Detailed mathematical approach

Problem Summary

Let \(A(n,k)\) be the number of length-\(n\) sequences \((a_1,\dots,a_n)\) such that every term divides \(n\) and

$a_1+\cdots+a_n \equiv -n \pmod{k}.$

The answer is required modulo \(10^9\). Since \(n\) can be enormous, the solution does not build sequences directly. Instead, it compresses the divisor set of \(n\) by residue modulo \(k\), then raises that residue distribution to the \(n\)-th power under cyclic convolution.

Mathematical Approach

Write \(D(n)=\{d\in\mathbb{Z}_{>0}: d\mid n\}\), and let \(M=10^9\). The key observation is that only residues modulo \(k\) matter, so the entire counting problem lives inside the cyclic group \(\mathbb{Z}_k\).

Step 1: Collapse the divisors into residue counts

For each residue \(r\in\{0,1,\dots,k-1\}\), define

$c_r=\left|\left\{d\in D(n): d\equiv r \pmod{k}\right\}\right|.$

The vector \(\mathbf{c}=(c_0,\dots,c_{k-1})\) records how many one-step choices land in each residue class. This loses no relevant information, because any future sum is reduced modulo \(k\).

Step 2: Concatenating choices becomes cyclic convolution

Suppose \(\mathbf{f}\) and \(\mathbf{g}\) describe the residue distributions of two independent blocks of choices. If the first block contributes residue \(i\) and the second contributes residue \(j\), then the combined residue is \(i+j \pmod{k}\). Therefore

$\left(\mathbf{f}\star\mathbf{g}\right)_t=\sum_{i=0}^{k-1} f_i\,g_{(t-i)\bmod k}.$

This is circular convolution on \(\mathbb{Z}_k\). After one pick the distribution is \(\mathbf{c}\); after two picks it is \(\mathbf{c}\star\mathbf{c}\); after \(m\) picks it is \(\mathbf{c}^{\star m}\).

Step 3: The target residue is fixed

The required sequences satisfy

$a_1+\cdots+a_n \equiv -n \pmod{k}.$

So the residue index to extract is

$t^\ast=(-n)\bmod k=\bigl(k-(n\bmod k)\bigr)\bmod k.$

Hence the desired count is

$A(n,k)=\left(\mathbf{c}^{\star n}\right)_{t^\ast}\pmod{M}.$

Step 4: Polynomial viewpoint in a quotient ring

Encode the residue vector as

$P(x)=\sum_{r=0}^{k-1} c_r x^r$

inside the ring \((\mathbb{Z}/M\mathbb{Z})[x]/(x^k-1)\). There, \(x^a x^b=x^{(a+b)\bmod k}\), so polynomial multiplication is exactly cyclic convolution. Therefore the coefficient of \(x^t\) in \(P(x)^n\) counts length-\(n\) divisor sequences whose sum is congruent to \(t \pmod{k}\). The answer is the coefficient of \(x^{t^\ast}\).

Step 5: Binary exponentiation removes the linear dependence on \(n\)

A direct dynamic program would perform one convolution per position and cost \(O(nk^2)\), which is impossible for the real input. Because convolution is associative, repeated squaring works: compute \(\mathbf{c}, \mathbf{c}^{\star 2}, \mathbf{c}^{\star 4}, \dots\), and combine only the powers corresponding to the set bits of \(n\). This reduces the main cost to \(O(k^2\log n)\).

Worked Example: \(A(3,4)=4\)

The divisors of \(3\) are \(1\) and \(3\), so modulo \(4\) the one-step vector is

$\mathbf{c}=(0,1,0,1).$

One more convolution gives

$\mathbf{c}^{\star 2}=(2,0,2,0),$

and the third choice gives

$\mathbf{c}^{\star 3}=(0,4,0,4).$

Since

$t^\ast=(-3)\bmod 4=1,$

the required component is \(4\). Equivalently, every term is either \(1\) or \(3\), and the total is \(1 \pmod 4\) exactly when an odd number of terms are \(3\), giving \(\binom31+\binom33=4\).

How the Code Works

The C++ and Java implementations first factor \(n\) by trial division and then enumerate every divisor recursively from the prime factorization. Those divisors are collapsed into a length-\(k\) count vector according to their residues modulo \(k\).

Next, the implementation performs circular convolution between two length-\(k\) vectors, always reducing arithmetic modulo \(10^9\). The compiled implementations use wider temporary accumulators and reduce periodically so that intermediate products remain safe.

Repeated squaring raises the one-step residue vector to the \(n\)-th convolution power, and the final answer is the component at residue \((-n)\bmod k\). The Python implementation returns the same value by delegating to the compiled solver, so the C++, Python, and Java implementations all follow the same mathematical method.

Complexity Analysis

Factoring \(n\) by trial division costs \(O(\sqrt{n})\) time in the worst case. If \(\tau(n)\) is the number of divisors of \(n\), then enumerating all divisors and filling the residue-count vector costs \(O(\tau(n))\) time and \(O(\tau(n))\) storage.

Each circular convolution of two vectors of length \(k\) costs \(O(k^2)\) time and \(O(k)\) additional working memory. Binary exponentiation uses \(O(\log n)\) such convolutions, so the full method runs in

$O\!\left(\sqrt{n}+\tau(n)+k^2\log n\right)$

time and uses

$O\!\left(k+\tau(n)\right)$

memory. For the actual input size, the convolution phase is the dominant cost.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=511
2. Circular convolution: Wikipedia — Circular convolution
3. Exponentiation by squaring: Wikipedia — Exponentiation by squaring
4. Divisor: Wikipedia — Divisor
5. Modular arithmetic: Wikipedia — Modular arithmetic