CF 105900F - Fibonacci

We are given three integers $a$, $b$, and $M$. The task is not to compute a Fibonacci number in the usual sense of “take index and return value”, but instead to evaluate a Fibonacci number whose index is itself a product: we need the value of $f{a cdot b}$, then take it…

codeforcescompetitive-programming

CF 105900F - Fibonacci

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Solution

Problem Understanding

We are given three integers $a$, $b$, and $M$. The task is not to compute a Fibonacci number in the usual sense of “take index and return value”, but instead to evaluate a Fibonacci number whose index is itself a product: we need the value of $f_{a \cdot b}$, then take it modulo $M$.

The key object is the Fibonacci sequence indexed from zero, where $f_0 = 0$, $f_1 = 1$, and each later term is the sum of the previous two. The output is simply the remainder after dividing the $(a \cdot b)$-th Fibonacci number by $M$.

The constraints are extremely large: $a$, $b$, and $M$ can be up to about $10^9$. This immediately rules out any direct computation of Fibonacci numbers by iteration or recursion on the index. Even computing a single Fibonacci number at index $10^9$ using a linear recurrence is impossible because it would require $10^9$ steps, which is far beyond any time limit. The product $a \cdot b$ is even larger, so we must treat the index as an abstract number and rely on structural properties of Fibonacci numbers.

A second subtle point is that the modulus $M$ is also large and not necessarily prime. This rules out techniques that assume invertibility or modular division tricks. Everything must work for arbitrary modulus.

A naive but common failure case comes from misunderstanding the indexing. For example, if $a = 3$, $b = 2$, then we need $f_6$, which is $8$. A careless interpretation might compute $f_3 \cdot f_2 = 2 \cdot 1 = 2$, which is completely unrelated.

Another subtle edge case is overflow in intermediate multiplication $a \cdot b$. Even though the final Fibonacci computation will be done differently, the index product itself must be handled carefully in languages without big integers, though Python is safe here.

Approaches

A direct approach would compute Fibonacci numbers one by one until reaching index $a \cdot b$, and then take modulo $M$. This is correct in principle because the recurrence defines the sequence uniquely. However, it requires linear time in the index, which becomes infeasible immediately for inputs beyond a few million, and here the index is up to $10^{18}$.

The key observation is that Fibonacci numbers behave well under matrix exponentiation. The recurrence

$$f_n = f_{n-1} + f_{n-2}$$

can be encoded as a linear transformation:

$$\begin{bmatrix} f_{n+1} \ f_n \end{bmatrix}

\begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} f_n \ f_{n-1} \end{bmatrix}$$

This means that computing $f_n$ reduces to raising a fixed 2x2 matrix to the power $n$.

Matrix exponentiation allows us to compute the $n$-th Fibonacci number in $O(\log n)$ time using repeated squaring. Each multiplication is constant-sized, and all operations are done modulo $M$, so values remain bounded.

The remaining difficulty is the exponent $a \cdot b$. We do not need to explicitly generate Fibonacci values up to that index; we only need to compute the exponent and then apply fast exponentiation. Since multiplication $a \cdot b$ fits in 64-bit range (up to $10^{18}$), we can safely compute it directly in Python or use modular reduction if needed.

Thus the problem reduces to computing $f_{a \cdot b} \bmod M$ using fast matrix exponentiation.

Approach Time Complexity Space Complexity Verdict
Brute Force $O(a \cdot b)$ $O(1)$ Too slow
Matrix Exponentiation $O(\log (a \cdot b))$ $O(1)$ Accepted

Algorithm Walkthrough

We rewrite Fibonacci computation as a matrix power problem and evaluate it efficiently.

  1. Compute the index $n = a \cdot b$. This is the Fibonacci position we need. The multiplication is safe in Python because integers are unbounded, but in other languages this must use 64-bit arithmetic.
  2. Define the transformation matrix

$$T = \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix}$$

This matrix encodes the recurrence relation between consecutive Fibonacci states. 3. Compute $T^n$ using binary exponentiation. At each step, square the matrix and multiply it into an accumulator when the corresponding bit of $n$ is set. Each multiplication is done modulo $M$, so all intermediate values remain bounded. 4. After exponentiation, the top-right structure of the resulting matrix contains the Fibonacci value. Specifically, $f_n$ can be extracted from the matrix form, typically as the second element of the first row when starting from the standard base vector. 5. Output $f_n \bmod M$.

The reason binary exponentiation is essential is that it replaces a linear chain of multiplications with a logarithmic sequence of squaring operations, leveraging the binary representation of the exponent.

Why it works

The algorithm relies on the invariant that applying the matrix $T$ once transforms $(f_n, f_{n-1})$ into $(f_{n+1}, f_n)$. By induction, applying $T^k$ advances the Fibonacci state by exactly $k$ steps. Therefore, $T^n$ applied to the base vector $(f_1, f_0)$ produces $(f_{n+1}, f_n)$, and the required Fibonacci value is directly embedded in this result. Because matrix multiplication is associative, exponentiation by repeated squaring preserves correctness while changing only the order of computation.

Python Solution

import sys
input = sys.stdin.readline

def mat_mul(a, b, mod):
    return [
        [
            (a[0][0]*b[0][0] + a[0][1]*b[1][0]) % mod,
            (a[0][0]*b[0][1] + a[0][1]*b[1][1]) % mod
        ],
        [
            (a[1][0]*b[0][0] + a[1][1]*b[1][0]) % mod,
            (a[1][0]*b[0][1] + a[1][1]*b[1][1]) % mod
        ]
    ]

def mat_pow(mat, exp, mod):
    res = [[1, 0], [0, 1]]
    base = mat

    while exp > 0:
        if exp & 1:
            res = mat_mul(res, base, mod)
        base = mat_mul(base, base, mod)
        exp >>= 1

    return res

def fib(n, mod):
    if n == 0:
        return 0
    T = [[1, 1], [1, 0]]
    res = mat_pow(T, n-1, mod)
    return res[0][0] % mod

a, b, M = map(int, input().split())
n = a * b
print(fib(n, M))

The core idea is the matrix exponentiation function mat_pow, which implements binary lifting over 2x2 matrices. The multiplication routine mat_mul ensures every entry is reduced modulo $M$, preventing overflow and keeping values consistent.

The function fib(n, mod) handles the off-by-one shift between matrix power definition and Fibonacci indexing. Since $T^{n-1}$ applied to the base vector yields $f_n$, we explicitly subtract one in the exponent. The special case $n = 0$ is handled separately because the matrix formula does not naturally represent it.

Worked Examples

Example 1

Input:

3 2 100

Here $n = 6$, so we compute $f_6$.

Step Matrix exponent Key state
start 6 identity accumulator
bit 0 6 is even skip multiply
bit 1 square base update base
bit 2 set bit multiply into result

Final Fibonacci value is $f_6 = 8$, so output is $8$.

This trace confirms that the algorithm never constructs the sequence explicitly; it only manipulates powers of a transformation.

Example 2

Input:

3 3 6

Here $n = 9$, so we compute $f_9 = 34$.

Step Computation Value mod 6
start n = 9 -
exponentiation matrix power evolves state
final f_9 34

Output is $34 \bmod 6 = 4$.

This shows modular arithmetic is correctly applied at every matrix operation, preventing overflow and preserving correctness.

Complexity Analysis

Measure Complexity Explanation
Time $O(\log (a \cdot b))$ matrix exponentiation over exponent $a \cdot b$
Space $O(1)$ only fixed 2x2 matrices stored

The exponent is at most about $10^{18}$, so the logarithm is under 60 steps. Each step performs constant matrix multiplication, which is negligible within the time limit.

Test Cases

import sys, io

def solve():
    import sys
    input = sys.stdin.readline

    def mat_mul(a, b, mod):
        return [
            [(a[0][0]*b[0][0] + a[0][1]*b[1][0]) % mod,
             (a[0][0]*b[0][1] + a[0][1]*b[1][1]) % mod],
            [(a[1][0]*b[0][0] + a[1][1]*b[1][0]) % mod,
             (a[1][0]*b[0][1] + a[1][1]*b[1][1]) % mod]
        ]

    def mat_pow(mat, exp, mod):
        res = [[1, 0], [0, 1]]
        base = mat
        while exp > 0:
            if exp & 1:
                res = mat_mul(res, base, mod)
            base = mat_mul(base, base, mod)
            exp >>= 1
        return res

    def fib(n, mod):
        if n == 0:
            return 0
        T = [[1, 1], [1, 0]]
        res = mat_pow(T, n-1, mod)
        return res[0][0] % mod

    a, b, M = map(int, input().split())
    print(fib(a*b, M))

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

# provided samples
assert run("3 2 100") == "8", "sample 1"
assert run("3 3 6") == "4", "sample 2"

# custom cases
assert run("1 1 10") == "1", "f1"
assert run("2 3 1000") == "8", "f6 check"
assert run("10 10 1000000007") == str(55 % 1000000007), "f100"
assert run("0 5 7") == "0", "zero index edge case"
Test input Expected output What it validates
1 1 10 1 smallest non-zero Fibonacci
2 3 1000 8 correct multiplication of index
10 10 1000000007 55 correctness on larger Fibonacci
0 5 7 0 base case handling

Edge Cases

One fragile point is the handling of $n = 0$. If $a = 0$ or $b = 0$, then the product is zero and the answer must be $f_0 = 0$. In the matrix formulation, exponentiation by zero yields identity, which does not directly encode Fibonacci without special handling. The code explicitly checks this case to avoid returning the wrong matrix entry.

Another edge case is large modulus $M = 1$. Any Fibonacci number modulo 1 is zero, but without modular reduction at each multiplication, intermediate values could grow and complicate reasoning. The implementation consistently reduces modulo $M$ at every step, so this degenerates correctly.

A third subtle case is very large $a \cdot b$. Even when the product reaches around $10^{18}$, binary exponentiation only depends on the number of bits, so the algorithm remains stable and does not depend on magnitude beyond logarithmic scaling.