MVCN2TST - Editorial


Div-2 Contest
Div-1 Contest

Author and Editorialist: Simon St James
Tester: Suchan Park




Persistence, Graph Theory, AVL Tree, Segment Tree


Given a rooted tree G, find the c_i^{\text{th}} element in a list L of pairs of nodes (u,v) that form a valid reparenting, with a specific ordering on L, and remove it from L, with each c_i being processed online. A valid reparenting is a (u,v) where severing u from its parent in G and adding an edge between u and v results in a tree.


The reparenting (u,v) is valid if and only if v is not a descendant of u. As we find each reparenting, we simulate its removal from L by mapping each c_i to point to the corresponding index X_i in the original L using e.g. an order-statistic tree, so the problem boils down to finding each requested element (u_i, v_i) at a given index X_i in the original L, online.

The solution is broken down into three “phases”, one for each of the three clauses in the ordering:

Phase One: Find u_i (the \textit{nodeToReparent}). For each node u in order, find the number of v's such that (u,v) is a valid reparenting, and form a prefix sum array from this. The first element in this array strictly greater than X_i gives our u_i, and can be found via binary search.

Phase Two: Find the \textit{newParentHeight} i.e. \textit{height}(v_i). We know u_i and need to find the Y_i^\textit{th} element in L that reparents u_i, where Y_i is easily computed. Using some simple observations and pre-computed lookups, we can compute the number of non-descendants of u_i with height less than or to any h (\textit{findNumNonDescendantsUpToHeight}(u_i, h)) in \mathcal{O}(\log N). Similarly to Phase One, we can then perform an Abstract Binary Search on h to find the correct \textit{height}(v_i).

Phase Three: Find the final v_i (\textit{newParent}). We know u_i and \textit{height}(v_i), and now just need to find v_i which is the Z_i^\text{th} node at height \textit{height}(v_i) that is not a descendant of u_i, where the index Z_i is easily computed. If we list all nodes with the given height in the order they are visited in a DFS (\textit{NHDFS}(h)) in another precomputation step, we see that the descendents of u_i at height \textit{height}(v_i) all lie in an easily-found interval (l,r) in \textit{NHDFS}(\textit{height}(v_i)). By forming persistent AVL Trees of the prefixes and suffixes of \textit{NHDFS}(h) for each h in the precomputation step, we can find in \mathcal{O}(1) a pair of AVL Trees representing all (sorted) nodes to the left of l and all to the right of r in \textit{NHDFS}(\textit{height}(v_i)), respectively. We can then adapt the standard algorithm for finding the Z_i^\text{th} element in two sorted arrays to work with AVL Trees and find the Z_i^\text{th} node not in (l,r) i.e. v_i.


As is often the case, my solution is rather clunkier than other people’s (the approach to Phase Three in particular), but I’m going to with my original solution regardless :slight_smile:

As mentioned in the Brief Explanation, the pair (u,v) is a valid reparenting if and only if v is not a descendent of u in the (rooted) tree G. The size of the list L of valid reparentings is \mathcal{O}(N^2); far too big to construct, let alone to erase from for all Q queries, so we have to be a bit more cunning!

Again as mentioned, we don’t actually remove elements from L; instead we track the indices of the elements of L that we’ve removed and use this information to map each new c_i to its corresponding index X_i in the original list L. Most people seem to use gcc’s internal __gnu_pbds::tree tree for this, though since I was writing a persistent AVL Tree anyway I used that to implement the tracking and mapping; see the \textit{IndexRemapper} class in my code.

Having sidestepped the issue of removing elements from L, the problem becomes “find the X_i^{\textit{th}} element in the original L, processing each X_i online”. I haven’t actually tried it, but I suspect that removing the requirement that the elements be found online would lead to a significantly easier problem.

Anyway, onto the first sub-problem, Phase One: finding u_i (\textit{nodeToReparent} in the code) of the remapped c_i, X_i, in the original list L, without constructing L!

There are no reparentings with u=1, so the first few elements of L are taken up by the valid reparentings that reparent node 2; then the next few are those that reparent node 3, etc. For a given u, the number of valid reparentings (u,v) that reparent u is simply the number of v such that v is not a descendent of u i.e. N-u.\textit{numDescendants}. We compute u.\textit{numDescendants} for all u in \mathcal{O}(N) in a precomputation step, and then create a prefix sum array \textit{CRPS} (\textit{numCanReparentToPrefixSum} in the code) such that \textit{CRPS}(u) is the total number of valid reparentings that reparent a node x with x \le u.

As an example, here is the L used in example test case 2 of the Problem, adjusted so that the indices are 0-relative:

$L$ for Example Testcase 2
index u v h(v)
0. 2 1 0
1. 2 5 1
2. 2 6 1
3. 2 7 1
4. 2 3 2
5. 2 4 2
6. 3 1 0
7. 3 2 1
8. 3 5 1
9. 3 6 1
10. 3 7 1
11. 3 4 2
12. 4 1 0
13. 4 2 1
14. 4 5 1
15. 4 6 1
16. 4 7 1
17. 4 3 2
18. 5 1 0
19. 5 2 1
20. 5 6 1
21. 5 7 1
22. 6 1 0
23. 6 2 1
24. 6 5 1
25. 6 7 1
26. 6 3 2
27. 6 4 2
28. 7 1 0
29. 7 2 1
30. 7 5 1
31. 7 6 1
32. 7 3 2
33. 7 4 2

and here is the table of \textit{CRPS}(u) for each u for this example, in order (u=1 is omitted):

u \textit{CRPS}(u)
2 6
3 12
4 18
5 22
6 28
7 34

If we look at the 17^\text{th} (0-relative!) element in L, we see that it reparents the node 4.
If we look at the 18^\text{th} element in L, we see that it reparents the node 5.
If we look at the 28^\text{th} element in L, we see that it reparents the node 7.
In general, hopefully the pattern is clear - the node reparented by the X_i^\text{th} element of L is the first u such that \textit{CRPS}(u) > X_i. Since \textit{CRPS} is non-decreasing, we can easily find this u using a binary search, and so find our u_i in \mathcal{O}(\log N), fulfilling Phase One.

Now that we know u_i, we can restrict our attention to the sub-list of L of reparentings that reparent u_i; our final desired parenting is at some index Y_i in this sublist. How do we find Y_i? In focussing on this sublist, we are ignoring all the first elements of L that reparent a node x < u_i, so we must subtract this number from X_i. By definition, this number is \textit{CRPS}(u_i - 1). The index Y_i is called \textit{numOfReparentingThatReparentsNode} in the code; I’ll be sticking with Y_i here, for obvious reasons :slight_smile:

So: for Phase Two, we need to find the height of v_i (\textit{newParentHeight} in the code), which is the height of v in the Y_i^\textit{th} valid reparenting that reparents u_i. In Phase One, we made a tally of all valid reparentings that reparented a node less than or equal to x, and found the first x such that this tally exceeded X_i; we do a similar trick here in finding the number of reparentings that reparent u_i to a node with height less than or equal to h, \textit{NDUH}(u_i, h) (\textit{findNumNonDescendantsUpToHeight} in the code) and find the first h such that \textit{NDUH}(u_i, h) exceeds Y_i. To work out how to compute \textit{NDUH}(u_i, h), consider the following schematic (click to animate):

  • The set of nodes to which we can reparent u_i consists of those in section A minus a set we’ll call \textit{AD}: the nodes in A that are descendents of u_i
  • \textit{AD} is equal to the set D of all descendents of u_i minus the set that we’ll call \textit{DH}: those nodes x\in D with x.\textit{height} > h
  • From the schematic, \textit{DH} is precisely the set of proper descendents of all y such that y\in D and y.\textit{height}=h.

Thus, \textit{NDUH}(u_i, h)=|A|-|AD|=|A|-|D|+|DH|. |A| and |D| are trivial to compute so it remains to compute this sum of proper descendents.

Now, in the precomputation step, we perform a DFS and use the common technique of logging, for each node x, the “time” at which we first visit x (x.\textit{dfsBeginVisit}) and the time at which we finish exploring x (x.\textit{dfsEndVisit}). We then form a list, for each h, of all nodes x with x.\textit{height} = h ordered by their \textit{dfsBeginVisit}, \textit{NHDFS}(h) (called \textit{nodesAtHeightInDFSOrder} in the code). \textit{NHDFS} will come in useful for Phase Three, too.

A well-known fact is that y is a descendant of x if and only if y.\textit{dfsBeginVisit} > x.\textit{dfsBeginVisit} and y.\textit{dfsEndVisit} < x.\textit{dfsEndVisit} and as a consequence, we see that the set of nodes at height h that are descendents of a node x form an interval in \textit{NHDFS}(h), and this interval [l, r] can be found via a binary search on \textit{NHDFS}(h) (see \textit{descendantRangeFor} in the code). To compute \textit{DH} we need to find the sum of the proper descendents of nodes in an interval of \textit{NHDFS}(h), which can be done by pre-computing prefix arrays (\textit{numProperDescendantsForNodeAtHeightPrefixSum} in the code).

So given a u_i and a height h, we can now compute \textit{NDUH}(u_i, h) in \mathcal{O}(\log N), and we need to find the first h such that this value exceeds Y_i. We could list all such \textit{NDUH} for each height h, but this would result in an \mathcal{O}(N^2) algorithm, so instead we use the fact that \textit{NDUH} is non-decreasing with h to perform an Abstract Binary Search on h, finally finding the object of Phase Two: \textit{height}(v_i).

Now onto Phase Three. We now know u_i and \textit{height}(v_i), and want to find v_i itself (called \textit{newParent} in the code), which is the Z_i^{\text{th}} element in the sublist of L that reparents u_i to a node with height \textit{height}(v_i). In restricting our attention to this sub-list, we are skipping all elements that reparent u_i to a parent with height less than \textit{height}(v_i), so Z_i = Y_i - \textit{NDUH}(u_i, \textit{height}(v_i) - 1). If \textit{height}(v_i)>0, of course :slight_smile: Z_i is called \textit{numOfReparentingForNodeAndNewHeight} in the code.

This is a slightly harder sub-problem to solve than Phase Two, but thankfully a bit easier to explain. Let H=\textit{NHDFS}(h). Recall that the set of nodes at height h that are descendents of u_i form an interval [l, r] in H. If we formed a list consisting of the first l - 1 elements of H and the last r - 1 elements of H, sorted it, and found the Z_i^\text{th} element in the sorted list, then we’d have our final v_i. This would be far too slow, however.

If we could get a sorted array of the first l-1 elements of H and a sorted array of the last r-1 elements, we could use the well-known Find the k^{\text{th}} Element of Two Sorted Arrays algorithm to find the required Z_i^\text{th} element, but obtaining these lists would still be too slow. We can do something similar, though, using persistent AVL Trees.

Recall that persistent (or versioned) data structures are structures that are in a certain sense immutable: when we appear to change them, we are in fact bumping the version number and making the change on the new version of the structure, leaving the previous version unchanged and accessible via the prior version number. For Persistent AVL Trees, we can still insert a value in \mathcal{O}(\log N) but with the added bonus that we can jump to any previous revision (one revision per insertion) of the tree in \mathcal{O}(1).

How does this help? Imagine that in our precomputation step we created for each h a persistent AVL Tree \textit{prefixesForHeight}(h). The first revision (version 0) of this tree is empty. We then insert the first value of H=\textit{NHDFS}(h) into the tree; version 1 contains the first element (the prefix of size one) of H, in sorted order. We then insert the second element of H; version 2 of the tree contains the first two elements (the prefix of size two) of H, again in sorted order. We continue until all elements of H are added, and then create a similar tree, \textit{suffixesForHeight}(h). Revision 0 of this tree is empty, and for revision one we add the last element of H (the suffix of size one). Then we get revision two by adding the last-but-one element of H, and so on until all elements have been added.

Now, this doesn’t give us the sorted array of the first l-1 elements of H or the last r-1 elements of H, but it does give us, in \mathcal{O}(1), a pair of trees representing this pair of sorted arrays. We can now adapt the “Find the k^{\text{th}} Element of Two Sorted Arrays” to work with AVL Trees instead of arrays to find the object of Phase Three, v_i (see \textit{findKthFromPair} in the code).

And that’s it!

Complexity Analysis

  • Precomputation: \textit{nodesAtHeightInDFSOrder}, \textit{numProperDescendantsForNodeAtHeightPrefixSum} and \textit{numCanReparentToPrefixSum} are \mathcal{O}(N) in total over all h; \textit{prefixesForHeight} and \textit{suffixesForHeight} are \mathcal{O}(N \log N)
  • Phase One - \mathcal{O}(N)
  • Phase Two - \mathcal{O}(\log N) calls to \textit{findNumNonDescendantsUpToHeight}, each of which is \mathcal{O}(\log N) due to the call to \textit{descendantRangeFor}
  • Phase Three - \textit{findKthFromPair} would be \mathcal{O}(\log N) if it were working with sorted arrays, but AVL Trees give it an extra \mathcal{O}(\log N).


Most people solved Phases One and Two in the same way, and most people used some form of persistence for Phase Three, but not many people used AVL Trees: instead, they opted for Persistent Segment Trees (see e.g. the Setter’s solution).

There were one or two solutions using Wavelet Trees, and a few using Merge Sort Trees.


Setter's Solution (C++)

Tester's Solution (C++)
    #include <bits/stdc++.h>
    #include <ext/pb_ds/assoc_container.hpp> // Common file 
    #include <ext/pb_ds/tree_policy.hpp> 
    typedef __gnu_pbds::tree<long long, __gnu_pbds::null_type, std::less<long long>, __gnu_pbds::rb_tree_tag, 
    namespace Input {
        const int BUFFER_SIZE = int(1.1e5);
        char _buf[BUFFER_SIZE + 10];
        int _buf_pos, _buf_len;
        char seekChar() {
            if(_buf_pos >= _buf_len) {
                _buf_len = fread(_buf, 1, BUFFER_SIZE, stdin);
                _buf_pos = 0;
            assert(_buf_pos < _buf_len);
            return _buf[_buf_pos];
        bool seekEof() {
            if(_buf_pos >= _buf_len) {
                _buf_len = fread(_buf, 1, BUFFER_SIZE, stdin);
                _buf_pos = 0;
            return _buf_pos >= _buf_len;
        char readChar() {
            char ret = seekChar();
            return ret;
        long long readLong(long long lb, long long rb) {
            char c = readChar();
            long long mul = 1;
            if(c == '-') {
                c = readChar();
                mul = -1;
            long long ret = c - '0';
            char first_digit = c;
            int len = 0;
            while(!seekEof() && isdigit(seekChar()) && ++len <= 19) {
                ret = ret * 10 + readChar() - '0';
            ret *= mul;
            if(len >= 2) assert(first_digit != '0');
            assert(len <= 18);
            assert(lb <= ret && ret <= rb);
            return ret;
        int readInt(int lb, int rb) {
            return readLong(lb, rb);
        void readEoln() {
            char c = readChar();
            assert(c == '\n');
            //assert(c == '\n' || (c == '\r' && readChar() == '\n'));
        void readSpace() {
            char c = readChar();
            assert(c == ' ');
    using namespace Input;
    const int MAX_N = 200000;
    const int MAX_SUM_N = 200000;
    const int MAX_Q = 200000;
    const int MAX_SUM_Q = 200000;
    class DisjointSet {
        std::vector<int> par;
        DisjointSet(int n): par(n+1) {
            for(int i = 1; i <= n; i++) par[i] = i;
        int get(int x) {
            return par[x] == x ? x : (par[x] = get(par[x]));
        bool merge(int x, int y) {
            x = get(x);
            y = get(y);
            par[x] = y;
            return x != y;
    namespace PersistentSegmentTree {
        struct node {
            int sum, left, right, ts;
        node tree[int(1.2e7)]; int NUM_NODES;
        int TIME;
        node& get_node(int x) {
            node& ret = tree[x];
            if(ret.ts < TIME) {
                ret = {0, 0, 0, TIME};
            return ret;
        node& new_node() {
            return get_node(++NUM_NODES);
        int insert(int ni, int nl, int nr, int x) {
            node &nd = get_node(ni);
            int ret = ++NUM_NODES;
            node &new_node = get_node(ret);
            new_node = nd;
            if(nl == nr) {
                new_node.sum += 1;
                return ret;
            int nm = (nl + nr) >> 1;
            if(x <= nm) {
                new_node.left = insert(nd.left, nl, nm, x);
            }else {
                new_node.right = insert(nd.right, nm+1, nr, x);
            new_node.sum = tree[new_node.left].sum + tree[new_node.right].sum;
            return ret;
        int get_sum (int ni, int nl, int nr, int x, int y) {
            if(ni == 0 || x > y) return 0;
            node &nd = get_node(ni);
            if(nl == x && nr == y) {
                return nd.sum;
            int nm = (nl + nr) >> 1;
            int ret = 0;
            if(x <= nm) {
                ret += get_sum(nd.left, nl, nm, x, std::min(y, nm));
            if(nm < y) {
                ret += get_sum(nd.right, nm+1, nr, std::max(x, nm+1), y);
            return ret;
        int get_kth(int nl, int nr, std::vector<int> nodes, std::vector<int> coefs, long long &k) {
            const int NUM_CONSIDERING_NODES = nodes.size();
            while(nl < nr) {
                int nm = (nl + nr) >> 1;
                int leftsum = 0;
                for (int i = 0; i < NUM_CONSIDERING_NODES; i++) {
                    if (!nodes[i]) continue;
                    auto &nd = get_node(nodes[i]);
                    if (!nd.left) continue;
                    auto &ndl = get_node(nd.left);
                    leftsum += ndl.sum * coefs[i];
                if (k < leftsum) {
                    for (int i = 0; i < NUM_CONSIDERING_NODES; i++) {
                        if (!nodes[i]) continue;
                        nodes[i] = get_node(nodes[i]).left;
                    nr = nm;
                } else {
                    k -= leftsum;
                    for (int i = 0; i < NUM_CONSIDERING_NODES; i++) {
                        if (!nodes[i]) continue;
                        nodes[i] = get_node(nodes[i]).right;
                    nl = nm+1;
            return nl;
        void clear() {
            NUM_NODES = 0;
    int sumN = 0, sumQ = 0;
    std::vector<int> gph[MAX_N+5];
    int depth[MAX_N+5];
    int subtreeSize[MAX_N+5];
    long long candidatePrefixSum[MAX_N+5];
    std::vector<int> ord;
    std::vector<int> nodes_with_depth[MAX_N+5], roots_with_depth[MAX_N+5];
    int L[MAX_N+5], R[MAX_N+5];
    int dfs (int u, int p) {
        depth[u] = depth[p] + 1;
        L[u] = ord.size();
        for (int v : gph[u]) if(v != p) subtreeSize[u] += dfs(v, u);
        R[u] = ord.size();
        return ++subtreeSize[u];
    int roots[MAX_N+5];
    void run() {
        int N = readInt(1, MAX_N);
        sumN += N;
        assert(sumN <= MAX_SUM_N);
        DisjointSet ds(N);
        for (int e = 0; e < N-1; e++) {
            int u = readInt(1, N);
            int v = readInt(1, N);
            bool merged = ds.merge(u, v);
        dfs(1, 0);
        for (int i = 1; i <= N; i++) {
            candidatePrefixSum[i] = candidatePrefixSum[i-1] + (N - subtreeSize[i]);
        for (int i = 1; i <= N; i++) {
            roots[i] = PersistentSegmentTree::insert(roots[i-1], 0, N, depth[ord[i-1]]);
        for (int d = 0; d <= N; d++) if(!nodes_with_depth[d].empty()) {
            for (int u : nodes_with_depth[d]) {
                int new_root = PersistentSegmentTree::insert(roots_with_depth[d].back(), 1, N, u);
        int Q = readInt(1, MAX_Q);
        sumQ += Q;
        assert(sumQ <= MAX_SUM_Q);
        new_tree usedPositions;
        const int MOD = 1'000'000'007;
        long long decryptionKey = 0;
        for (long long pow2 = 1, pow3 = 1, qid = 0; qid < Q; qid++) {
            (pow2 *= 2) %= MOD;
            (pow3 *= 3) %= MOD;
            long long encryptedChoice = readLong(0, (1ll << 36));
            long long decryptedChoice = (encryptedChoice ^ decryptionKey) - 1;
            long long target = -1;
                long long low = decryptedChoice;
                long long high = std::min(decryptedChoice + qid, candidatePrefixSum[N]);
                while (low <= high) {
                    long long mid = (low + high) / 2;
                    long long currentPosition = mid - usedPositions.order_of_key(mid+1);
                    if (currentPosition < decryptedChoice) {
                        low = mid+1;
                    } else {
                        target = mid;
                        high = mid-1;
                if (target < 0) {
                    printf("qid = %d, dc = %lld %lld\n", qid, decryptedChoice, candidatePrefixSum[N]);
                assert(target >= 0);
            // find target-th position from list
            // candidatePrefixSum[u-1] <= target < candidatePrefixSum[u]
            int u = std::upper_bound(candidatePrefixSum, candidatePrefixSum + N + 1, target) - candidatePrefixSum;
            target -= candidatePrefixSum[u-1];
            int d = PersistentSegmentTree::get_kth(
                0, N, 
                {roots[N], roots[R[u]], roots[L[u]-1]},
                {+1, -1, +1},
            int pl = std::distance(
                    nodes_with_depth[d].begin(), nodes_with_depth[d].end(), u,
                    [&](const int &a, const int &b) { return L[a] < L[b]; }
            int pr = std::distance(
                    nodes_with_depth[d].begin(), nodes_with_depth[d].end(), ord[R[u]-1],
                    [&](const int &a, const int &b) { return L[a] < L[b]; }
            int v = PersistentSegmentTree::get_kth(
                1, N,
                {roots_with_depth[d].back(), roots_with_depth[d][pr], roots_with_depth[d][pl]},
                {+1, -1, +1},
            assert(target == 0);
            assert(v > 0);
            assert(depth[v] == d);
            (decryptionKey += pow2 * u + pow3 * v) %= MOD;
        printf("%lld\n", decryptionKey);
        // initialize global vars
        for (int u = 0; u <= N; u++) {
            depth[u] = 0;
            subtreeSize[u] = 0;
            L[u] = R[u] = 0;
            candidatePrefixSum[u] = 0;
    int main() {
    #ifdef IN_MY_COMPUTER
        freopen("", "r", stdin);
        int T = readInt(1, 1000);
        while(T--) {
        return 0;

All diagrams and animations were created using the wonderful manim library. High-res MP4 renders can be found here.

How do i optimise this to pass second subtask? Calculating valid reparentings and sorting alone costs O(n^2logn). I’ve used lca on all pairs to check valid reparentings. For querying i’ve maintained a segment tree which keeps track of the deleted pair index i from the sequence.

1 Like

As explained, this approach is really untractable and you need to be more cunning :slight_smile:

Have a read of the Editorial; I know it’s a lot of text, but it should hopefully point you in the right direction :slight_smile:


Oh, sorry @lemme_try - I misread and didn’t notice that you asked for the second subtask. There’s a much faster way of deciding if a reparenting is valid - see the \textit{dfsBeginVisit} stuff in the Editorial. With an \mathcal{O}(N) pre-processing step, you can decide if one node is a descendent of the other in just \mathcal{O}(1)! :slight_smile:

Next problem that i got 15 points with correct complexity:
Next time add subtask with with smaller n please…

I looked to “Setter’s solution” to understand what i could do better. And i see 700 line code with some weird data structure and even time limit is less than 2 * max_time_of_this_solution. You should know that there are solutions using different approach with worst constant factor…

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I used Fenwick tree in phase 3

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My TODO list now includes reading the editorial in detail, I’ll do it as soon as I can, but in the meantime, I was courius about your alernative solution (without centroids decomp) and tried a bamboo tree (just one line) with all positions with 1 coin and I wasn’t able to get your alternate code to work, maybe I did something wrong? @ss

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Can you provide the precise test input you used?

A simple O(Nsqrt(N) ) algo worked for me…{simple being a relative word here}

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Sounds interesting - can I see it? :slight_smile:

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here it is

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I think this is why there are so many solutions…

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Would you say, perhaps, that the time limits aren’t tight enough? XD

Ya, we can maybe say that but 2*10^5 provoked me to try O(N sqrt(N)) if it would be maybe 10^6 then I would have thought something better!!

1 Like

… and this, everyone, is why being a Problem Setter sucks :slight_smile:


Moreover, I didn’t even try moving coins 2 properly coz I saw not many people are able to solve it…!! maybe you should update the difficulty level of moving coins 2

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I tried increasing stack size but didn’t work

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Sorry, is this for MOVCOIN2 or MVCN2TST? :slight_smile:


Ah, obviously MOVCOIN2 from context, but we’re in the wrong thread :slight_smile:

Hmmm … I can indeed reproduce the error - not sure which one is correct!


Wait, no I can’t - I was using the wrong binary before :slight_smile: Not sure what’s going on there; I’m getting:


for both of them. What OS and compiler are you using? Can you increase the stack size still further?

I tried both windows and linux but don’t worry, if it run in your system is by sure something of my configuration, anyway, someday I have to make peace with centroid decomposition

1 Like

Can you please tell what is the role of the sqrt(n) and your solution is based on which algorithm