Problem can be solved using dp with optimizations based on operations with formal power series. After some conversions the following formula can be obtained:

Calculate this expression and output coefficient at $$$x^k$$$.

This task can be solved using dynamic programming. State $$$dp[i][j]$$$ is count of sets of size $$$j$$$ that can be construct using first $$$i$$$ types of candy. Transitions in dynamic programming: $$$dp[i + 1][j + q \cdot b_i] = dp[i + 1][j + q \cdot b_i] + dp[i][j], (0 \le q \le a_i, j + q \cdot b_i \le k)$$$. Answer is $$$dp[n][k]$$$. This solution have complexity $$$O(n\cdot k^2)$$$ and doesn't fit in time limit. All transitions for fixed $$$i$$$ can be done in $$$O(k)$$$ using prefix sums. Thus, complexity is $$$O(n\cdot k)$$$ but it's slow too.

Let's try speed up solution using multiplying polynomials. Let

than

From this recursive expression, we can deduce that

Coefficient of the $$$k$$$-th degree $$$A_n(x)$$$ is answer to the problem. Multiplying polynomials of this kind is not convenient to calculate. Notice that

than expression for calculating $$$A_n(x)$$$ can be rewritten as follows:

This expression can be solved using technique for solving subset sum problem. Let's perform the following transformations:

We use the Taylor series formula to calculate logarithm: $$$log(1 + x) = -\sum_{t=1}^\infty \frac{(-1)^t\cdot x^t}{k}$$$. Thus, we get:

Let's rewrite this expression modulo $$$x^{k + 1}$$$ (the coefficients at large powers are not of interest to us):

Let $$$cnt_j$$$ is the count of $$$i$$$ that $$$j = (1 + a_i) \cdot b_i$$$. Then the formula takes the form:

Polynomial $$$R(x)$$$ can be calculated as well:

where $$$cnt_j^{\prime}$$$ is the count of $$$i$$$ that $$$j = b_i$$$.

Polynomials $$$L(x)$$$ and $$$R(x)$$$ can be calculated in time $$$O(n + \sum_{j=1}^{k}\frac{k}{j}) = O(n + k\cdot log(k))$$$.

Let $$$P(X) = L(x) - R(x) = \sum_{i=0}^{k}p_i \cdot x^i$$$. We need to calculate $$$G(x) = exp(P(x)) = \sum_{i=0}^{k}g_i \cdot x^i$$$ to get answer. There are many ways to calculate the exponent of the polynomial. For example we can differentiate $$$G(x)$$$: $$$G^{\prime}(x) = G(x) \cdot P^{\prime}(x)$$$ and get recursive expression for coefficients:

This expression for each $$$i$$$ can be calculated using technique ''meet-in-the-middle'' and fast Fourier transform with complexity $$$O(k\cdot log^2(k))$$$.

Answer to the problem is the coefficient of the $$$k$$$-th degree of the computed polynomial. Complexity of the solution is $$$O(n + k\cdot log^2(k))$$$.

If you know the location of the box with one candy and the box with two candies, you can compare two boxes in one query, and sort the array ($$$a < b$$$ if and only if $$$a + 1 < b + 2$$$ for different $$$a, b$$$)

Given 4 boxes, we can always eliminate at least one box for not being the box with one / two candies, in three queries (asking all possible pairings)

Therefore, we can start with every box, select a quadruplet and eliminate the box selected by the process above. We will end up with three boxes, such that the boxes with one and two candies are among them. I will leave it to you to figure out how to eliminate the last box. After that you sort and finish.