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: Time Limit Exceeded)

#include<stdio.h>
int main(){
  int n,h[100],i,j,max=0,min,sum=0;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&h[i]);
  }
  for(i=1;i<n;i++){
    if(h[i]>=h[i-1]){
      max=i;
    }
    else {
      sum+=h[max];
      for(j=i-1;j<n;j++){
        if(h[j]<h[j-1]){
          min=j;
        }
        else {
          sum-=h[min];
          i=j-1;
          break;
        }
      }
    }
  }
  if(h[n-1]>h[n-2]){
    sum+=h[max];
  }
  printf("%d",sum);
}