Submission #443896

#	Time	Username	Problem	Language	Result	Execution time	Memory
443896		zaneyu	Xoractive (IZhO19_xoractive)	C++14	100 / 100	8 ms	456 KiB

Xoractive

#include "interactive.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<ll,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll INF64=1e18+1;
const int INF=0x3f3f3f3f;
const int MOD=1e9+9;
const ld PI=acos(-1);
const ld eps=1e-3;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
inline ll mult(ll a,ll b){
    if(a>=MOD) a%=MOD;
    if(b>=MOD) b%=MOD;
    return (a*b)%MOD;
}
inline ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
vector<int> arr[7];
vector<int> guess(int n) {
    int a0=ask(1);
    REP(j,7){
        vector<int> v;
        REP1(i,n-1){
            if(i&(1<<j)) v.pb(i+1);
        }
        if(!sz(v)){
            continue;
        }
        vector<int> a=get_pairwise_xor(v);
        v.pb(1);
        vector<int> b=get_pairwise_xor(v);
        multiset<int> bb;
        for(auto x:b) bb.insert(x);
        for(auto x:a){
            bb.erase(bb.find(x));
        }
        for(auto x:bb){
            arr[j].pb(x^a0);
        }
    }
    vector<int> ans;
    map<int,bool> in;
    for(int i=n-1;i>=1;i--){
        map<int,int> cnt;
        int z=0;
        REP(j,7){
            if(i&(1<<j)){
                for(auto x:arr[j]){
                    cnt[x]++;
                }
                ++z;
            }
        }
        for(auto x:cnt){
            if(x.s==2*z and !in.count(x.f)){
                in[x.f]=1;
                ans.pb(x.f);
                break;
            }
        }
    }
    ans.pb(a0);
    reverse(ALL(ans));
    return ans;
}

• Subtask 1: 6 / 6
• Subtask 2: 94 / 94