Submission #1192553

#	Time	Username	Problem	Language	Result	Execution time	Memory
1192553		stefanopulos	XOR (IZhO12_xor)	C++20	100 / 100	74 ms	23368 KiB

xor

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ldb;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;

#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for(auto& a : x)
 
#define sz(a) (int)(a).size()
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 250005; 

int n, X;
int niz[mxN];

int pref[mxN];

const int L = 30;

int idx = 0;
int mx[20 * mxN];
int lipa[20 * mxN][2];
void add(int A, int pos){
    int tr = 0;
    fb(i,L - 1,0){
        int a = ((A >> i) & 1);
        if(!lipa[tr][a])lipa[tr][a] = ++idx;
        tr = lipa[tr][a]; mx[tr] = max(mx[tr], pos);
    }
}

int query(int A){
    int tr = 0, ans = 0;
    fb(i,L - 1,0){
        int a = ((A >> i) & 1);
        int b = ((X >> i) & 1);

        if(b == 1){
            if(!lipa[tr][a ^ 1])return ans;
            tr = lipa[tr][a ^ 1];
        }
        else{
            if(lipa[tr][a ^ 1])ans = max(ans, mx[lipa[tr][a ^ 1]]);
            if(!lipa[tr][a])return ans;
            tr = lipa[tr][a];
        }

    }
    return max(mx[tr], ans);
}

int main(){
    cin.tie(0)->sync_with_stdio(0);

    cin >> n >> X;
    ff(i,1,n)cin >> niz[i];

    ff(i,1,n)pref[i] = pref[i - 1] ^ niz[i];

    int pos = 0, duz = 0;
    fb(i,n,1){
        add(pref[i], i);

        int gde = query(pref[i - 1]);
        if(duz <= gde - i + 1){
            duz = gde - i + 1;
            pos = i;
        }

    }

    cout << pos << " " << duz << '\n';

    return 0;
}
/*

4 6
6 1 2 4

4 6
0 0 0 6

// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 100 / 100