若 $b_1 > 0$ 或者 $b_n < 0$ 那么肯定不是最优解，因为此时把 $b_1$ 或者 $b_n$ 设置成 0 不会使得结果更差，同时也能够保持数组递增。如果 $b_i - a_i > b_{i - 1}$，此时也不是最优解，此时把 $b_i$ 变成 $b_i - a_i$ 不会使得结果更差，也能够保持数组递增。猜测最终的数组 $b$ 中必有一个元素是 $0$（如何证明？），然后枚举为 $0$ 的位置即可。

根据上面的分析，最终的数组满足 $b_1 \le 0, b_n \ge 0$，由于数组 $b$ 是严格递增的，所以其中必定有一个元素是 $0$。

// Date: Fri Nov 24 19:17:58 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 5010;
int t, n, a[N];

int main(void) {
#ifdef _DEBUG
  freopen("1667a.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  while (cin >> n) {
    For1(i, 1, n) { cin >> a[i]; }

    if (n == 1) {
      cout << "0\n";
      continue;
    }

    ll res = -1;

    For1(i, 1, n) {
      ll sum = 0, pre = 0;

      Rof1(j, 1, i - 1) {
        if (pre == 0) {
          pre = a[j];
          sum++;
        } else {
          sum += pre / a[j];
          if (pre % a[j] == 0) {
            pre += a[j];
            sum++;
          } else {
            pre = pre - (pre % a[j]) + a[j];
            sum++;
          }
        }
      }

      pre = 0;
      For1(j, i + 1, n) {
        if (pre == 0) {
          pre = a[j];
          sum++;
        } else {
          sum += pre / a[j];
          if (pre % a[j] == 0) {
            pre += a[j];
            sum++;
          } else {
            pre = pre - (pre % a[j]) + a[j];
            sum++;
          }
        }
      }

      if (res == -1)
        res = sum;
      else
        res = min(res, sum);
    }

    cout << res << '\n';
  }

  return 0;
}