제출 #58687

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
58687		Benq	Gift (IZhO18_nicegift)	C++14	26 / 100	1470 ms	153792 KiB

nicegift

#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 = 1000001;

int N,K;
vpl A;
map<vi,ll> ret;

void ins(vi v, ll oc) {
    for (auto a: v) {
        A[a].f -= oc;
        if (A[a].f < 0) {
            cout << -1;
            exit(0);
        }
    }
    sort(all(v));
    assert(sz(v) == K);
    ret[v] += oc;
}

void input() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    srand(135195);
    cin >> N >> K; A.resize(N);
    F0R(i,N) {
        // A[i].f = rand() % 3+1;
        // cout << A[i].f << " ";
        cin >> A[i].f;
        A[i].s = i+1;
    }
    // cout << "\n";
    sort(all(A)); reverse(all(A));
}

void tri(int l, int r, int ind, ll ti) {
    vi v;
    F0R(i,l) v.pb(i);
    FOR(i,ind,ind+K-l) v.pb(l+(i%(r-l+1)));
    ins(v,ti);
}

int l,m,r;

void solve() {
    l = m = r = K-1;
    while (1) {
        while (l && A[l-1].f == A[l].f) l--;
        while (m+1 < N && A[m+1].f == A[m].f) m ++;
        r = m;
        while (r+1 < N && A[r+1].f == A[m].f-1) r ++;
        
        ll ti = INF;
        if (r+1 != K && l != 0) ti = (A[l-1].f-A[l].f)/(r+1-K); // A[l-1] - ti*(r-l+1) >= A[l]-ti*(K-l)
        ti = min(ti,(A[r].f-(r == N-1 ? 0 : A[r+1].f))/(K-l));
        if (ti) FOR(i,l,r+1) tri(l,r,i,ti);
        
        while (!(l != 0 && A[l-1].f == A[l].f) && !(r != N-1 && A[r+1].f < A[r].f)) {
            if (A[0].f == 0) return;
            vi v;
            F0R(i,l) { v.pb(i); }
            for (; sz(v) < K; ) { v.pb(m--); if (m < l) m = r; }
            ins(v,1);
        }
        
        /*cout << l << " " << r << "\n";
        cout << "A " << (l != 0 && A[l-1].f != A[l].f) << "\n";
        cout << "B " << (r != N-1 && A[r+1].f < A[r].f) << "\n";
        F0R(i,N) cout << A[i].f << " ";
        exit(0);*/
    }
}

void print() {
    cout << sz(ret) << "\n";
    for (auto a: ret) {
        cout << a.s << " ";
        for (int i: a.f) cout << A[i].s << " ";
        cout << "\n";
    }
}

int main() {
    input();
    solve();
    print();
}

/* 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: 7 / 7
• Subtask 2: 0 / 11
• Subtask 3: 0 / 12
• Subtask 4: 19 / 19
• Subtask 5: 0 / 51