=1;f<N+1;f++){
			if(H[f-1]==0&&H[f]==1){sum+=1;
			}
		}
		//printf("\n");
	
	}
	printf("%d",sum);
}