Consider each binary bit independently.

Think about a greedy approach.

Try some special cases, such as $$$r=c=2$$$, $$$n=r$$$, or $$$m=c$$$, etc.

In terms of systems of equations, the given problem fixes for us the parities of at most (and exactly) $$$(n-r+1)\cdot(m-c+1)$$$ cells (by giving us a system of equations to solve with that many equations). What this means is that, if we are to fill an upper border of thickness $$$r-1$$$ and a left border of thickness of $$$c-1$$$, then every cell from that point onward is determined uniquely by some equation (Tip: imagine how this pans out for a $$$2\times2$$$ matrix; after filling the first row and left column arbitrarily, cell (2,2) is uniquely determined, then (2,3), and so on).

Therefore, the answer is $$$2^{nm - (n-r+1)(m-c+1)}$$$.

Try working through some cases by hand. You may notice that whenever there is more than one non-zero number, the first player wins. Then think about why this is true.

How does everyone’s contribution change when you give a gift to someone?

Consider determining all the numbers from back to front.

When you determine a number, subtract $$$1$$$ from all previous values that are greater than this number. This way, all $$$p$$$ values decrease by one, while all $$$s$$$ values remain unchanged. Now you only need to solve the problem of determining the last number.

By symmetry, all vertices are equivalent. Therefore, we only need to calculate the number of graphs in which a specific set is successful, and then multiply by $$$\binom{n}{m}$$$.

Let the nodes in the set be key nodes and the remaining nodes be non-key nodes. Consider the condition for the set to be successful and use the Principle of Inclusion-Exclusion to handle these conditions.

You might need to calculate the following: $$$n$$$ vertices form a set of disjoint cycles without self-loops. If there are $$$k$$$ cycles, the contribution is $$$(-1)^k$$$. Find the sum of contributions over all possible configurations.

The answer is $$$1-n$$$, which you can try to prove.

Use the Binomial Theorem to simplify the sum.

Notice that the problem can be transformed as follows: the $$$i$$$-th edge appears during the time interval $$$[\mathrm{low}_i, w_i]$$$. For each vertex, compute the maximum value of the departure time plus the weight of the destination vertex.

Consider doing the transitions from back to front.

We only need to support the following operations:

• Add and delete edges;
• Query the maximum value in a connected component;
• Take the max of all weights in a connected component with a given number.

Consider modifying the dynamic graph connectivity approach.

Consider the case where $$$l_i=0$$$.

Just Consider the most straightforward DP approach, and then try to find some patterns.

Try transforming the modulo $$$2$$$ counting DP into a game and solve it.

You should already know the answer for $$$l_i=0$$$; try using the Principle of Inclusion-Exclusion to count.

Enumerate which element is strictly greater than half of the total sum.

You should already know the $$$O(nm/w)$$$ algorithm. Consider that, under modulo $$$2$$$, we have $$$(1+x^k)^2=1+x^{2k}$$$.