## 1) Abridged problem statement

For each test case, you are given a positive integer **N** (as a decimal string, potentially very large). Count how many **positive integers X** satisfy:

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

where **R(X)** is the decimal representation of **X** reversed (leading zeros dropped in the reversed number, e.g., \(R(150)=51\)).  
Input ends with a line containing a single `0`. For each test case, output the count (as an integer with no leading zeros).

---

## 2) Detailed editorial (how the solution works)

### Key observations

Let \(X\) be written in decimal with digits \(x_0 x_1 \dots x_{m-1}\) (left to right), so \(x_0 \neq 0\).  
Then \(R(X)\) has digits \(x_{m-1}\dots x_1 x_0\) (but note reversing can drop trailing zeros of X; that is naturally handled if we treat X as an integer with no leading zeros).

When adding \(X + R(X)\), digit positions pair up:

- The **leftmost** digit of the sum depends on \(x_0 + x_{m-1}\) plus a carry coming from the *inside*.
- The **rightmost** digit depends on \(x_{m-1} + x_0\) plus a carry coming from the *outside*.

So it’s natural to process digits in **pairs from the outside in**:
- pair \((x_L, x_R)\) where \(L\) goes from left to center and \(R\) goes from right to center.

Carries are the tricky part because:
- The carry into the **right end** (least significant digit) is the usual carry from adding the previous (more inner) digits when going from right to left.
- But if we process pairs from outside inward, we need to account for both:
  - carry propagating from the right side (least significant side) inward/outward,
  - and what carry we must “expect” on the left side to make the left digit match.

### Two length cases

Let \(n = |N|\).

When adding \(X + R(X)\), the result can have either:
1. **Same length as X** (no extra carry creating a new most significant digit).
2. **One more digit than X** (a carry creates a leading `1` in N).

Equivalently, for a given N, there are two scenarios for X:

- **Case A (no outer carry):** \(|X| = |N|\). Then there is **no** carry “beyond” the leftmost digit. This is modeled by starting with `expecting_carry_initial = 0` on the left side and processing the full string N.

- **Case B (leading 1 created):** \(|X| = |N|-1\) and \(N\) must start with `'1'` (because the only possible leading digit created by a carry in decimal addition is 1). In this case, we drop that leading `'1'` from N and solve on \(N' = N[1:]\) while starting with `expecting_carry_initial = 1`.

The final answer is:
- solutions for Case A, plus
- solutions for Case B (only if \(N[0]='1'\) and \(n>1\)).

### DP state definition

For a given target string **S** (either N or N without its leading 1), we process positions:

- `left` from 0 upward
- `right` from |S|-1 downward

At each step we choose digits:
- \(a = x_{left}\)
- \(b = x_{right}\)
with the constraint \(a \neq 0\) if `left == 0` (no leading zeros in X).

If `left == right` (middle digit in odd length), then we must have \(a=b\).

We maintain a DP over two binary carry-like values:

- `exp_carry` (0/1): what carry we are **expecting on the left digit** of S at the current outer position due to addition of the current pair and inner carries.
- `carry_r` (0/1): the carry coming **from the right side** into the current right digit (the normal addition carry for the least significant direction).

DP meaning (as implemented):
- `dp[exp_carry][carry_r] = number of ways` to choose already-processed outer pairs so that they match the already-fixed outer digits of S.

Transition when processing a pair:

Let:
- `d_left` = digit S[left]
- `d_right` = digit S[right]
- sum = a + b

**Right digit constraint (least-significant-side equation):**
\[
(sum + carry\_r) \bmod 10 = d\_right
\]
and the next carry from the right becomes:
\[
new\_carry\_r = \left\lfloor\frac{sum + carry\_r}{10}\right\rfloor
\]

**Left digit constraint (most-significant-side equation):**

At the left digit, what you “see” is affected by the carry coming from the inner side; in this outside-in DP, that inner-side carry is represented by the current `exp_carry` and must be consistent.

The code rearranges the condition into computing what carry should be expected for the next inner layer:

\[
new\_exp\_carry = d\_left + 10 \cdot exp\_carry - sum
\]

This value must be either 0 or 1, otherwise impossible. Intuition: it encodes whether the inner addition must produce a carry into this left digit to make the digit match.

**Middle digit handling:**
If `left == right`, there is only one digit \(a=b\), and we must have the carry coming from the right equal to the carry expected on the left (they meet at the center). The code enforces:
- only accept if `new_carry_r == exp_carry`.

### Finalization

After processing all pairs, we’re left with dp over carry states. Valid completions require the “expected carry” and the “right carry” to match appropriately at the end; the code sums:
\[
dp[0][0] + dp[1][1]
\]

### Big integers

The number of solutions can be enormous (for very long N), so the solution uses a custom `bigint` implementation to store counts.

### Complexity

Let \(L = |S|\). For each of ~\(L/2\) positions, we iterate over:
- 2 `exp_carry` * 2 `carry_r` states,
- digits `a` in 0..9 (with one restriction),
- digits `b` in 0..9 (or just `a` in the middle).

So it’s \(O(L \cdot 100)\) transitions, very fast; the heavier cost is big integer addition, but still feasible.

---

## 3) Provided C++ solution with detailed comments (line-by-line style)

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

// Pretty-print a pair as "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Read a pair from input
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Read an entire vector from input (assumes size already set)
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) in >> x;
    return in;
}

// Print a vector with spaces
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) out << x << ' ';
    return out;
}

// We store big integers in base 1e9 (fits in 32-bit int with carry in 64-bit).
const int base = 1000000000;
const int base_digits = 9;

// Big integer struct capable of +, -, *, /, % etc.
// In this problem we mostly need: construction from int, +, +=, isZero, printing.
struct bigint {
    vector<int> z;  // digits in little-endian: z[0] is least significant block
    int sign;       // +1 or -1

    bigint() : sign(1) {}
    bigint(long long v) { *this = v; }
    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) { sign = v.sign; z = v.z; }

    // Assign from a 64-bit integer
    void operator=(long long v) {
        sign = 1;
        if(v < 0) { sign = -1; v = -v; }
        z.clear();
        for(; v > 0; v /= base) z.push_back(v % base);
    }

    // Addition
    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v; // start from v and add this->z into it
            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) res.z.push_back(0);
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) res.z[i] -= base;
            }
            return res;
        }
        // If signs differ, reduce to subtraction
        return *this - (-v);
    }

    // Subtraction
    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            // If |this| >= |v|, do normal subtraction
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) res.z[i] += base;
                }
                res.trim();
                return res;
            }
            // Otherwise result is negative
            return -(v - *this);
        }
        // Different signs -> addition
        return *this + (-v);
    }

    // Multiply by small int
    void operator*=(int v) {
        if(v < 0) { sign = -sign; v = -v; }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) z.push_back(0);
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
        }
        trim();
    }

    bigint operator*(int v) const { bigint res = *this; res *= v; return res; }

    // Division with remainder: returns (q, r)
    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = (int)a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = (int)b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) { r += b; --d; }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    // Integer square root (not needed for this problem)
    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);
        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0);
                if(r1 >= 0) { r = r1; break; }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }
    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    // Divide by small int
    void operator/=(int v) {
        if(v < 0) { sign = -sign; v = -v; }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const { bigint res = *this; res /= v; return res; }

    int operator%(int v) const {
        if(v < 0) v = -v;
        int m = 0;
        for(int i = (int)z.size() - 1; i >= 0; --i)
            m = (z[i] + m * (long long)base) % v;
        return m * sign;
    }

    // Convenience compound ops
    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    // Comparisons
    bool operator<(const bigint& v) const {
        if(sign != v.sign) return sign < v.sign;
        if(z.size() != v.z.size()) return z.size() * sign < v.z.size() * v.sign;
        for(int i = (int)z.size() - 1; i >= 0; i--)
            if(z[i] != v.z[i]) return z[i] * sign < v.z[i] * sign;
        return false;
    }
    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const { return !(*this < v) && !(v < *this); }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    // Remove leading zero blocks and normalize sign for zero
    void trim() {
        while(!z.empty() && z.back() == 0) z.pop_back();
        if(z.empty()) sign = 1;
    }

    // Check if value == 0
    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    // Unary minus
    bigint operator-() const { bigint res = *this; res.sign = -sign; return res; }

    // Absolute value
    bigint abs() const { bigint res = *this; res.sign *= res.sign; return res; }

    // Convert to long long (unsafe if too large; not used here)
    long long longValue() const {
        long long res = 0;
        for(int i = (int)z.size() - 1; i >= 0; i--) res = res * base + z[i];
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    // Read decimal string into bigint
    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') sign = -sign;
            ++pos;
        }
        for(int i = (int)s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++)
                x = x * 10 + s[j] - '0';
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s; stream >> s; v.read(s); return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) stream << '-';
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i)
            stream << setw(base_digits) << setfill('0') << v.z[i];
        return stream;
    }

    // Base conversion helper used by multiplication
    static vector<int> convert_base(const vector<int>& a, int old_digits, int new_digits) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * 10;
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) res.pop_back();
        return res;
    }

    typedef vector<long long> vll;

    // Karatsuba multiplication for large bigint*bigint
    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++)
                for(int j = 0; j < n; j++)
                    res[i + j] += a[i] * b[j];
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) a2[i] += a1[i];
        for(int i = 0; i < k; i++) b2[i] += b1[i];

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
        for(int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];

        for(int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
        for(int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
        for(int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
        return res;
    }

    // Full bigint multiplication
    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) a.push_back(0);
        while(b.size() < a.size()) b.push_back(0);
        while(a.size() & (a.size() - 1)) { a.push_back(0); b.push_back(0); }

        vll c = karatsubaMultiply(a, b);

        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

string N;

// Reads one test case; exits the program on EOF
void read() {
    if(!(cin >> N)) exit(0);
}

// Count solutions for a given target string S and initial expected carry on the left
bigint count_solutions(const string& S, int expecting_carry_initial) {
    int n = (int)S.size();
    if(n == 0) return 0;

    // dp[exp_carry][carry_r] = number of ways so far
    vector<vector<bigint>> dp(2, vector<bigint>(2, 0));
    dp[expecting_carry_initial][0] = 1; // initially no carry from the rightmost side

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

        vector<vector<bigint>> ndp(2, vector<bigint>(2, 0));

        for(int exp_carry = 0; exp_carry < 2; exp_carry++) {
            for(int carry_r = 0; carry_r < 2; carry_r++) {
                if(dp[exp_carry][carry_r].isZero()) continue;

                // Choose digit a = X[left]
                for(int a = 0; a <= 9; a++) {
                    // X cannot have leading zeros
                    if(left == 0 && a == 0) continue;

                    // Choose digit b = X[right] (must equal a if middle digit)
                    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-side digit constraint
                        int right_val = sum + carry_r;
                        if(right_val % 10 != d_right) continue;
                        int new_carry_r = right_val / 10;

                        // Compute what carry we must expect for the next inner layer on the left
                        int new_exp_carry = d_left + 10 * exp_carry - sum;
                        if(new_exp_carry < 0 || new_exp_carry > 1) continue;

                        if(is_middle) {
                            // At the center, the carry from right and expected carry must meet
                            if(new_carry_r == exp_carry) {
                                ndp[0][0] += dp[exp_carry][carry_r];
                            }
                        } else {
                            // Normal transition inward
                            ndp[new_exp_carry][new_carry_r] += dp[exp_carry][carry_r];
                        }
                    }
                }
            }
        }

        dp = ndp;
        left++;
        right--;
    }

    // Valid completions require the two carry notions to match at the end
    bigint ans = 0;
    for(int c = 0; c < 2; c++) ans += dp[c][c];
    return ans;
}

void solve() {
    // Termination condition in the statement: a single "0" ends input.
    if(N == "0") exit(0);

    bigint ans = 0;

    // Case A: |X| = |N|, no extra leading carry
    ans += count_solutions(N, 0);

    // Case B: |X| = |N|-1, meaning N must start with '1' from the outer carry
    if(N.size() > 1 && N[0] == '1') {
        ans += count_solutions(N.substr(1), 1);
    }

    cout << ans << "\n";
}

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

    for(int test = 1;; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

## 4) Python solution (same idea, with detailed comments)

```python
import sys

# Python's int is arbitrary precision, so we don't need a custom bigint.

def count_solutions(S: str, expecting_carry_initial: int) -> int:
    """
    Count number of positive integers X such that X + reverse(X) == S (as digits),
    under a particular initial 'expected carry' on the left side.

    expecting_carry_initial:
        0 -> corresponds to case |X| == |S|
        1 -> corresponds to case where original N had an extra leading '1' carry
    """
    n = len(S)
    if n == 0:
        return 0

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

    left, right = 0, n - 1
    while left <= right:
        d_left = ord(S[left]) - ord('0')
        d_right = 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

                # choose a = X[left]
                for a in range(10):
                    # X is positive and has no leading zeros
                    if left == 0 and a == 0:
                        continue

                    # choose b = X[right]; if middle, must be the same digit
                    if is_middle:
                        b_range = (a,)
                    else:
                        b_range = range(10)

                    for b in b_range:
                        s = a + b

                        # Right digit must match:
                        # (a + b + carry_r) % 10 == digit at S[right]
                        right_val = s + carry_r
                        if right_val % 10 != d_right:
                            continue
                        new_carry_r = right_val // 10  # 0 or 1

                        # Left digit constraint transformed into:
                        # new_exp_carry = d_left + 10*exp_carry - (a+b)
                        new_exp_carry = d_left + 10 * exp_carry - s
                        if new_exp_carry < 0 or new_exp_carry > 1:
                            continue

                        if is_middle:
                            # At the middle, the carry coming from the right
                            # must equal the carry expected on the left.
                            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: X has same length as N
    ans = count_solutions(N, 0)

    # Case B: X has length len(N)-1, only possible if N begins with '1'
    if len(N) > 1 and N[0] == '1':
        ans += count_solutions(N[1:], 1)

    return ans


def main():
    out_lines = []
    for line in sys.stdin:
        N = line.strip()
        if not N:
            continue
        if N == "0":
            break
        out_lines.append(str(solve_one(N)))
    sys.stdout.write("\n".join(out_lines))

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

---

## 5) Compressed editorial

Process digit pairs \((x_L, x_R)\) of \(X\) from outside to center. Each pair determines both the left digit and right digit of \(N = X + R(X)\), but involves carries from both ends. Use DP with two binary carry states:

- `exp_carry` = carry expected into the current left digit (from inner positions),
- `carry_r` = carry into the current right digit (usual carry from less significant side).

For each outer position `(left,right)` and each state, try digits `a` and `b` (with no leading zero for `a` at left=0; and `a=b` in the middle). Enforce:
- `(a+b+carry_r) % 10 == N[right]`, set `new_carry_r = (a+b+carry_r)//10`,
- `new_exp_carry = N[left] + 10*exp_carry - (a+b)` must be 0 or 1.

Transition inward accordingly; at the middle require `new_carry_r == exp_carry`. Sum final `dp[c][c]`.

Answer is sum of two cases:
1) \(|X|=|N|\): run DP on `N` with initial `exp_carry=0`.
2) \(|X|=|N|-1\): only if `N` starts with `1`, run DP on `N[1:]` with initial `exp_carry=1`.

Use big integers for the count (Python int works; C++ uses custom bigint).