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 9.2.1) (judge verdict: Runtime Error)

#include<stdio.h>
#include<stdlib.h>

int main(void){
  int N;
  int *p;
  
  scanf("%d",&N);
  p = (int *)malloc(sizeof(int)*N);
  
  char *s;
  gets(s);
  
  int ip=0;		//0<=ip<N
  int i=0;
  int j=0;
  int max=-1;
  int sum=0;
  char *num="\0";
  for(i=0; s[i]!='\0'||ip<N; i++){
    if(s[i]!=' '){
      num[j]=s[i];
      j++;
    }
    else{
      num[j]='\0';
      p[ip]=atoi(num);
      if(p[ip]>max){
        max=p[ip];
      }
      else{
        sum+=(max-p[ip]);
      }
      ip++;
      num="\0";
      j=0;
    }
  }
  
  printf("%d\n",sum);
  
 return 0; 
}