| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 1165796 | monkey133 | Weird Numeral System (CCO21_day1problem2) | C++20 | 0 ms | 0 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;
vector<ll> hh[5000005];
ll ans[5000005];
set<ll> found;
ll dfs(ll x, ll layer = 0)
{
if (found.count(x)) return -1;
found.insert(x);
for (ll u: hh[(x%k + k)%k])
{
if (x==u)
{
ans[layer] = u;
return layer + 1;
}
else if ((res = dfs((x - u)/k, layer + 1))!=-1)
{
ans[layer] = u;
return res;
}
}
return -1;
}
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();
ll x = dfs(a);
if (x!=-1)
{
for (ll i=x-1; i>=0; --i) cout << ans[i] << ' ';
cout << endl;
}
else {cout << "IMPOSSIBLE\n";}
}
return 0;
}