Submission #1268254

#	Time	Username	Problem	Language	Result	Execution time	Memory
1268254		iamarman	XOR (IZhO12_xor)	C++20	100 / 100	1749 ms	34204 KiB

xor

#include<bits/stdc++.h>

using namespace std;



struct Node {
    int child[2];
    int cnt;
    int anyIndex; // stores an example index of a number passing through this node
    Node(){ child[0]=child[1]=-1; cnt=0; anyIndex=-1; }
};

struct Trie {
    vector<Node> t;
    int B;
    Trie(int bits=31){ B = bits; t.emplace_back(); } // use 31..0 bits by default

    void clear(){ t.clear(); t.emplace_back(); }

    void insert(int val, int idx){
        int cur = 0;
        for(int i = B; i >= 0; --i){
            int b = (val >> i) & 1;
            if(t[cur].child[b] == -1){
                t[cur].child[b] = t.size();
                t.emplace_back();
            }
            cur = t[cur].child[b];
            t[cur].cnt++;
            t[cur].anyIndex = idx; // remember an index that goes through here
        }
    }

    // returns pair<maxxor, index_of_prefix_used_in_trie>
    pair<int,int> query_with_index(int x){
        int cur = 0;
        if(cur == -1) return {0, -1};
        int ans = 0;
        for(int i = B; i >= 0; --i){
            if(cur == -1) break;
            int b = (x >> i) & 1;
            int want = !b;
            if(t[cur].child[want] != -1 && t[t[cur].child[want]].cnt > 0){
                ans |= (1<<i);
                cur = t[cur].child[want];
            } else if(t[cur].child[b] != -1 && t[t[cur].child[b]].cnt > 0){
                cur = t[cur].child[b];
            } else {
                // no path
                cur = -1;
                break;
            }
        }
        int idx = (cur == -1 ? -1 : t[cur].anyIndex);
        return {ans, idx};
    }
};


void solve()
{
    int n, X;
    if(!(cin >> n >> X)) return;
    vector<int> vec(n+1,0);
    for(int i=1;i<=n;i++){
        cin >> vec[i];
        vec[i] ^= vec[i-1];
    }

    Trie trie(29); // 0..29 bits is enough for <=1e9, change to 31 if you prefer
    int bestLen = 0;
    int bestL = -1, bestLenActual = -1;

    auto ok = [&](int len)->bool{
        trie.clear();
        trie.insert(0, 0); // prefix a[0] at index 0
        for(int i = len; i <= n; ++i){
            trie.insert(vec[i-len], i-len);
            auto pr = trie.query_with_index(vec[i]);
            if(pr.first >= X){
                // pr.second is the prefix index j used; subarray is j+1 .. i
                bestL = pr.second + 1;
                bestLenActual = i - pr.second;
                return true;
            }
        }
        return false;
    };

    int lo = 1, hi = n+1;
    while(hi - lo > 1){
        int mid = lo + (hi - lo) / 2;
        if(ok(mid)) lo = mid;
        else hi = mid;
    }

    // re-run to get concrete indices for the maximal length lo
    if(ok(lo)){
        // print start and actual length (>= lo)
        cout << bestL << " " << bestLenActual << "\n";
    } else {
        // no subarray found at all
        cout << -1 << "\n";
    }

}
int main()
{ 
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    int t=1;
   // cin>>t;
    for(int i=1;i<=t;i++)
    {
         //cout<<"Case "<<i<<": ";
        solve();
    }
}

• Subtask 1: 100 / 100