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>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include <time.h>
typedef long long llint;
typedef long double ldouble;
int compare(const void *a, const void *b){return *(llint *)a - *(llint *)b;}



int main(void){
    llint n,i,water=0,base=0,k=0;
    scanf("%lld",&n);
    llint h[n+1];
    for(i=1;i<n+1;i++){
        scanf("%lld",&h[i]);
    }
    h[0]=0;
    for(i=0;i<n;i++){
        if(h[i]<=h[i+1]) k=h[i+1]-base;
        if(h[i]>h[i+1]){
            base=h[i+1];
            water+=k;
            k=0;
        }
    }
    water+=k;

    printf("%lld\n",water);
    return 0;
}