<|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.

442. X + R(X) = N
Time limit per test: 0.75 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



Ruslan is crazy about counting numbers and solving problems. His favourite pastime is to make up a problem and solve it by himself. Some time ago he heard about a very interesting problem: given the positive integer N, you have to say whether such X that X + R(X) = N exists or not, where X is a positive integer, and R(X) is the number X written backwards. Then, Ruslan has decided that this task is elementary, so he didn't start solving it, but made up a more difficult problem instead.

You are given the positive integer number N. How many positive integer numbers X are there, that X + R(X) = N?

R(X) is the number X written backwards. For example: R(123) = 321 R(150) = 51

Input
Input will consist of multiple test cases. Each case will be a single line containing number N (). A line with a single zero terminates the input.

Maximum size of input is  bytes.

Output
Output for each test case should consist of a single integer on a line, indicating the number of numbers X satisfying the condition. Do not output leading zeros.

Example(s)
sample input
sample output
1
2
11
13
14003
767513456469789456166547987979741366664879441
0 
0
1
1
0
60
0

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

Given a positive integer **N** (as a decimal string; can be very large), count how many **positive integers X** satisfy:

\[
X + R(X) = N
\]

where **R(X)** is X written backwards in decimal (leading zeros are not kept, e.g. \(R(150)=51\)).  
Input has multiple test cases; a line containing `0` ends input.  
For each test case output the count (no leading zeros).

---

## 2) Key observations

1. **Digits interact in mirrored pairs.**  
   If \(X\) has digits \(x_0 x_1 \dots x_{m-1}\) (left to right), then \(R(X)\) has digits \(x_{m-1}\dots x_1 x_0\).  
   In the sum \(X + R(X)\), the **leftmost digit** depends on \(x_0 + x_{m-1}\) plus a carry from inside; the **rightmost digit** depends on the same pair \(x_{m-1}+x_0\) plus a carry from the right.

2. **The result length is either \(|X|\) or \(|X|+1\).**  
   - If there is **no final carry** beyond the most significant digit, then \(|X| = |N|\).
   - If there **is** a carry beyond the most significant digit, then \(N\) must start with `'1'`, and \(|X| = |N|-1\).

3. **Carries can be managed with a small DP.**  
   When processing digit pairs from the **outside in**, we need to track:
   - `carry_r` ∈ {0,1}: the usual carry coming into the current **right** digit (least significant side).
   - `exp_carry` ∈ {0,1}: what carry the **inner digits must generate** into the current **left** digit to match \(N\). (An “expected carry from inside”.)

4. **Counts can be huge.**  
   The number of valid X can be enormous for very long N, so use big integers:
   - Python: built-in `int`
   - C++: use `boost::multiprecision::cpp_int` (simpler than custom bigint)

---

## 3) Full solution approach

We solve each test case by summing two independent cases:

### Case A: \(|X| = |N|\) (no extra leading carry)
Run a DP on the full string `N` with initial `exp_carry = 0`.

### Case B: \(|X| = |N|-1\) (an extra leading carry created the first digit)
This is only possible if `N[0] == '1'` and `|N| > 1`.  
Drop that leading `'1'` and run DP on `N[1:]` with initial `exp_carry = 1`.

---

### DP for a given target string `S` and initial expected carry

Let `S` be the digits we want to match (either `N` or `N[1:]`).

We process indices:
- `left` from 0 upward
- `right` from len(S)-1 downward
so each step handles the pair `(left, right)`.

State:
- `dp[exp_carry][carry_r]` = number of ways to choose the already-processed outer digits of X such that the outer digits of the sum match.

Transition:
Choose digits:
- `a = X[left]`
- `b = X[right]`
Constraints:
- If `left == 0`, then `a != 0` (X is positive and has no leading zeros).
- If `left == right` (middle of odd length), then `a == b`.

Let:
- `dL = digit(S[left])`
- `dR = digit(S[right])`
- `sum = a + b`

**Right digit equation (normal addition):**
\[
(sum + carry_r) \bmod 10 = dR
\]
\[
new\_carry\_r = \left\lfloor \frac{sum + carry_r}{10} \right\rfloor
\]

**Left digit must match `dL`, given we currently “expect” `exp_carry` from inner digits.**  
The following rearrangement yields the expected carry for the next inner layer:
\[
new\_exp\_carry = dL + 10 \cdot exp\_carry - sum
\]
`new_exp_carry` must be either 0 or 1, otherwise impossible.

**Middle position handling (`left == right`):**
At the center, carries must be consistent from both sides; enforce:
- `new_carry_r == exp_carry`
and then we finish (collapse to a terminal state).

Finish:
After processing all pairs, valid completions are those where both carry notions match:
\[
answer = dp[0][0] + dp[1][1]
\]

Complexity:  
For length \(L\), there are \(O(L)\) positions; each checks at most 100 digit pairs and 4 carry states → very fast.

---

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

```cpp
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>

using namespace std;
using boost::multiprecision::cpp_int;

/*
  Count solutions for a fixed target digit string S, under a chosen initial
  expected carry on the left side.

  expecting_carry_initial:
    0 -> corresponds to case |X| == |S| (no extra leading carry)
    1 -> corresponds to case where original N had an extra leading '1' carry,
         and we are solving for S = N without that leading '1'.
*/
static cpp_int count_solutions(const string& S, int expecting_carry_initial) {
    int n = (int)S.size();
    if (n == 0) return 0;

    // dp[exp_carry][carry_r]
    cpp_int dp[2][2] = {};
    dp[expecting_carry_initial][0] = 1; // initially no carry coming into the rightmost digit

    int left = 0, right = n - 1;
    while (left <= right) {
        int dL = S[left]  - '0';
        int dR = S[right] - '0';
        bool is_middle = (left == right);

        cpp_int ndp[2][2] = {};

        for (int exp_carry = 0; exp_carry <= 1; exp_carry++) {
            for (int carry_r = 0; carry_r <= 1; carry_r++) {
                cpp_int ways = dp[exp_carry][carry_r];
                if (ways == 0) continue;

                // choose a = X[left]
                for (int a = 0; a <= 9; a++) {
                    // no leading zeros for X
                    if (left == 0 && a == 0) continue;

                    // choose b = X[right]
                    int b_lo = is_middle ? a : 0;
                    int b_hi = is_middle ? a : 9;

                    for (int b = b_lo; b <= b_hi; b++) {
                        int sum = a + b;

                        // Right digit constraint:
                        int right_val = sum + carry_r;
                        if (right_val % 10 != dR) continue;
                        int new_carry_r = right_val / 10; // 0 or 1 (since max 9+9+1=19)

                        // Expected carry for next inner layer on the left:
                        int new_exp_carry = dL + 10 * exp_carry - sum;
                        if (new_exp_carry < 0 || new_exp_carry > 1) continue;

                        if (is_middle) {
                            // At the center, carries must be consistent:
                            // the carry produced on the right must equal the carry expected on the left.
                            if (new_carry_r == exp_carry) {
                                // DP ends here; put all valid ways into a terminal bucket.
                                ndp[0][0] += ways;
                            }
                        } else {
                            ndp[new_exp_carry][new_carry_r] += ways;
                        }
                    }
                }
            }
        }

        // Move inward
        memcpy(dp, ndp, sizeof(dp));
        left++;
        right--;
    }

    // Final consistency: sum dp[c][c]
    return dp[0][0] + dp[1][1];
}

static cpp_int solve_one(const string& N) {
    // Case A: |X| == |N|, no extra leading carry
    cpp_int ans = count_solutions(N, 0);

    // Case B: |X| == |N|-1, possible only if N begins with '1' (the extra carry)
    if (N.size() > 1 && N[0] == '1') {
        ans += count_solutions(N.substr(1), 1);
    }
    return ans;
}

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

    string N;
    while (cin >> N) {
        if (N == "0") break;
        cout << solve_one(N) << "\n";
    }
    return 0;
}
```

---

## 5) Python implementation (detailed comments)

```python
import sys

def count_solutions(S: str, expecting_carry_initial: int) -> int:
    """
    Count number of positive integers X such that X + reverse(X) == S
    (where S is a decimal string), given an initial expected carry on the left.

    expecting_carry_initial:
      0 -> case |X| == |S|
      1 -> case original N had a leading carry, and S is N without its first '1'
    """
    n = len(S)
    if n == 0:
        return 0

    # dp[exp_carry][carry_r]
    dp = [[0, 0], [0, 0]]
    dp[expecting_carry_initial][0] = 1  # no carry coming into the rightmost digit initially

    left, right = 0, n - 1
    while left <= right:
        dL = ord(S[left]) - ord('0')
        dR = ord(S[right]) - ord('0')
        is_middle = (left == right)

        ndp = [[0, 0], [0, 0]]

        for exp_carry in (0, 1):
            for carry_r in (0, 1):
                ways = dp[exp_carry][carry_r]
                if ways == 0:
                    continue

                for a in range(10):
                    # X cannot start with 0
                    if left == 0 and a == 0:
                        continue

                    # b is free unless we're at the middle of an odd length
                    if is_middle:
                        b_iter = (a,)
                    else:
                        b_iter = range(10)

                    for b in b_iter:
                        s = a + b

                        # Right digit constraint:
                        right_val = s + carry_r
                        if right_val % 10 != dR:
                            continue
                        new_carry_r = right_val // 10  # 0 or 1

                        # Left digit constraint transformed:
                        new_exp_carry = dL + 10 * exp_carry - s
                        if new_exp_carry < 0 or new_exp_carry > 1:
                            continue

                        if is_middle:
                            # Center: carries must meet consistently
                            if new_carry_r == exp_carry:
                                ndp[0][0] += ways
                        else:
                            ndp[new_exp_carry][new_carry_r] += ways

        dp = ndp
        left += 1
        right -= 1

    # Final consistency: sum dp[c][c]
    return dp[0][0] + dp[1][1]


def solve_one(N: str) -> int:
    # Case A: same length
    ans = count_solutions(N, 0)

    # Case B: one less digit, only if leading digit is '1'
    if len(N) > 1 and N[0] == '1':
        ans += count_solutions(N[1:], 1)

    return ans


def main() -> None:
    out = []
    for line in sys.stdin:
        N = line.strip()
        if not N:
            continue
        if N == "0":
            break
        out.append(str(solve_one(N)))
    sys.stdout.write("\n".join(out))

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

This DP pattern (outside-in with two 0/1 carry states) is the core idea: it avoids constructing X explicitly and works even when N has hundreds (or thousands) of digits.