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: Runtime Error)

#include<stdio.h>
#include<stdbool.h>
int main(void){
  int n=0;
  int h[n];
  int mizuyariCount=0;
  int l=0;
  int r=n;
  bool allZero=false;
  
  //本数指定
  scanf("%d",&n);
  for(int i=0;i<n;i++){
    //要素ごとの高さ指定
    scanf("%d",&h[i]);
  }
  
  //ここからmizuyari を Count
  while(allZero==true){
    
  	// l の指定
  	for(int j=0;j<n;j++){
    	if(h[n]!=0){
      	  l=j;
          break;
    	}
    }
      // r の指定
      for(int k=0;k<n;k++){
        if(h[n]==0){
          r = k -1;
        }
      }
 
    //0でない部分で水やり    
    for(l;l<r;l++){
      h[l] = h[l]-1;
    }
    mizuyariCount++;
  }
  printf("%d",mizuyariCount);
  return 0;
  }