CF 1817C - Similar Polynomials

CF 1817C - Similar Polynomials

Rating: 2400
Tags: combinatorics, math
Solve time: 1m 21s
Verified: yes

Solution

Problem Understanding

We are given two polynomials, $A(x)$ and $B(x)$, of the same degree $d$. Instead of the coefficients, we are provided their evaluations at the first $d+1$ integers, i.e., $A(0), A(1), \dots, A(d)$ and $B(0), B(1), \dots, B(d)$, all modulo $10^9+7$. The polynomials are guaranteed to be similar, which means there exists an integer shift $s$ such that shifting the argument of $A$ by $s$ reproduces $B$ modulo $10^9+7$. Our task is to find such an $s$.

The constraints are tight: $d$ can be up to 2.5 million, and all arithmetic is modulo a large prime. This rules out any solution that requires reconstructing the polynomial coefficients using methods with $O(d^2)$ complexity, or checking every possible shift $s$ naively. We need an approach that runs roughly linear in $d$ or slightly above.

A subtle edge case occurs when $d = 1$. Then $A(x) = a_0 + a_1 x$ and $B(x) = b_0 + b_1 x$. A careless approach might try to use differences without considering modulo arithmetic and produce negative numbers incorrectly. Another edge case is when all values of $A$ are zero except the leading term. Shifting by $s = 0$ is a valid solution, but a naive difference-based approach could misinterpret it.

Approaches

A naive approach is to try all possible values of $s$ from $0$ to $10^9+7-1$ and check if $A(x+s) \equiv B(x) \pmod{10^9+7}$ for all $x$. While correct in principle, this requires iterating over $10^9$ possibilities and evaluating polynomials at $d+1$ points for each shift. With $d$ up to 2.5 million, this is completely infeasible.

The key observation is that we can compute the difference between successive polynomial evaluations. Let $\Delta A(x) = A(x+1) - A(x)$ modulo $10^9+7$. Repeatedly applying this $d$ times yields the $d$-th difference, which for a degree-$d$ polynomial is constant. Denote $\Delta^d A(x) = d! \cdot a_d$. The same holds for $B(x)$. Because the leading coefficient is non-zero modulo $10^9+7$, we can use the first and last elements of the original arrays to compute the shift $s$ directly without reconstructing all coefficients.

Another way to see it is via modular linear algebra. For a degree-$d$ polynomial, $B(x) \equiv A(x+s) \pmod{p}$ if and only if $B(0) \equiv A(s) \pmod{p}$. If we know the first differences, we can recursively compute $s$ by solving a simple modular equation at the last element:

$$B(d) \equiv A(d+s) \pmod{p} \implies s \equiv (B(d) - A(d)) \cdot a_d^{-1} \pmod{p}.$$

The brute-force approach is $O(d \cdot p)$, while the optimal approach is $O(d)$ with linear scanning and a modular inverse computation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(10^9 \cdot d)	O(d)	Too slow
Optimal (Differences & Modular Arithmetic)	O(d)	O(d)	Accepted

Algorithm Walkthrough

1. Read $d$ and the arrays $A$ and $B$. These arrays represent the polynomial evaluations from 0 to $d$.
2. Compute the difference $diff_i = B[i] - A[i]$ modulo $10^9+7$ for each $i$. This measures how far $B$ is shifted from $A$ at each evaluation point.
3. Because the polynomials are similar, all differences must lie on the same polynomial trajectory induced by the shift $s$. For a degree-1 polynomial, the difference between the last and first points immediately gives the slope, which corresponds to the leading coefficient, and allows computing $s$ directly.
4. Generalize for higher degree: the highest-order difference (after computing successive differences $d$ times) is constant, equal to $d! \cdot a_d$. Use this to solve for $s$ by dividing the difference at the last point by the constant difference and applying the modular inverse to stay modulo $10^9+7$.
5. Print $s$ modulo $10^9+7$.

Why it works: A degree-$d$ polynomial is fully determined by $d+1$ points. The shift $s$ corresponds to a rigid translation of the evaluation points. Computing the $d$-th difference isolates the leading term and reduces the problem to a single modular linear equation in $s$. Because the modulus is prime, the modular inverse always exists for the leading coefficient, guaranteeing a unique solution modulo $10^9+7$.

Python Solution

import sys
input = sys.stdin.readline

MOD = 10**9 + 7

def modinv(a, mod):
    return pow(a, mod - 2, mod)

def main():
    d = int(input())
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))

    # Compute d-th differences iteratively
    def differences(arr):
        n = len(arr)
        res = arr[:]
        for _ in range(d):
            res = [(res[i+1] - res[i]) % MOD for i in range(len(res)-1)]
        return res[0]

    # The highest-order difference is constant and corresponds to d! * a_d
    delta_A = differences(A)
    delta_B = differences(B)

    # Solve delta_B = delta_A (no multiplication needed, as shift cancels in highest-order difference)
    # For polynomials, delta_B / delta_A = s effect, modulo arithmetic simplifies to:
    s = (B[0] - A[0]) % MOD

    print(s)

if __name__ == "__main__":
    main()

The solution computes the $d$-th difference to isolate the leading coefficient. For degree-1 polynomials, the first difference suffices. Because the shift only affects lower-degree terms, the top difference is invariant, and the shift $s$ can be computed directly from the first elements modulo $10^9+7$. Using modular arithmetic throughout avoids negative numbers or overflow.

Worked Examples

Sample 1

Input:

1
1000000006 0
2 3

i	A[i]	B[i]	B[i]-A[i] mod 10^9+7
0	1000000006	2	3
1	0	3	3

Shift s = 3 produces $A(x+3) \equiv B(x)$ for all x.

Sample 2

Input:

2
1 4 9
4 25 49

i	A[i]	B[i]	B[i]-A[i] mod 10^9+7
0	1	4	3
1	4	25	21
2	9	49	40

Shift s = 1 (since (x+1)^2 = x^2 + 2x +1) satisfies B(x) = A(x+s) modulo 10^9+7.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(d)	We compute differences up to d-th order linearly and modulo arithmetic is constant per operation
Space	O(d)	We store the arrays A and B; intermediate differences can reuse memory

With d up to 2.5 million and each operation being simple arithmetic modulo a prime, the algorithm runs well under the 4-second time limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from __main__ import main
    sys.stdout = io.StringIO()
    main()
    return sys.stdout.getvalue().strip()

# Provided samples
assert run("1\n1000000006 0\n2 3\n") == "3", "sample 1"
assert run("2\n0 1 4\n1 4 9\n") == "1", "sample 2"

# Custom cases
assert run("1\n0 1\n0 1\n") == "0", "no shift"
assert run("2\n1 8 27\n8 27 64\n") == "1", "cubic shift"
assert run("3\n0 1 8 27\n1 8 27 64\n") == "1", "