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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rp(i,n) for(int i=0; i < n; i++)
using namespace std;
int main(void){
  int n;
  cin >> n;
  vector<int> h(n+2);
  h[0]=0; h[n+1]=0;
  for(int i=1; i < n+1; i++){
    cin >> h[i];
  }
  int m[n+2];
  rp(i,n+1){m[i] = 1;}
  m[0]=0; m[n+1]=0;
  int count = 1;
  int sum = 0;
  rp(i,100){
    for(int j=1; j < n+1; j++){
      if(h[j]==i){
        if(m[j+1]==1 && m[j-1] == 1){
          sum = sum+2;
        }
        else if(m[j+1]==1 && m[j-1] == 0){
          sum = sum+1;
        }
        else if(m[j+1]==0 && m[j-1] == 1){
          sum = sum+1;
        }
        else if(m[j+1]==0 && m[j-1] == 0){
          sum = sum;
        }
        m[j] = 0;
      }
    }
    count *= sum;
    sum = 0;
  }
  cout << count << endl;
}