Submission #1165803

#TimeUsernameProblemLanguageResultExecution timeMemory
1165803monkey133Weird Numeral System (CCO21_day1problem2)C++20
0 / 25
52 ms117824 KiB
#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll k,q, d, m, a,res;
vector<ll> hh[5000005], ans;
set<ll> found;


bool dfs(ll x, ll layer = 0)
{
	if (found.count(x)) return 0;
	found.insert(x);
	for (ll u: hh[(x%k + k)%k])
	{
	  if (x==u)
	  {
	    ans.pb(u);
	    return 1;
	  }
		else if (dfs((x - u)/k, layer + 1))
		{
		  ans.pb(u);
			return 1;
		}
	}
	return 0;
	
}


int main()
{
	cin >> k >> q >> d >> m;
	for (ll i=1; i<=d; ++i) 
	{
		cin >> a;
		hh[(a%k + k)%k].pb(a);
	}
	while (q--)
	{
		cin >> a;
		found.clear();
		ans.clear();
		if (dfs(a))
		{
			for (ll u: ans) cout << u << ' ';
			cout << endl;
		}
		else {cout << "IMPOSSIBLE\n";}
		
	}
	return 0;
}

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