A Certain Unscientific Repeated Prefix Sum: Solution¶

Method and contributions¶

The problem setter's solution expands the polynomial in rising-binomial polynomials, so each prefix sum increments a basis index. Codex (GPT-6) reduced conversion to quadratic time and linear space through a recurrence between powers, and used batch inverses for adjacent binomial coefficients. Use std.cpp. The setter's monomial-by-monomial triangular elimination has a complete explanation and alternative.cpp, also supporting the full range.

Work modulo the prime \(p=2147483647\) . Define

\[ B_0(x)=1,\qquad B_j(x)=\binom{x+j-1}{j}\quad(j\ge1) \]

Replacing repeated summation by binomial evaluation¶

The adjacent-difference identity for binomial coefficients gives, for every \(j\ge0\) ,

\[ \sum_{x=1}^{n}B_j(x)=B_{j+1}(n) \]

For \(j=0\) , this is simply the sum of \(n\) ones. If the input polynomial is \(P(x)=\sum_{j=0}^{k}w_jB_j(x)\) , applying \(m\) prefix sums gives

\[ \sum_{j=0}^{k}w_j\binom{n+m+j-1}{m+j}\pmod p \]

No recursion with \(m\) levels or array of length \(n\) is needed.

Quadratic basis conversion¶

The basis satisfies

\[ xB_j(x)=(j+1)B_{j+1}(x)-jB_j(x) \]

Keep the basis coefficients of the current monomial \(x^i\) in Current , initially \(x^0=B_0(x)\) . Add \(a_i\) times those coefficients to \(w\) . To obtain the next power, process \(j=i+1,i,\ldots,1\) and set

\[ Current_j\leftarrow j(Current_{j-1}-Current_j)\pmod p \]

Then set \(Current_0=0\) . Descending order preserves the previous power's coefficients on the right, and the previously absent highest coefficient is zero. Each degree scans a linear number of entries, giving \(O(k^2)\) work with two arrays of length \(k+1\) .

Binomial coefficients and batch inverses¶

First evaluate

\[ C_0=\binom{n+m-1}{m}=\frac{n(n+1)\cdots(n+m-1)}{m!} \]

Multiply numerator and denominator separately, then invert the denominator with one binary exponentiation. This takes \(O(m+\log p)\) time. Continue with

\[ C_j=C_{j-1}\frac{n+m+j-1}{m+j} \]

For denominators \(m+1,\ldots,m+k\) , compute prefix products \(D_j\) . Invert \(D_k\) once, then work backward. The current inverse product times \(D_{j-1}\) is \((m+j)^{-1}\) ; multiplying it by \(m+j\) obtains the preceding inverse product. All inverses therefore take \(O(k+\log p)\) time.

The largest \(n+m+k-1\) is \(2010000299<p\) , so every positive factorial factor involved is below the modulus and invertible. Products between modular reductions fit signed 64-bit integers because \(p^2<2^{63}\) .

Correctness and complexity¶

The multiplication-by-\(x\) identity proves inductively that Current represents each monomial. Weighted accumulation gives the complete input polynomial. Each prefix sum increments the basis index, so \(m\) sums yield exactly the displayed binomial expression. Products, inverses, and adjacent ratios evaluate those binomial coefficients correctly, proving the output correct.

The reference takes \(O(m+k^2+\log p)\) time and \(O(k)\) auxiliary space. The triangular program takes \(O(m+k^3+k\log p)\) time and \(O(k^2)\) space, providing a direct alternative interpretation of the conversion. The reference has a clear advantage in both conversion time and space.