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>
#include<string.h>
int main(void){
    int N,h[100],H[100],i,a,count=0,all=0;
    scanf("%d",&N);
    for(i=0;i<N;i++){
        scanf("%d",&h[i]);
        H[i]=0;
    }

    for(i=0;i<N;i++){
        while(1){
            if(h[i]==0){
                break;
            }
            H[i]++;
            count++;
            all++;
            if(h[i]==H[i]){
                
                if(H[i]<h[i+1]){
                    H[i+1]+=count;
                }else if(H[i]>h[i+1]){
                    H[i+1]=h[i+1]-1;
                    count=h[i+1]-1;
                    all--;
                    if(i==N-1){
                        all++;
                    }
                }
                break;
                
            }
        }
    }

    
    printf("%d",all);
    
    return 0;
}