983B - XOR-pyramid - CodeForces Solution


dp *1800

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C++ Code:

#include <bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
// #define ordered_set tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag, tree_order_statistics_node_update>

typedef long long int ll;
typedef long double ld;

const ld eps = 1e-9;
const int mod = 1000000007;
const ll inf = 1e18;
const int N1 = 2e5 + 10;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    // mt19937 rng((unsigned int)chrono::steady_clock::now().time_since_epoch().count());
    // auto start = chrono::high_resolution_clock::now();

    // code here. dont use mbinpow,binpow,right, fact and isprime as function names
    // (1/b)%M is (b^(M-2))%M

    ll n;
    cin>>n;

    vector<ll> v(n);
    for (ll i = 0; i < n;i++){
        cin >> v[i];
    }

    vector<vector<ll>> dp(n, vector<ll>(n, 0));
    vector<vector<ll>> mx(n, vector<ll>(n, 0));
    for (ll i = 0; i < n;i++){
        dp[i][i] = v[i];
        mx[i][i] = v[i];
    }

    for (ll len = 1; len < n;len++){
        for (ll i = 0; i < n - len;i++){
            ll j = i + len;
            dp[i][j] = dp[i][j - 1] ^ dp[i + 1][j];
            mx[i][j] = max(dp[i][j], max(mx[i][j - 1], mx[i + 1][j]));
        }
    }

    ll q;
    cin >> q;

    while(q--){
        ll l, r;
        cin >> l >> r;
        l--, r--;

        cout << mx[l][r] << '\n';
    }
}


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