Core Logic and Mathematical Principles
In previous chapters, we tackled standard digit DP, specialized number theory, and tree composite structures. In the ultimate selection of provincial competitions, NOI, and even ACM contests, digit DP reaches its highest dimensional form: the cross-domain integration of base-free dynamic transitions and matrix multiplication (Matrix Exponentiation).
The core characteristic of such problems is: the given range $N$ is extremely large (e.g., $N \le 10^{18}$ or even $N \le 10^{100}$ given in string form), while requiring extremely macro and high-order global statistics on digit features (e.g., counting the total number of valid configurations after executing a certain digit state transition matrix $K$ times for all numbers between $1$ and $N$).
To overcome this ultimate threshold, we must completely rewrite the “digit-by-digit recursion” of digit DP into an algebraic network and introduce the following two ultimate techniques:
1. Unified Algebraic Transformation of Arbitrary Base (Base-B) Digit Space
Standard digit DP is often limited to decimal (18.1) or binary (bit compression). However, when faced with “in base $B$, the digits satisfy a certain intricate recursive relationship,” we can no longer rely on fixed numerical constants. We must abstract the base $B$ as an algebraic operator. For any base $B$ number, its transition from high digits to low digits can be uniformly represented as a transition grid of a base state machine:
$$X' = X \cdot B + d \quad (0 \le d < B)$$
This abstraction allows us to seamlessly cross all counting barriers from binary (Boolean algebra) to hexadecimal (high-dimensional compact space) simply by adjusting the base $B$ and the dimensions of the transition matrix without modifying the underlying logic.
2. Matrix Acceleration of Linear Digit Transitions
When the memoization search of digit DP progresses under limit = false (free state), you will find that: the transition method from pos to pos - 1 is completely isomorphic and invariant for all positions. Since the transition rules form a constant linear system, we can compress the entire set of digit state transition equations when limit = false into a state transition matrix $M$.
If the lower digits need to advance freely $P$ steps, traditional DFS requires $O(P \times \text{States})$ iterations. However, through matrix exponentiation, we can directly push a long segment of “free digit intervals” through matrix multiplication at a very low complexity of $O(\text{States}^3 \log P)$. This process of elevating “topological search” to “pure algebraic matrix exponentiation” is the most perfect ultimate form of high-order dynamic programming.
State Design and Algorithm Derivation
Taking the classic top problem “Counting High-Dimensional Free Base Digit Sequences” as an example: in base $B$, count how many numbers in the interval $[1, N]$ satisfy that no two adjacent digits are the same in their base $B$ representation. The number $N$ can have up to $10^5$ digits (given in the form of a large number array), with $B \le 100$.
Due to the astonishing length of $N$ reaching $10^5$, any traditional $O(\text{len})$ DFS will face enormous constant pressure even if it runs just once, let alone the entanglement of multiple states. We must utilize matrix acceleration to batch evaporate effective digits.
1. Constructing the Free State Matrix
Let the state vector be $\vec{V} = [v_0, v_1, \dots, v_{B-1}]^T$, where $v_d$ represents the number of valid configurations when the current digit is filled with the number $d$. According to the problem requirements, the current digit cannot be the same as the previous digit. Therefore, the size of the linear transition matrix $M$ from the previous digit to the current digit is $B \times B$:
$$M[i][j] = \begin{cases} 1 & i \neq j \\ 0 & i = j \end{cases}$$
If $P$ consecutive digits are completely in the “free sky” without limit restrictions, then the overall contribution of these $P$ digits' state transitions is directly equivalent to:
$$\vec{V}_{\text{final}} = M^P \cdot \vec{V}_{\text{init}}$$
2. Matrix Divide-and-Conquer Algorithm with limit Constraints
Since the high digits of $N$ still have precise upper limit constraints, we cannot directly run fast exponentiation on the entire graph. The standard examination strategy is: scan each digit of $N$ from high to low. As long as the current digit does not reach the upper limit (i.e., a digit smaller than the current digit of $N$ is filled), all lower digits will be instantly liberated. We immediately perform matrix exponentiation on the remaining lower digit lengths to directly harvest all valid counts of this large branch in one go.
C++ Standard Source Code (NOIP/Provincial Selection Ultimate Matrix Digit Template)
The following source code demonstrates how to perfectly intertwine large number decomposition, matrix exponentiation, and digit constraints to achieve instant computation for ultra-long numbers of up to $10^5$ digits.
#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;
const int MOD = 1e9 + 7;
int B; // Base radix
string N_str; // Ultra-long upper limit number N
int digits[100005];
int len;
// Linear algebra components: Standard matrix class
struct Matrix {
int n;
vector<vector<long long>> mat;
Matrix(int n) : n(n), mat(n, vector<long long>(n, 0)) {}
static Matrix identity(int n) {
Matrix res(n);
for (int i = 0; i < n; ++i) res.mat[i][i] = 1;
return res;
}
Matrix operator*(const Matrix& other) const {
Matrix res(n);
for (int i = 0; i < n; ++i) {
for (int k = 0; k < n; ++k) {
if (!mat[i][k]) continue;
for (int j = 0; j < n; ++j) {
res.mat[i][j] = (res.mat[i][j] + mat[i][k] * other.mat[k][j]) % MOD;
}
}
}
return res;
}
};
// Core algebraic operator: Matrix exponentiation
Matrix matrix_pow(Matrix base, long long p) {
Matrix res = Matrix::identity(base.n);
while (p > 0) {
if (p & 1) res = res * base;
base = base * base;
p >>= 1;
}
return res;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// Read base B and ultra-long string N (assuming input is already in B-base character representation)
if (!(cin >> B >> N_str)) return 0;
len = N_str.length();
for (int i = 0; i < len; ++i) {
// Convert characters from high to low into digits
if (N_str[i] >= '0' && N_str[i] <= '9') digits[i] = N_str[i] - '0';
else digits[i] = N_str[i] - 'A' + 10;
}
// 1. Preprocess the base transition matrix M
Matrix M(B);
for (int i = 0; i < B; ++i) {
for (int j = 0; j < B; ++j) {
if (i != j) M.mat[i][j] = 1; // Adjacent cannot be the same
}
}
long long total_ans = 0;
// 2. Independently count all valid numbers with lengths less than len (these numbers cannot exceed N due to their shorter length)
// This is equivalent to solving the free count after leading zeros are stripped
for (int L = 1; L < len; ++L) {
// The highest digit cannot be 0, there are B-1 ways to fill
// The remaining L-1 digits can be directly solved through matrix exponentiation
if (L == 1) {
total_ans = (total_ans + B - 1) % MOD;
} else {
Matrix T = matrix_pow(M, L - 1);
// Assume the highest digit fills a certain non-zero number, calculate the sum of all possibilities for lower digits
long long one_digit_sum = 0;
for (int j = 0; j < B; ++j) {
one_digit_sum = (one_digit_sum + T.mat[0][j]) % MOD; // Symmetry of the matrix
}
total_ans = (total_ans + one_digit_sum * (B - 1)) % MOD;
}
}
// 3. Core divide-and-conquer: Precisely match valid numbers of length equal to len constrained by N
int pre_digit = -1; // Record the physical hard limit digit passed down from high digits
bool is_valid_prefix = true; // Mark whether the prefix is valid up to the current high digit
for (int i = 0; i < len; ++i) {
if (!is_valid_prefix) break; // If the high digit prefix no longer satisfies “different adjacent digits”, the lower digits lose all discussion significance, directly intercept
int limit_up = digits[i];
// Enumerate the digits that can be filled in the current position
for (int d = (i == 0 ? 1 : 0); d < limit_up; ++d) {
// Intercept conflicts with the adjacent high digit
if (i > 0 && d == pre_digit) continue;
// As long as the current digit fills a number d smaller than the upper limit, the remaining lower digits (len - 1 - i) will instantly gain complete freedom
int rem_len = len - 1 - i;
if (rem_len == 0) {
total_ans = (total_ans + 1) % MOD; // Touching the bottom, producing 1 isolated solution
} else {
Matrix T = matrix_pow(M, rem_len);
// Currently filled d, propagate from d
long long current_branch_sum = 0;
for (int j = 0; j < B; ++j) {
current_branch_sum = (current_branch_sum + T.mat[d][j]) % MOD;
}
total_ans = (total_ans + current_branch_sum) % MOD;
}
}
// Force synchronize the current digit to the upper limit digit of N, continue to push down
if (i > 0 && limit_up == pre_digit) {
is_valid_prefix = false; // N itself has already self-collided illegally at this position, all subsequent lower digits are completely sealed
}
pre_digit = limit_up;
}
// 4. Final isolated special case: N itself is also a possible valid number
if (is_valid_prefix) {
total_ans = (total_ans + 1) % MOD;
}
cout << total_ans << "\n";
return 0;
}NOIP/Provincial Selection Practical Pitfall Guide
1. Ignoring the illegality of the large number itself leading to lower digit pollution (Prefix Self-Collision Bug)
- Phenomenon: In the divide-and-conquer process of moving down digit by digit, if the upper limit number $N$ itself has already violated the constraints at high digits (for example, $N = 3325$, where two consecutive
3s appear in the hundreds and thousands places), if the program does not intercept, continuing to enumerate digits smaller than2in the tens and units places and adding matrix exponentiation will result in counting illegal numbers like3310,3312. - Avoidance Method: It is essential to introduce a boolean sentinel like
is_valid_prefixin the process of locking down each digit. Once it is found that adjacent digits of $N$ have already crossed the red line, immediately executebreakto completely truncate subsequent lower digit branches' accumulation.
2. Ignoring the loss of valid shorter numbers with lengths less than $N$ (Short Digit Loss)
- Phenomenon: When $N = 1000$, shorter numbers like
99,5are obviously valid solutions. Matrix exponentiation generally can only handle fixed-length topological recursions. If you directly start applying fast exponentiation from the first digit, those shorter numbers with lengths less thanlenwill be missed, leading to WA (Wrong Answer). - Avoidance Method: Before the core divide-and-conquer loop, a separate discrete loop must be opened to collect contributions of all valid numbers with lengths from $1$ to $ ext{len}-1$ through matrix exponentiation (as shown in step 2 of the source code).