Submission #58493

#	Time	Username	Problem	Language	Result	Execution time	Memory
58493		Benq	Nice sequence (IZhO18_sequence)	C++14	6 / 100	4 ms	860 KiB

sequence

#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 T;

vi tri(int N, int M) { // N > M 
    if (N % M == 0) {
        vi z; F0R(i,N-1) z.pb(1);
        return z;
    }
    int r = N%M;
    // int res = 2*(N-1)-(N%M); // middle is positive
    vi v;
    F0R(i,N-1) {
        if (i % M < r) v.pb(1);
        else v.pb(-1);
    }
    vi res = vi(v.begin()+r,v.end()); reverse(all(res));
    res.insert(res.end(),all(v));
    
    pi x = {r,M-r}; // r*A - (M-r)*B > 0 --> r*A > (M-r)*B --> A/B > (M-r)/r 
    pi y = {0,0}; F0R(i,N) if (res[i] == 1) y.f ++; else y.s ++; 
    
    pi z = {y.s+x.s,y.f+x.f};
    F0R(i,sz(res)) {
        if (res[i] == 1) res[i] *= z.f;
        else res[i] *= z.s;
    }
    return res;
    /*cout << "AH " << N << " " << M << "\n";
    for (int i: res) cout << i << " ";
    exit(0);
    return {};*/
}

vi flip(vi z) {
    F0R(i,sz(z)) z[i] *= -1;
    return z;
}

vi triN(int N, vi z) {
    if (sz(z) < N) return z;
    int tmp = 0; F0R(i,N) tmp += z[i];
    if (tmp > 0) z = flip(z);
    return z;
}

vi triM(int N, vi z) {
    if (sz(z) < N) return z;
    int tmp = 0; F0R(i,N) tmp += z[i];
    if (tmp < 0) z = flip(z);
    return z;
}

void solve() {
    int N,M; cin >> N >> M;
    vi z = tri(max(N,M),min(M,N));
    
    z = triN(N,z);
    z = triM(M,z);
    
    cout << sz(z) << "\n";
    for (int i: z) cout << i << " ";
    cout << "\n";
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> T;
    F0R(i,T) solve();
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 6 / 6
• Subtask 2: 0 / 9
• Subtask 3: 0 / 14
• Subtask 4: 0 / 15
• Subtask 5: 0 / 14
• Subtask 6: 0 / 18
• Subtask 7: 0 / 24