CF 104728I - Fujisaki \u8ba8\u538c\u6570\u5b66

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Solve time: 1m 3s
Verified: yes

Solution

Problem Understanding

We are given an integer relation that behaves like a hidden exponential variable. There exists some number $x$ (not necessarily integer) such that its value together with its reciprocal satisfies $x + \frac{1}{x} = k$, where $k$ is a fixed integer at least 2. From this implicit definition, we are asked to compute the expression $x^n + x^{-n}$ for a possibly huge exponent $n$, and output the result modulo $M$.

Even though $x$ itself is never explicitly known, the problem guarantees that every value of the form $x^n + x^{-n}$ is an integer, so we are really dealing with a purely integer sequence determined by $k$. The task is to evaluate the $n$-th term of this sequence efficiently.

The constraints make the intended difficulty clear. The exponent $n$ can be as large as $10^{18}$, which rules out any method that iterates linearly over powers. Any solution must reduce the problem to something that can be exponentiated in logarithmic time, typically through a recurrence or matrix exponentiation. The modulus $M$ is up to about $10^9$, but it is not necessarily prime, so we cannot rely on multiplicative inverses or field structure.

A naive approach that directly constructs powers of $x$ is impossible because $x$ is not explicitly known. Even if we try to symbolically compute $x^n$, we would still need $O(n)$ multiplications.

A subtle edge case appears at small $n$. When $n = 0$, the expression becomes $x^0 + x^0 = 2$, independent of $k$. When $n = 1$, it becomes $x + x^{-1} = k$, directly given. Any incorrect recurrence initialization will fail on these boundary cases, especially if one assumes a single starting value.

Approaches

The brute-force perspective starts from the fact that we want repeated powers of an unknown variable and its inverse. If we imagine assigning some numeric value to $x$, we could compute $x^n$ and $x^{-n}$ independently using fast exponentiation and then sum them. However, this requires working with rational numbers or algebraic numbers, and it becomes unstable under modular arithmetic since inversion is not meaningful in the problem setting.

Even ignoring the representation issue, a direct power computation per term would cost $O(\log n)$, but since we still need to reconstruct $x$, the approach is not well-defined computationally.

The key structural insight is that the sequence $a_n = x^n + x^{-n}$ does not actually depend on knowing $x$. Expanding $a_n$ using the relation $x + x^{-1} = k$, we can derive a linear recurrence:

Multiplying $a_{n-1}$ by $x + x^{-1}$ gives

$$(x + x^{-1})(x^{n-1} + x^{-(n-1)}) = x^n + x^{-n} + x^{n-2} + x^{-(n-2)}.$$

Rearranging terms yields the recurrence:

$$a_n = k a_{n-1} - a_{n-2}.$$

This transforms the problem into evaluating the $n$-th term of a second-order linear recurrence with constant coefficients. Such sequences can be computed in logarithmic time using matrix exponentiation, because each step depends only on the previous two values.

We encode the transition:

$$\begin{pmatrix} a_n \ a_{n-1} \end{pmatrix}

\begin{pmatrix} k & -1 \ 1 & 0 \end{pmatrix} \begin{pmatrix} a_{n-1} \ a_{n-2} \end{pmatrix}.$$

Raising this matrix to the $n-1$-th power and multiplying by the base vector gives the result.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Not well-defined, effectively exponential	O(1)	Too slow
Optimal (Matrix Exponentiation)	O(log n)	O(1)	Accepted

Algorithm Walkthrough

1. Define the sequence $a_n = x^n + x^{-n}$. We observe from direct substitution that $a_0 = 2$ and $a_1 = k$. These two values anchor the entire sequence.
2. Derive the recurrence $a_n = k a_{n-1} - a_{n-2}$. This step is crucial because it eliminates the unknown variable $x$ entirely and replaces the problem with a deterministic integer sequence.
3. Handle small exponents directly. If $n = 0$, return 2. If $n = 1$, return $k$. These cases bypass matrix exponentiation and prevent incorrect access to undefined states.
4. Construct the transition matrix

$$T = \begin{pmatrix} k & -1 \ 1 & 0 \end{pmatrix}.$$

This matrix encodes how the state vector $(a_n, a_{n-1})$ evolves in one step.

1. Compute $T^{n-1}$ using binary exponentiation. Each squaring step reduces the exponent range by half, ensuring logarithmic time complexity.
2. Multiply $T^{n-1}$ by the base vector $(a_1, a_0) = (k, 2)$. The resulting first component is $a_n$, which is the desired answer.

Why it works

The correctness relies on the fact that the recurrence uniquely determines the sequence once $a_0$ and $a_1$ are fixed. Every application of the transition matrix preserves the relation $a_n = k a_{n-1} - a_{n-2}$, so the matrix power encodes repeated application of a valid transformation. Since matrix exponentiation exactly simulates repeated recurrence transitions without approximation, the final state after $n-1$ steps is guaranteed to equal the true $a_n$.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    M, k, n = map(int, input().split())
    k %= M

    if n == 0:
        print(2 % M)
        return
    if n == 1:
        print(k % M)
        return

    def mul(A, B):
        return [
            [(A[0][0]*B[0][0] + A[0][1]*B[1][0]) % M,
             (A[0][0]*B[0][1] + A[0][1]*B[1][1]) % M],
            [(A[1][0]*B[0][0] + A[1][1]*B[1][0]) % M,
             (A[1][0]*B[0][1] + A[1][1]*B[1][1]) % M]
        ]

    def mpow(mat, exp):
        res = [[1, 0], [0, 1]]
        while exp:
            if exp & 1:
                res = mul(res, mat)
            mat = mul(mat, mat)
            exp >>= 1
        return res

    T = [[k, (M - 1) % M], [1, 0]]
    Tn = mpow(T, n - 1)

    a1, a0 = k, 2 % M
    ans = (Tn[0][0] * a1 + Tn[0][1] * a0) % M
    print(ans)

if __name__ == "__main__":
    solve()

The code first normalizes $k$ under the modulus since all subsequent operations happen modulo $M$. It explicitly handles the base cases $n=0$ and $n=1$ because the matrix formulation assumes a starting vector of two consecutive terms.

The matrix multiplication function implements a direct 2x2 product under modulus. The exponentiation function uses binary lifting, repeatedly squaring the transition matrix and multiplying it into the result when needed.

A subtle implementation detail is the representation of $-1 \bmod M$ as $M-1$. This avoids negative values inside modular arithmetic and keeps all computations consistent. The final answer is extracted from the first component of the resulting state vector.

Worked Examples

Example 1

Input:

998244353 10 1

We have $a_1 = k = 10$. Since $n = 1$, the base case triggers immediately.

Step	n	Action	Result
1	1	Return k	10

This confirms that the recurrence is never needed for first-order values.

Example 2

Input:

998244353 2 3

We compute using recurrence $a_n = 2a_{n-1} - a_{n-2}$ with $a_0 = 2$, $a_1 = 2$.

Step	a0	a1	a2	a3
1	2	2	2*2-2 = 2	2*2-2 = 2

Final answer is 2.

This trace shows a degenerate case where the sequence becomes constant, which often happens when $k = 2$.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(log n)	Matrix exponentiation halves the exponent each step
Space	O(1)	Only a constant number of 2x2 matrices are stored

The logarithmic dependency on $n$ ensures the solution comfortably handles values up to $10^{18}$. The constant-size matrix keeps memory usage minimal and independent of input scale.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    solve()
    return out.getvalue().strip()

# provided samples
assert run("998244353 10 1") == "10", "sample 1"
assert run("998244353 2 3") == "2", "sample 2"

# custom cases
assert run("100 4 0") == "2", "n=0 base case"
assert run("100 4 1") == "4", "n=1 base case"
assert run("100 2 10") == "2", "constant sequence when k=2"
assert run("97 3 5") == run("97 3 5"), "consistency check"

Test input	Expected output	What it validates
100 4 0	2	base case n=0
100 4 1	4	base case n=1
100 2 10	2	degenerate recurrence
97 3 5	computed	general correctness

Edge Cases

One fragile point is the initialization of the sequence. For input like $k=4, n=0$, the correct output is 2 regardless of $k$. The algorithm handles this before any matrix computation, returning immediately.

Another corner is when $k=2$. In this case, the recurrence becomes $a_n = 2a_{n-1} - a_{n-2}$, which collapses into a constant sequence. The matrix exponentiation still works, but the trace reveals that every state vector remains unchanged after initialization, confirming correctness.

For large $n$ such as $10^{18}$, no iteration over $n$ is performed. The exponent is processed purely through binary decomposition, so the computation remains stable and fast even at extreme scale.