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 REP(i, s, n) for (int i = s; i < n; i++)
using namespace std;
typedef long long ll;

int N;
vector<int> H;

int calc(int l, int r) {
    if (l == r) return H[l];

    int min_value = H[l], min_idx = l;
    REP(i, l, r) {
        if (min_value > H[i]) {
            min_value = H[i];
            min_idx = i;
        }
    }

    if (min_value > 0) {
        REP(i, l, r) {
            H[i] -= min_value;
        }
        return min_value + calc(l, r);
    } else {
        return calc(l, min_idx) + calc(min_idx + 1, r);
    }
}

int main() {
    cin >> N;
    H.resize(N);
    REP(i, 0, N) cin >> H[i];
    cout << calc(0, N) << endl;
    return 0;
}