<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

342. Reihenfolge
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



Consider positive integers A and B. Your task is to represent A as the algebraic sum of integer powers of B with the minimal possible number of terms. In other words, A = s1 Bk1 + s2 Bk2 +... + sn Bkn, where si = -1 or si = 1, ki are integers and n should be minimized.

Input
The first line of the input contains positive integer A written without leading zeroes. A contains no more than 3000 digits. Second line contains integer B (1 ≤ B ≤ 106).

Output
Print one integer number n.

Example(s)
sample input
sample output
1120
10
4

<|response|>
## 1) Abridged problem statement (concise)

Given a huge positive integer **A** (up to 3000 decimal digits) and an integer **B** (1 ≤ B ≤ 10⁶), represent:

\[
A = \sum_{i=1}^{n} s_i \, B^{k_i}, \quad s_i \in \{-1, +1\},\ k_i \in \mathbb{Z}
\]

Minimize the number of terms **n**, and output this minimum.

---

## 2) Key observations

1. **Group equal exponents ⇒ coefficients per power**  
   Any expression can be regrouped as:
   \[
   A = \sum_k c_k B^k,\quad c_k\in\mathbb{Z}
   \]
   And each coefficient \(c_k\) needs at least \(|c_k|\) terms of \(\pm B^k\).  
   So minimizing the number of terms is equivalent to minimizing:
   \[
   n = \sum_k |c_k|
   \]

2. **Negative exponents are unnecessary in an optimal solution**  
   Negative powers \(B^{-t}\) create fractions. To still sum to integer \(A\), fractional parts must cancel, which never beats “carrying” that contribution to nonnegative exponents.  
   Therefore, an optimal solution exists with **only \(k \ge 0\)**.

3. **So the task becomes: find a signed-digit base‑B expansion minimizing sum of absolute digits**  
   Write \(A\) in base \(B\): \(A=\sum d_i B^i,\ 0\le d_i<B\).  
   At each digit you may keep digit \(r\) or replace it by \(r-B\) (negative) and carry +1 upward.

4. **Carry is only 0 or 1** (for \(B\ge 2\))  
   When processing digit \(d_i\) with incoming carry \(c\in\{0,1\}\), let \(t=d_i+c\).  
   - If you choose digit \(t\bmod B = r\), next carry is \(q=\lfloor t/B\rfloor\in\{0,1\}\).
   - If you choose digit \(r-B\) (only if \(r\ne 0\)), next carry becomes \(q+1\). This only occurs when \(q=0\), so the next carry is still **1**.

5. **Special case \(B=1\)**  
   Every power \(1^k=1\). The only way to sum to \(A>0\) with minimal terms is \(A\) times “+1”.  
   Answer: **n = A**.

---

## 3) Full solution approach

### Step A: Handle \(B=1\)
Print \(A\) (as a big integer). Done.

### Step B: Convert A to base B digits (for \(B\ge 2\))
Compute digits \(d_0,d_1,\dots,d_{L-1}\) (least significant first) using repeated:
- `d = A % B`
- `A = A / B`
This needs big-integer division by an int (fast enough).

### Step C: Dynamic Programming over digits with carry
Define:

- `dp[i][c]` = minimal cost (minimal number of terms) to represent digits from position `i` upward, given incoming carry `c` (0 or 1) added to digit `i`.

Base case (past most significant digit, i = L):
- `dp[L][0] = 0`
- `dp[L][1] = 1` (a leftover carry corresponds to one extra \(+B^L\) term)

Transition for i = L-1 down to 0:

Let `t = d[i] + c`  
Let `r = t % B`, `q = t / B`

Two choices:

1. Use digit `r` (nonnegative):
   - cost = `r`
   - next carry = `q`
   - candidate = `r + dp[i+1][q]`

2. Use digit `r - B` (negative), only if `r != 0`:
   - cost = `|r-B| = B-r`
   - next carry = `q+1` (which is 1 here)
   - candidate = `(B-r) + dp[i+1][q+1]`

Take minimum.

Answer is `dp[0][0]`.

### Complexity
Let \(L\) be number of base‑\(B\) digits of \(A\).  
- Base conversion: \(O(L)\) big-int `/` and `%` by int  
- DP: \(O(L)\)  
Memory: \(O(L)\) or \(O(1)\) with rolling variables.

---

## 4) C++ implementation (detailed comments)

```cpp
#include <bits/stdc++.h>
using namespace std;

/*
  We need big integer only for:
  - reading A (up to 3000 digits)
  - repeated division by int B (<= 1e6) to extract base-B digits
  - printing A when B==1

  Below is a compact bigint supporting:
  - construction from string
  - isZero()
  - operator%(int)
  - operator/=(int)
  - stream output
*/

struct BigInt {
    static const int BASE = 1000000000; // 1e9 per limb
    static const int BASE_DIGS = 9;

    vector<int> a; // little-endian limbs (a[0] least significant)

    BigInt() {}
    BigInt(const string& s) { read(s); }

    void trim() {
        while (!a.empty() && a.back() == 0) a.pop_back();
    }

    bool isZero() const { return a.empty(); }

    void read(const string& s) {
        a.clear();
        // s is non-negative in this problem
        for (int i = (int)s.size(); i > 0; i -= BASE_DIGS) {
            int x = 0;
            int l = max(0, i - BASE_DIGS);
            for (int j = l; j < i; j++) x = x * 10 + (s[j] - '0');
            a.push_back(x);
        }
        trim();
    }

    // remainder by int v
    int operator%(int v) const {
        long long rem = 0;
        for (int i = (int)a.size() - 1; i >= 0; --i) {
            rem = (rem * BASE + a[i]) % v;
        }
        return (int)rem;
    }

    // divide by int v (v>0)
    void operator/=(int v) {
        long long rem = 0;
        for (int i = (int)a.size() - 1; i >= 0; --i) {
            long long cur = a[i] + rem * BASE;
            a[i] = (int)(cur / v);
            rem = cur % v;
        }
        trim();
    }

    friend ostream& operator<<(ostream& os, const BigInt& x) {
        if (x.a.empty()) return os << 0;
        os << x.a.back();
        for (int i = (int)x.a.size() - 2; i >= 0; --i) {
            os << setw(BASE_DIGS) << setfill('0') << x.a[i];
        }
        return os;
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string A_str;
    int B;
    cin >> A_str >> B;

    BigInt A(A_str);

    // Special case B=1:
    // All powers are 1, so A = (+1)+(+1)+... exactly A times, minimal.
    if (B == 1) {
        cout << A << "\n";
        return 0;
    }

    // Convert A to base-B digits (least significant first)
    vector<int> d;
    BigInt x = A;
    while (!x.isZero()) {
        d.push_back(x % B);
        x /= B;
    }
    int L = (int)d.size();

    // DP: dp[i][carry], carry in {0,1}
    // We'll store dp for all i to keep code straightforward (O(L) memory).
    vector<array<long long, 2>> dp(L + 1);

    // Base case after the top digit
    dp[L][0] = 0;
    dp[L][1] = 1; // leftover carry => one extra +B^L term

    // Process from most significant digit down to least
    for (int i = L - 1; i >= 0; --i) {
        for (int carry = 0; carry <= 1; ++carry) {
            int t = d[i] + carry;
            int r = t % B;
            int q = t / B; // 0 or 1

            // Option 1: use digit r
            long long best = (long long)r + dp[i + 1][q];

            // Option 2: use digit r-B (negative) if r != 0
            // cost = B-r, next carry becomes q+1 (which is 1 in this branch)
            if (r != 0) {
                best = min(best, (long long)(B - r) + dp[i + 1][q + 1]);
            }

            dp[i][carry] = best;
        }
    }

    cout << dp[0][0] << "\n";
    return 0;
}
```

---

## 5) Python implementation (detailed comments)

```python
import sys

def solve() -> None:
    A_str = sys.stdin.readline().strip()
    B = int(sys.stdin.readline().strip())

    A = int(A_str)  # Python supports arbitrary precision integers

    # B = 1: only term values are ±1, so to get positive A minimally we use A times +1
    if B == 1:
        print(A)
        return

    # Convert A to base-B digits (least significant first)
    digits = []
    x = A
    while x > 0:
        digits.append(x % B)
        x //= B

    L = len(digits)

    # dp_next[c] = dp[i+1][c] while iterating i downward
    # Base: dp[L][0] = 0, dp[L][1] = 1
    dp_next0, dp_next1 = 0, 1

    # Process from most significant digit to least
    for i in range(L - 1, -1, -1):
        di = digits[i]

        # Compute dp[i][0]
        t = di
        r = t % B
        q = t // B  # always 0 because di < B
        best0 = r + (dp_next0 if q == 0 else dp_next1)
        if r != 0:
            # choose digit r-B => cost B-r, carry becomes 1
            best0 = min(best0, (B - r) + dp_next1)

        # Compute dp[i][1]
        t = di + 1
        r = t % B
        q = t // B  # 0 or 1
        best1 = r + (dp_next0 if q == 0 else dp_next1)
        if r != 0:
            # negative digit branch, next carry = q+1 (only possible when q==0)
            best1 = min(best1, (B - r) + dp_next1)

        dp_next0, dp_next1 = best0, best1

    # answer is dp[0][0]
    print(dp_next0)

if __name__ == "__main__":
    solve()
```

This DP produces the minimum possible number of ± powers of B needed to sum to A within the given constraints.