Problem 520: Simbers

Mathematical Foundation

Theorem 1 (Parity Bitmask Characterization). A positive integer $m$ is a simber if and only if, letting $d_1 d_2 \cdots d_L$ be its decimal representation, the XOR-parity vector

equals the zero vector, where $e_j$ is the standard basis vector with a 1 in position $j$ and $\oplus$ denotes componentwise XOR (equivalently, addition modulo 2).

Proof. The $j$-th component of the mask is $\sum_{i=1}^{L}[d_i = j] \bmod 2$, which is 0 if and only if digit $j$ appears an even number of times. Thus $\text{mask} = \mathbf{0}$ if and only if every digit appears an even number of times. $\square$

Theorem 2 (Digit DP Correctness). The DP with states $(\text{pos}, \text{mask}, \text{tight}, \text{started})$ correctly counts all integers in $[1, N]$ that are simbers.

Proof. We construct each integer $m \leq N$ digit-by-digit from the most significant position.

• Position: $\text{pos} \in \{0, 1, \ldots, L-1\}$ where $L$ is the number of digits of $N$.
• Mask: $\text{mask} \in \{0,1\}^{10}$ tracks the parity of each digit’s occurrence count so far.
• Tight: A boolean indicating whether the prefix constructed so far matches $N$ exactly (constraining future digits).
• Started: A boolean indicating whether a nonzero digit has been placed (to handle leading zeros correctly: leading zeros should not flip the mask for digit 0).

Transitions: At position $\text{pos}$, choose digit $d \in \{0, \ldots, \ell\}$ where $\ell = d_{\text{pos}}$ if tight, else $\ell = 9$.

• If not started and $d = 0$: proceed with mask unchanged, still not started.
• Otherwise: set $\text{started} = \text{true}$, $\text{mask} \leftarrow \text{mask} \oplus (1 \ll d)$.

Base case: At $\text{pos} = L$, the number is a simber iff $\text{started} = \text{true}$ and $\text{mask} = 0$.

Each integer $m \in [0, N]$ corresponds to exactly one path through the DP. The started flag ensures $m = 0$ is not counted (or counted separately if needed). $\square$

Lemma 1 (State Space Bound). The number of distinct DP states is at most $L \times 2^{10} \times 2 \times 2 = 4096L$.

Proof. There are $L$ positions, $2^{10} = 1024$ possible masks, 2 values for tight, and 2 values for started. $\square$

Lemma 2 (Simplification for $N = 10^n$). When $N = 10^n$, we can separately count simbers with exactly $\ell$ digits for $\ell = 1, 2, \ldots, n$. For exactly $\ell$ digits, the leading digit $d_1 \in \{1, \ldots, 9\}$ and the remaining $\ell - 1$ digits are in $\{0, \ldots, 9\}$, with the constraint that the final mask is $\mathbf{0}$.

Proof. Since $10^n$ itself is not a simber (digit 1 appears once), $Q(n)$ counts simbers in $[1, 10^n - 1]$, which is the union of simbers with $\ell$ digits for $\ell = 1, \ldots, n$. For a fixed digit length, there is no upper-bound constraint (all $\ell$-digit numbers are $< 10^n$), so the tight constraint is trivially released after the first digit, and we can use the combinatorial formula directly. $\square$

Editorial

A simber is a number where every digit appears an even number of times. Count simbers up to 10^n using digit DP with parity bitmask. We count simbers in [1, 10^n - 1]. We then count L-digit numbers with all-even digit parities. Finally, such that the parity mask becomes 0.

Pseudocode

Count simbers in [1, 10^n - 1]
For each digit length L = 1, ..., n:
Count L-digit numbers with all-even digit parities
dp[mask] = number of ways to fill remaining positions
such that the parity mask becomes 0
Process position by position
Count L-digit simbers
Position 0: digit d1 in {1, ..., 9}
Positions 1..L-1: digit in {0, ..., 9}
DP over positions, tracking mask
After choosing d1:

Complexity Analysis

• Time: For each digit length $L$ (from 1 to $n$), the DP processes $L$ positions with $2^{10} = 1024$ mask states and 10 digit choices per state. Total: $O(n \cdot n \cdot 1024 \cdot 10) = O(10240\, n^2)$. This can be optimized to $O(n \cdot 1024 \cdot 10)$ by processing all digit lengths simultaneously.
• Space: $O(2^{10}) = O(1024)$ for the DP table (using rolling arrays).

Answer

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

// Digit DP for counting simbers up to N
// Parity bitmask: bit i = 1 means digit i appeared odd times

string num;
int L;
map<tuple<int,int,int,int>, long long> memo;

long long dp(int pos, int mask, int tight, int started) {
    if (pos == L) {
        return (started && mask == 0) ? 1 : 0;
    }
    auto key = make_tuple(pos, mask, tight, started);
    auto it = memo.find(key);
    if (it != memo.end()) return it->second;

    int limit = tight ? (num[pos] - '0') : 9;
    long long total = 0;
    for (int d = 0; d <= limit; d++) {
        int nt = tight && (d == limit);
        if (!started && d == 0) {
            total += dp(pos + 1, mask, nt, 0);
        } else {
            total += dp(pos + 1, mask ^ (1 << d), nt, 1);
        }
    }
    return memo[key] = total;
}

long long count_simbers(long long N) {
    num = to_string(N);
    L = num.size();
    memo.clear();
    return dp(0, 0, 1, 0);
}

int main() {
    // Q(10^n) for n = 1..15
    for (int n = 1; n <= 15; n++) {
        long long N = 1;
        for (int i = 0; i < n; i++) N *= 10;
        long long q = count_simbers(N);
        cout << "Q(10^" << n << ") = " << q << endl;
    }
    return 0;
}

"""
Problem 520: Simbers
A simber is a number where every digit appears an even number of times.
Count simbers up to 10^n using digit DP with parity bitmask.
"""

from functools import lru_cache

def count_simbers_up_to(N: int):
    """
    Count simbers in [1, N] using digit DP.
    Parity bitmask: bit i = 1 means digit i has appeared odd number of times.
    A simber has mask = 0 at the end.
    """
    digits = [int(d) for d in str(N)]
    L = len(digits)

    @lru_cache(maxsize=None)
    def dp(pos, mask, tight, started):
        if pos == L:
            return 1 if (started and mask == 0) else 0
        limit = digits[pos] if tight else 9
        total = 0
        for d in range(0, limit + 1):
            new_tight = tight and (d == limit)
            if not started and d == 0:
                total += dp(pos + 1, mask, new_tight, False)
            else:
                new_mask = mask ^ (1 << d)
                total += dp(pos + 1, new_mask, new_tight, True)
        return total

    result = dp(0, 0, True, False)
    dp.cache_clear()
    return result

def count_simbers_brute(N: int):
    """Brute-force for verification."""
    count = 0
    for n in range(1, N + 1):
        s = str(n)
        if all(s.count(d) % 2 == 0 for d in set(s)):
            count += 1
    return count

# Verify on small inputs
for N in [10, 100, 1000, 10000]:
    dp_ans = count_simbers_up_to(N)
    bf_ans = count_simbers_brute(N)
    assert dp_ans == bf_ans, f"Mismatch at N={N}: {dp_ans} vs {bf_ans}"
    print(f"Q(N={N}): {dp_ans}")

# Compute Q(10^n) for n = 1..15
print("\nQ(10^n) for various n:")
results = {}
for n in range(1, 16):
    N = 10**n
    q = count_simbers_up_to(N)
    results[n] = q
    print(f"  Q(10^{n}) = {q}")