Submission #636854

#	Time	Username	Problem	Language	Result	Execution time	Memory
636854		saayan007	Beautiful row (IZhO12_beauty)	C++14	100 / 100	931 ms	205508 KiB

beauty

#include <bits/stdc++.h>
using namespace std;

/* higher types */
typedef int64_t ll;
typedef long double ld;

/* pairs */
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;

/* vectors */
typedef vector<int> vi;
typedef vector<ll> vl;

typedef vector<pi> vpi;
typedef vector<pl> vpl;

/* loops */
#define fur(i, a, b) for(ll i = a; i <= (ll)b; ++i)
#define ruf(i, a, b) for(ll i = a; i >= (ll)b; --i)

/* shortforms */
#define fr first 
#define sc second
#define pb push_back
#define mp make_pair

#define nl "\n"
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()

/* constants */
const ll infl = numeric_limits<int64_t>::max();
const int infi = numeric_limits<int>::max();
const int modi = 1e9 + 7;
const ll modl = 1e9 + 7;

ll three(ll x)
{
    ll ans = 0;
    while(x > 0)
    {
        if(x%3 == 1)
            ++ans;
        x /= 3;
    }
    return ans;
}

void solve()
{
    ll n;
    cin >> n;
    vl a(n), tw(n), th(n);
    fur(i, 0, n-1)
    {
        cin >> a[i];
        tw[i] = __builtin_popcountll(a[i]);
        th[i] = three(a[i]);
    }

    bool adj[n][n];
    fur(i, 0, n-1)
        fur(j, 0, n-1)
            adj[i][j] = (tw[i] == tw[j] || th[i] == th[j]);

    vector<vl> dp((1 << n), vl(n, 0));
    fur(i, 0, n - 1)
        dp[1 << i][i] = 1;
    fur(i, 1, (1 << n) - 1)
    {
        fur(j, 0, n - 1)
        {
            if(!(i&(1<<j)))
                continue;
            fur(k, 0, n - 1)
            {
                if(k == j || !(i & (1 << k)) || !adj[k][j])
                    continue;
                dp[i][j] += dp[i - (1 << j)][k];
            }
        }
    }

    ll ans = 0;
    fur(i, 0, n - 1)
        ans += dp[(1 << n) - 1][i];
    cout << ans << nl;
}

int main() 
{
    ios::sync_with_stdio(0);
    cin.tie(0);

	ll t = 1;
    // cin >> t;
    while(t--)
    {
        solve();
    }
}

• Subtask 1: 100 / 100