Submission #887996

#	Time	Username	Problem	Language	Result	Execution time	Memory
887996		Ahmed_Solyman	Klasika (COCI20_klasika)	C++14	33 / 110	98 ms	52820 KiB

klasika

/*
In the name of Allah
made by: Ahmed_Solyman
*/
#include <bits/stdc++.h>
#include <ext/rope>
 
using namespace std;
using namespace __gnu_cxx;
#pragma GCC optimize("-Ofast")
#pragma GCC optimize("-O1")
//-------------------------------------------------------------//
typedef long long ll;
typedef unsigned long long ull;
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define PI acos(-1)
#define lb lower_bound
#define ub upper_bound
#define endl '\n'
#define all(v) v.begin(),v.end()
#define allr(v) v.rbegin(),v.rend()
#define sum_to(n) (n*(n+1))/2
#define pb push_back
#define pf push_front
#define fil(arr,x) memset(arr,x,sizeof(arr))
const ll mod=1e9+7;
int dx[8]={0,1,0,-1,1,1,-1,-1};
int dy[8]={1,0,-1,0,1,-1,-1,1};
//-------------------------------------------------------------//
ll lcm(ll a,ll b)
{
    return (max(a,b)/__gcd(a,b))*min(a,b);
}
void person_bool(bool x)
{
    cout<<(x?"YES":"NO")<<endl;
}
const int N=6e6+5;
int trie[N][2];
int id=1;
void insert(int x){
    int p=0;
    for(int i=29;i>=0;i--){
        bool g=x & (1<<i);
        if(~trie[p][g])p=trie[p][g];
        else p=trie[p][g]=id++;
    }
}
int solve(int x){
    int p=0,ret=0;
    for(int i=29;i>=0;i--){
        bool g=x & (1<<i);g^=1;
        if(~trie[p][g])p=trie[p][g],ret|=(1<<i);
        else p=trie[p][g^1];
    }
    return ret;
}
int main()
{
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    #ifndef ONLINE_JUDGE
  ////  freopen("input.in", "r", stdin);
  ////  freopen("output.out", "w", stdout);
    #endif
    fast
    fil(trie,-1);
    int q;cin>>q;
    vector<int>d(q+1);
    int v=2;
    insert(0);
    while(q--){
        string s;cin>>s;
        if(s=="Add"){
            int u,w;cin>>u>>w;
            d[v]=d[u]^w;
            insert(d[v++]);
        }
        else{
            int a,b;cin>>a>>b;
            cout<<solve(d[a])<<endl;
        }
    }
    return 0;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 22
• Subtask 3: 33 / 33
• Subtask 4: 0 / 44