Assume $$$s$$$ and $$$t$$$ are sorted. Let $$$dp[k][mask]$$$ be the weighted sum over all matchings of $$${t_1, t_2, \ldots t_k}$$$, to $$$ { s_i: i \in mask } $$$, where you need to multiply by $$$\frac{1}{(t_j - s_i)!}$$$ for matching $$$t_j$$$ to $$$s_i$$$.

Now, what is the number of possible states for a given $$$k$$$? $$$\binom{n}{k}$$$ is an obvious upper bound. Another upperbound is $$$\binom{N + k - n}{k}$$$, where $$$N = 33$$$ is the number of available positions. (To see this, note that the number of states is maximized when $$$s_i = i$$$ for all $$$i$$$ and $$$t_i = N - n + i$$$ for all i). So, for a given $$$k$$$, the number of states is at most $$$\binom{\min(n, N + k - n)}{k} \leq \binom{\frac{N+k}{2}}{k}$$$. Over various $$$k$$$, this quantity is maximized at $$$k \approx N / \phi$$$, and hence the number of states for a given $$$k$$$ never exceed $$$\binom{24}{15} \approx 1.3 \times 10^6$$$

Although not required, you can get rid of the map based (high constant or log) factors by a two pointers-ish approach. More details can be seen in the code:

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Let $$$U$$$ be the set of positive integers $$$< N$$$ , coprime to $$$N$$$. $$$S(N) = \sum_{u \in U} \frac{1}{u^2} = \sum_{v \in U} v^2 \pmod N $$$, as there is a one to one matching between elements and their inverses.

Now, we only want to sum over $$$v$$$, that have $$$gcd(v, N) = 1$$$. So, applying inclusion exclusion, we have:

Let $$$N = 2^{e_2} 3^{e_3} n$$$, where $$$n$$$ is coprime to $$$6$$$. Clearly, $$$S(N) = 0 \pmod n$$$. Also, for any $$$p \in { 2, 3}$$$, such that $$$e_p \neq 0$$$,

This is because, only the $$$Nd / 6$$$ term would matter, and $$$\sum_{d | N} d \mu(d) = \displaystyle \prod_{\text{prime } q \vert N} (1 - q)$$$.

Call a prime $$$q$$$ invalid if $$$q = 1 \pmod {3}$$$. If $$$N$$$ is a power of $$$2$$$, $$$s(N) = N/2 \pmod {N}$$$. Else, if $$$3 \nmid N$$$ or if an invalid prime divides $$$N$$$, then $$$S(N) = 0 \pmod {N}$$$. Otherwise, $$$ S(N) = \frac{N}{2} - \frac{N}{6} (-1)^{\omega(N)} \pmod N $$$.

So, once we fix the power of $$$3$$$ in $$$N$$$, we need to iterate over powers of all primes that are $$$2$$$ modulo $$$3$$$. The number of such numbers can be huge. However, we can iterate over $$$\frac{x}{lpf(x)}$$$, where $$$lpf(x)$$$ denotes the largest prime factor of $$$x$$$. This part is probably better explained in the code.

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Fiendish Five

Note that, the sum remains the same modulo 5. Since, we eventually want all numbers to be the same, we need that the initial sum is also divisible by 5. Let the numbers be $$$a_1, a_2, a_3, a_4, a_5$$$. Let $$$b_i = a_i - \sum_{j=1}^{5} a_i / 5$$$. Then, each operation is equivalent to adding $$$1$$$ to two indices and subtracting $$$1$$$ from two other indices. Hence, each operation can be represented by a column vector with one $$$0$$$, two $$$-1$$$'s and two $$$1$$$'s. There are $$$30$$$ such column vectors, but they exist in pairs since for each column vector $$$v$$$, $$$-v$$$ is also considered in these 30 vectors. Hence, we need to consider only $$$15$$$. Now, we need to select a subset of these $$$15$$$ columns, such that a given vector can be represented as an integer weighted sum of those. (Here, we allow negative weights as well, because again the negation of each vector is also a valid operation). There are two important observations:

• It is always possible to achieve the objective in $$$4$$$ operations. Hint: At least $$$3$$$ of the $$$5$$$ numbers must have the same parity.

• Any $$$3$$$ of the $$$15$$$ column vectors are linearly independent. Hence, if the answer is $$$< 4$$$, we can simply iterate over all subsets of size $$$\leq 3$$$ and find the unique solution corresponding to that subset if it exists, and see if that is integral. An optimization in this step is that we can preprocess the gaussian elimination for the corresponding matrices and store the sequence of row operations, and the final matrix in RRE form, and at the time of queries, just apply those operations on the target vector.

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