Submission #48919

#	Time	Username	Problem	Language	Result	Execution time	Memory
48919		Benq	Carnival (CEOI14_carnival)	C++14	100 / 100	15 ms	600 KiB

carnival

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

int n, numCol, col[151];
vi dis;

bool in(vi posi, int x) {
    cout << sz(posi)+1 << " ";
    for (int i: posi) cout << i << " ";
    cout << x << endl;
    int z; cin >> z;
    return z == sz(posi);
}

void ad(int x) {
    vi posi = dis;
    while (sz(posi) > 1) {
        vi POSI = vi(posi.begin(),posi.begin()+sz(posi)/2); 
        if (in(POSI,x)) posi = POSI;
        else posi = vi(posi.begin()+sz(posi)/2,posi.end());
    }
    if (in(posi,x)) col[x] = col[posi[0]];
    else {
        dis.pb(x);
        col[x] = ++numCol;
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    FOR(i,1,n+1) ad(i);
    cout << "0 "; 
    FOR(i,1,n+1) cout << col[i] << " ";
    cout << endl;
}

// read the question correctly (is y a vowel? what are the exact constraints?)
// look out for SPECIAL CASES (n=1?) and overflow (ll vs int?)

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 20 / 20
• Subtask 5: 20 / 20