By contradiction, we can prove that for an undirected graph, each edge belongs to at most one cycle if and only if there are at most two non-intersect paths between any two vertices. In addition, the maximum flow problem and the minimum cut problem are equivalent, so we only need to consider how to cut some edges with the lowest sum of capacities so that the source vertex $$$s$$$ is not connected to the sink vertex $$$t$$$.

Let's take a look at the $$$s$$$-$$$t$$$ cut problem. If there are at least two edges in one cycle that are cut, we can figure out by adjustment that at most two of edges in this cycle should be cut, since there are at most two non-intersect paths between $$$s$$$ and $$$t$$$. In addition, if there are many cycles where some edges of each cycle are cut, it can be known by adjustment and contradiction that, under the optimal strategy, at most one cycle can contain edges to be cut.

One observation is that if there is at least one edge in a cycle should be cut in an optimal way, there must be exactly two edges to be cut, and one of them must be the edge with the smallest capacity. Hence, we can remove the edge with the smallest capacity in each cycle and then add its capacity to the other edges in the cycle, which won't change the maximum flow between any two vertices in value.

The reduced graph forms a tree. If we add these edges in descending order of its capacity from scratch, we can determine $$$\mathrm{flow}(s, t) = w$$$ $$$(s \in S, t \in T)$$$ for the two sets of vertices $$$S$$$ and $$$T$$$ that are firstly connected after joining the edges with capacities no less than $$$w$$$.

We can maintain these sets of vertices by the disjoint set union and count in each set the number of vertices whose bits in some fixed position are $$$1$$$, and then it is easy to calculate the answer by enumerating each bit. Time complexity is $$$\mathcal{O}(m \log n)$$$.

Note that $$$\mathrm{flow}(s, t)$$$ can be $$$2 \times 10^9$$$, which means the answer may exceed $$$2^{63}$$$. To reach the peak value, we can just construct a circle with edges of the same weight $$$10^9$$$.

When enumerating bits, we can treat the lowest $$$(\log n)$$$ bits and other bits differently, which explains why the time complexity is $$$\mathcal{O}(m \log n)$$$ rather than $$$\mathcal{O}(m \log w)$$$.

For each permutation of positions, the minimum number of swaps equals to $$$n$$$ minus the number of disjoint cycles in the permutation.

We can solve the problem by enumerating all the $$$n$$$-permutations. For each permutation, we only need to check if it can be reached within $$$k$$$ swaps as swapping the same position is permitted.

Time complexity is expected to be $$$\mathcal{O}(m!)$$$, where $$$m = \left \lfloor \log_{10} n \right \rfloor + 1$$$. In order to achieve that, we can maintain some information about chains and cycles so that we can modify and retrieve efficiently during backtracking.

For each permutation of positions, the minimum number of swaps equals to $$$n$$$ minus the number of disjoint cycles in the permutation.

We can precalculate the minimum number of steps to obtain a permutation from the original one, and then enumerate permutations and update the answer. Time complexity is $$$\mathcal{O}\left(\sum_{i = 1}^{9}{(i \cdot i!)} + T m!\right)$$$, where $$$m = \left \lfloor \log_{10} n \right \rfloor + 1$$$.

Some solution cannot handle the case $$$n = 10^9$$$, but it can be solved by hand.

Many greedy solutions can be hacked even by random tests, so... the dataset is almost randomized. :P

As we know, the Möbius inversion formula can be expressed as

where $$$\mu$$$ is the Möbius function and the sums extend to all positive divisors $$$d$$$ of $$$n$$$.

Similarly, we can obtain the following formula.

If we apply this formula, the problem can be solved in $$$\mathcal{O}(2^{\omega(n)} n)$$$, where $$$\omega(n)$$$ means the number of distinct prime factors of $$$n$$$. To prove that, you need to know what the Möbius function is, and observe that $$$f(n)$$$ in this problem is a polynomial with only two non-zero terms. In addition, $$$\sum_{d | n}{\varphi(d)} = n$$$.

By the way, $$$\omega(n) \leq 6$$$ when $$$n \leq 10^5$$$, because $$$2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 = 510510$$$.

Q: Why does Fast Fourier Transform get a verdict of TLE? It seems $$$\log n \sim 2^{\omega(n)}$$$ when $$$n \leq 10^5$$$.

A: FFT implies big constant factor. Also, the above solution does not require floating point operation.

The union is difficult to compute but the intersection is easy, so let's apply $$$|S_u \cup S_v| = |S_u| + |S_v| - |S_u \cap S_v|$$$, where $$$S_x$$$ is the set of vertices whose distances to $$$x$$$ are no more than $$$w$$$.

Note that the tree is not changed, so you can precalculate the structure of its centroid decomposition, in order to calculate $$$|S_u|$$$ online.

Since the restricted distances are the same, if $$$S_u$$$ and $$$S_v$$$ have a non-empty intersection, their intersection will equal to the set of vertices whose distances to the middle point of the path between $$$u$$$ and $$$v$$$ are no more than $$$\left(w - \frac{dist(u, v)}{2}\right)$$$, which can be counted similarly. To avoid many discussions, you can add extra vertices at the middle points of edges.

Time complexity and space complexity are both $$$\mathcal{O}(n \log n)$$$. However, some solution may lose out due to cache-unfriendly behaviors. One way to ease that is to discuss the situations instead.

Note that the cutting areas do not intersect, which implies the different cutting areas do not affect each other, so we can reduce the problem into calculating angles of triangles. Cosine theorem would be helpful.

Anyway, you can paste your geometry coding template to solve, but it seems like overkill.

Though the data range is small, we should be careful calling functions like acos(x) where $$$x$$$ should be ranged in $$$[-1, 1]$$$, or otherwise NaN will give you a lesson.

The statement is a bit different from the above, however, the meanings are equivalent based on Dilworth's theorem.

We may be familiar to the case that $$$p_i = 2$$$ for all $$$i$$$, which is equivalent to find a maximal subset of the power set of some set $$$S$$$ such that none of the sets is a strict subset of any other. Besides, the conclusion, also known as Sperner's theorem, is that the size of the maximal subset is $$${|S| \choose \left \lfloor |S| / 2 \right \rfloor}$$$.

A generalization of Sperner's theorem could show the answer to this problem is the number of points satisfying $$$\sum_{i}{x_i}$$$ is equal to $$$M = \left \lfloor \frac{1}{2} \sum_{i}{(p_i + 1)} \right \rfloor$$$. The proof is similar so we omit it.

Then, as we solve many counting problems in the form of $$$\sum_{i}{x_i} = M~(1 \leq x_i \leq p_i)$$$, we can use inclusion-exclusion principle to get the answer

where $$$J = {1, 2, \ldots, n}$$$.

The remaining part is straightforward. We can regard $$${M - 1 - \sum_{i \in I}{p_i} \choose n - 1}$$$ as a low-degree polynomial of $$$\sum_{i \in I}{p_i}$$$ and use the meet-in-the-middle approach to split $$$I$$$ into two two halves almost evenly. Time complexity is $$$\mathcal{O}(2^{n / 2} n^2)$$$.

Queries and changes don't appear in turns, so we can solve the problem offline. We can build a sparse table to postpone changes like lazy propagation on segment tree.

For each operation $$$(l, r, v)$$$, let $$$d$$$ be $$$\left \lfloor \log_2(r - l + 1) \right \rfloor$$$, and replace this operation by two operations $$$(l, l + 2^d - 1, v)$$$ and $$$(r - 2^d + 1, r, v)$$$. Then each operation is able to be recorded in the sparse table. Also, it is convenient that when there are two operations covering the same interval, we can just keep the one with a higher value. After that, we split each recorded operation into two smaller ones in decreasing order of length. We do it repeatedly until the length of intervals is $$$1$$$, and then we get the answer.

Time complexity is $$$\mathcal{O}(m + n \log n)$$$ and space complexity is $$$\mathcal{O}(n \log n)$$$.

We can apply radix sort carefully and cache-friendly so that we are able to enumerate operations in decreasing order of value. Then using some trick to hit each position almost once may lead to success.

However, due to cache-unfriendly behaviors, there was only one team that used it to get accepted. We do not recommend using this solution because it requires optimization techniques.

Due to the randomness of operations, there are several approaches to pass.

For example, if we assume the number of useful operations is almost $$$\min(m, 2 n)$$$, we can use a linear approach to find operations with the largest $$$\min(m, 2 n)$$$ values, and then do whatever you want to solve this problem.

Another example is using the trick mentioned in Segment tree beats, created by Syloviaely. Amazingly, during the contest, there were tremendous Chinese teams using this approach to pass, even though many of them didn't comprehend its amortized complexity analysis.

You can write a typical DP to solve, but notice that the more complex your solution is, the more tiresome retrieving will be.

One elegant way is to enumerate the value range $$$[x, y]$$$ that needs to be reversed. In doing so, we only need to find find the longest subsequence in the form of $$$0^* 1^* \ldots x^* y^* (y - 1)^* \ldots x^* y^* (y + 1)^* \ldots 9^*$$$, where $$$k^*$$$ represents any non-negative times of occurrences of integer $$$k$$$. It also makes retrieving easy to implement.

This idea has been noticed in the tutorial for 933A - Извилистые движения, created by visitWorld. Time complexity is $$$\mathcal{O}(D^3 n)$$$, where $$$D = 10$$$ is the maximal number of distinct values.

Let's solve an easier version first. In this version, we are given $$$N, {R_i}, K$$$ and we have to choose $$$N$$$ integers $$$x_1, x_2, \ldots, x_N$$$ such that $$$x_i \in [0, R_i]$$$ for all $$$i$$$, and the bitwise XOR of them equals to $$$K$$$. Similar problems can be found at Crystals — POI 2006 and Stone Game — Hackerrank.

In case that $$$x_i \in [0, 2^m - 1]$$$ for all $$$i$$$, we know the problem is related to XOR equation on bits. Furthermore, when $$$(N - 1)$$$ ones of these variables are fixed, the rest one is determined. So if any solution exists, the number of solution is always $$$2^{m (N - 1)}$$$, no matter what $$$K$$$ is.

If we apply digit DP on bits, we can classify all the situations into different groups by the highest digit satisfying that there exists at least one variable $$$x_j$$$ which is strictly less than $$$R_i$$$ on this digit. Then, we know the lower bits of $$$x_j$$$ can be arbitrary, so we can leave its lower bits first and determine that by equation after other variables are enumerated and fixed. It is not difficult to design a digit DP to count. Time complexity for this part is $$$\mathcal{O}(N \log R)$$$.

Now let's turn back to the harder version, where $$$N$$$ becomes bigger and the lower bound of $$$x_i$$$ is limited. First, we can apply the inclusion-exclusion principle to eliminate the extra lower bound, in other words, we reduce it into many states in which either $$$x_i \in [0, R]$$$ or $$$x_i \in [0, L - 1]$$$ is satisfied. Then, let's take a look at the digit DP. We can record the sign caused by the inclusion-exclusion principle, and thus boost the same transition of DP using fast power algorithm on matrix multiplications. Time complexity is $$$\mathcal{O}((8^3 \log N + Q) \log R)$$$.

Oops... It seems like I've missed an important part — why matrix multiplications can reduce $$$2^N$$$ states into $$$2$$$ states. On each bit, we keep the higher bits as the same as the upper bounds, use DP to ensure at least one variable $$$x_j$$$ exists and determine XOR sum on this bit, and then determine the lower bits by XOR equation. The only difference among states is due to the upper bounds of the higher bits when we apply the inclusion-exclusion principle. Fortunately, their bitwise XOR sums differ when the parities of the numbers of upper bounds that are $$$R$$$ differ, so we can record the parity, which is equivalent to the sign caused by the inclusion-exclusion principle, to reduce the states.

Based on the fact observed in case that $$$x_i \in [0, 2^m - 1]$$$, many solutions can get accepted.

For example, let $$$dp(i, j)$$$ be the number of sequences $$$x_1, x_2, \ldots, x_i$$$ such that their XOR sum is $$$j$$$. We can observe for each $$$i$$$, values of $$$dp(i, j)$$$ can be split into several intervals with respect to $$$j$$$ such that values in each interval are the same. Maintaining these intervals can get accepted, as the number of intervals seems to be $$$\mathcal{O}(\log R)$$$, which is not relevant to $$$N$$$.

Tester quailty: If we exploit this property more deeply, we can get the exact formula and solve the problem in time complexity $$$\mathcal{O}((10 \log N + Q) \log R)$$$.

A simpler version can be found at GuessTheSubstring -- TCO 2011 Semifinal 1 500, created by misof. The harder version increases the upper bound of $$$n$$$ and requires high precision.

Before introducing our linear solution, we have to confess that it is too hard to create a dataset against solutions in time complexity $$$\mathcal{O}(n \log n)$$$. Why are judging machines so fast? If anyone has ideas to improve the dataset, please share in comments.

The process of guessing essentially constructs a decision tree, which is equivalent to make the Huffman encoding for all substrings of $$$A$$$. What makes things difficult is that there can be $$$\mathcal{O}(n^2)$$$ distinct substrings, so firstly we need to reduce the number of different initial states. Note that we only care about the number of the occurrences rather than what it is, and we know the maximal number is at most $$$n$$$, so if we can categorize substrings by their frequency, we may simulate the Huffman tree's construction quickly.

There are several approaches to count all the frequency of substrings in linear time. One approach is to build the suffix array through SA-IS algorithm, and then build the compressed suffix tree from this suffix array. Counting on the suffix tree is easy. However, due to the character set size, suffix automaton in space complexity $$$\mathcal{O}(n \Sigma)$$$ may fall down.

After counting, let's try to calculate the cost of building the Huffman tree in linear time. In this process, we need to merge two states with the lowest frequency into a new state repeatedly until there remains only one state. Note that the number of new state's occurrences is equal to the sum of the old two.

Denote $$$c_i$$$ as the number of states that occur $$$i$$$ times $$$(i = 1, 2, \ldots, n)$$$. For the states with the number of occurrences $$$\leq n$$$, we can speed up the consecutive same actions at once, for example, merge at most $$$\frac{c_{i}}{2}$$$ pairs of states that occur $$$i$$$ times. For the states with the number of occurrences $$$> n$$$, we can conclude the number of such states $$$< \frac{n}{2}$$$, as $$$\sum_{i}{(i \cdot c_i)}$$$ is always equal to $$$\frac{n (n + 1)}{2}$$$. We can use a first-in-first-out queue to maintain and merge these states so that this part is finished linearly.

The above explication also shows that you don't have to discuss the occurrences in cases, but only utilize the trick of a queue, or two queues, to implement. By the way, it seems this idea has occurred in 700D - Код Хаффмана на отрезке, created by GlebsHP and Endagorion, however, few people were able to take advantage of it.

When talking about data structures of suffices, we usually refer to suffix array, suffix tree and suffix automaton.

More constantly, we build suffix array by prefix doubling with radix sort in time complexity $$$\mathcal{O}(n \log n)$$$, and build suffix automaton or suffix tree in time complexity $$$\mathcal{O}(n C)$$$ or $$$\mathcal{O}(n \log \Sigma)$$$, where $$$C$$$ is the constant caused by cache miss, as the space complexity is a bit large, $$$\mathcal{O}(n \Sigma)$$$. In usual, we don't care about $$$C$$$, because $$$\Sigma$$$ is fairly small, like the size of the classical Latin alphabet, $$$26$$$.

Sometimes, problems require more efficiency to build some of them offline. In this case, we may have to build suffix array linearly and transform between any two of these structures. The transform is easy, for example, suffix array with LCP array can help construct the suffix tree, and suffix links of suffix automaton forms the suffix tree of reversed string. The linear construction for suffix array can be DC3 or SA-IS.

This is a classic problem of Pólya enumeration theorem, which is similar to Dodecahedron -- Petrozavodsk Winter Camp 2008, Ural SU Contest. Both are not hard after applying this theorem.

By applying that, we can only focus on the number of ways to set colors for orbits meeting the coloring conditions. For a situation in which every $$$d$$$ faces should have the same color, we can count $$$dp(i, j)$$$ as the number of ways such that the first $$$i$$$ colors have been used to color $$$j$$$ faces without breaking any condition.

There are only $$$4$$$ types of equivalence classes, so the time complexity is $$$\mathcal{O}(4 F^2 n)$$$, where $$$F = 60$$$ is the number of faces.

By the way, Pólya enumeration theorem requires division. In order to do it, and do it only once, in modulo $$$p$$$, you can calculate the other parts in modulo $$$|G| \cdot p$$$, where $$$|G| = 60$$$ is the total number of equivalence classes, and then divide it by $$$|G|$$$ directly. During that process, you should be careful with $$$64$$$-bit integer overflow.

The number of polyhedral rotation group is quite small. For example, chiral tetrahedral symmetry of order $$$12$$$ is the rotation group of regular tetrahedron, chiral octahedral symmetry of order $$$24$$$ is the rotation group of cube and octahedron, and chiral icosahedral symmetry of order $$$60$$$ is the rotation group of icosahedron and dodecahedron. Once we want to solve a problem concerning the isomorphism of 3D rotation, we can try to apply the above groups.

Let's first take a look at an easier version, where only binary operators $$$+$$$, $$$-$$$ are permitted. In this case, we can ignore parentheses by expanding them. Then, we can regard each operator as a sign of the variable behind it and the sign of the first variable is always positive, which can help us ignore the order of variables. In that way, we can conclude the number of different expressions is $$$(2^n - 1)$$$, as each sign must be either positive or negative, and we only need to make at least one positive sign.

Based on the above version, we can get a similar conclusion in case that only $$$*$$$, $$$/$$$ are permitted. But when at least three binary operators are permitted, things become more complex. Now let's discuss the case that $$$+$$$, $$$*$$$, $$$/$$$ (and with parentheses) are permitted, because to solve the original problem, we have to solve this task first.

In this case, parentheses cannot be expanded easily, but we can split the expression into priority levels. In each level, only $$$+$$$ is permitted or only $$$*$$$, $$$/$$$ are permitted. Each expression in a level $$$k$$$ is regarded as a single variable, or several expressions in the level $$$(k - 1)$$$ joined by operators in the level $$$k$$$.

We split expressions in this way so that we can better understand and ignore the order of an expression's successors. Also, the operators in the ancestor level cannot affect the successor levels. Hence, we can design a somewhat slow DP to count. Let $$$f(n)$$$ be the number expressions having $$$n$$$ variables such that its successors, if exists, is joined by $$$+$$$, and let $$$g(n)$$$ be the same for $$$*$$$, $$$/$$$. We have

where $$${n \choose x_1, x_2, \ldots, x_k}$$$ means $$$\frac{n!}{x_1! x_2! \ldots x_k!}$$$.

The transition is to enumerate $$$k$$$ unordered successors with their number of variables and relabel these $$$n$$$ variables, however, it's easy to understand but not easy to implement, so let's take a look at their exponential generating function.

Define that $$$F(x) = \sum_{n \geq 1}{\frac{f(n) x^n}{n!}}$$$, $$$G(x) = \sum_{n \geq 1}{\frac{g(n) x^n}{n!}}$$$. We have

where $$$\exp(A) = \sum_{k \geq 0}{\frac{A^k}{k!}}$$$.

We can calculte $$$\exp(A)$$$ based on a fact that $$$B = \exp(A) \Leftrightarrow B' = A' \exp(A) = A' B$$$. Together with that, we can desgin a DP to maintain the first $$$n$$$ terms of $$$F(x)$$$, $$$G(x)$$$, $$$\exp(F(x))$$$, $$$\exp(2 F(x))$$$, $$$\exp(G(x))$$$, in time complexity $$$\mathcal{O}(n^2)$$$. But if you are familiar with divide and conquer optimization with fast convolution in modulo integers, you can speed the process into $$$\mathcal{O}(n \log^2 n)$$$. The optimization is mentioned in the last paragraph of the tutorial for 438E - Ребенок и двоичное дерево, created by vfleaking. By the way, I've tried to apply the Newton-Iteration like approach, but I failed. If anyone comes up with a better solution, please share in comments.

Finally, let's focus on the original problem. The minus operator in a level can influence operators in the successor levels which may lead to duplicated counting, for example, $$$a + b / (c - d)$$$ and $$$a - b / (d - c)$$$ are equivalent, but if we keep the aforementioned counting approach, we will count them twice. It's not difficult to observe that if we regard an expression and its opposite (i.e. the expression such that their sum is zero) are equivalent, the counting will work. The transition is like

But there still exists a corner case that an expression's opposite may be invalid. In that case, the minus operator must not be involved in the expression, and how to count that has already been discussed. Therefore, the answer is twice the number of expressions that ignore the sign, minus the number of expressions that ignore the minus operator.

During the contest, I've made a trial. Before starting, I uploaded the first $$$300$$$ terms to OEIS, when there is no formula about this problem. Then, when I was randomly checking teams' submissions, I found about ten teams who had submitted the table of these terms.

I don't know what to say, but only hope the sport of programming competition won't be banned at some time in the future.