Palindrome Addition

Let $F(x)=\sum_{i=0}^{N}c_ix^i$. If $N=0$, then $F$ itself is palindromic, so we can output it directly. Assume $N\ge1$ from now on.

Construct coefficient arrays $a_0,\cdots,a_N$ and $b_0,\cdots,b_{N-1}$, and define

$$g(x)=\sum_{i=0}^{N}a_ix^i,\qquad h(x)=\sum_{i=0}^{N-1}b_ix^i.$$

Start with

$$a_0=a_N=c_N.$$

For $i=0,1,\cdots,\left\lfloor\frac{N-1}{2}\right\rfloor$, perform the following assignments in order, all modulo $M$:

$$b_i=c_i-a_i,\qquad b_{N-1-i}=b_i,$$ $$a_{i+1}=c_{N-1-i}-b_i,\qquad a_{N-1-i}=a_{i+1}.$$

Each iteration fixes the two outer coefficients of $b$ and then the next two outer coefficients of $a$. After the loop, $a_i=a_{N-i}$ for all $i$, so $g$ is palindromic, and $b_i=b_{N-1-i}$ also holds. Substituting the assignments coefficient by coefficient gives

$$F=g+h.$$

If $h=0$, output only $g$. If $h\ne0$ and $b_0\ne0$, then $b_{N-1}=b_0\ne0$, so $h$ has degree $N-1$ and is itself palindromic. In this case output $g,h$.

The remaining case is $h\ne0$ and $b_0=0$. Let $k$ be the smallest index with $b_k\ne0$. By symmetry, the first and last $k$ coefficients of $b$ are all zero and $b_{N-1-k}=b_k\ne0$. Define

$$d=N-1-2k,$$ $$r(x)=\sum_{j=0}^{d}b_{j+k}x^j.$$

Then $r$ is a nonzero palindromic polynomial and

$$h=x^kr.$$

Now define

$$s=r+x^kr.$$

Since $r$ is palindromic of degree $d$, the coefficients of $s$ are symmetric around degree $d+k$, so $s$ is palindromic as well. The polynomial $-r$ is also palindromic, and

$$s+(-r)=x^kr=h.$$

Therefore

$$F=g+s+(-r),$$

which uses at most three palindromic polynomials.

The leading coefficient of $g$ is $c_N\ne0$. In the two-polynomial case, the leading coefficient of $h$ is $b_0\ne0$. In the last case, the construction also gives nonzero leading coefficients for $r$, $s$, and $-r$. Hence every printed polynomial satisfies the definition in the statement.

Every coefficient is processed only a constant number of times, so the time complexity for one test case is $O(N)$ and the extra memory usage is $O(N)$. Over the whole input, the time complexity is $O\!\left(\sum(N+1)\right)$. Only addition and subtraction are used, so $M$ does not need to be prime.

Solution written by GPT5.6