제출 #58695

#제출 시각아이디문제언어결과실행 시간메모리
58695BenqGift (IZhO18_nicegift)C++14
7 / 100
1602 ms190272 KiB
#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; template<class T, int SZ> struct BIT { T bit[SZ+1]; BIT() { memset(bit,0,sizeof bit); } void upd(int k, T val) { // add val to index k for( ;k <= SZ; k += (k&-k)) bit[k] += val; } void upd(int l, int r, T val) { l ++, r ++; upd(l,val); upd(r+1,-val); } T query(int k) { k ++; T temp = 0; for (;k > 0;k -= (k&-k)) temp += bit[k]; return temp; } }; BIT<ll,MX> B; int N,K; vpl A, pre[MX]; map<vi,int> ret; void input() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> N >> K; A.resize(N); F0R(i,N) { // A[i].f = 2*(rand()%1000)+2; // cout << A[i].f << " "; cin >> A[i].f; A[i].s = i+1; } sort(all(A)); reverse(all(A)); F0R(i,sz(A)) B.upd(i,i,A[i].f); } void ins(vi v, ll oc = 1) { sort(all(v)); ret[v] += oc; } int l,m,r; void solve() { l = m = r = K-1; while (1) { // if (B.query(0) == 0) return; while (l && B.query(l-1) == B.query(l)) l--; while (m+1 < N && B.query(m+1) == B.query(m)) m ++; r = m; while (r+1 < N && B.query(r+1) == B.query(m)-1) r ++; ll ti = INF; if (r+1 != K && l != 0) ti = (B.query(l-1)-B.query(l))/(r+1-K); // A[l-1] - ti*(r-l+1) >= A[l]-ti*(K-l) ti = min(ti,(B.query(r)-(r == N-1 ? 0 : B.query(r+1)))/(K-l)); if (ti) { pre[l].pb({r,ti}); B.upd(l,r,-ti*(K-l)); B.upd(0,l-1,-ti*(r-l+1)); } while (!(l != 0 && B.query(l-1) == B.query(l)) && !(r != N-1 && B.query(r+1) >= B.query(l)-1)) { if (B.query(N-1) < 0) { cout << -1; exit(0); } if (B.query(0) == 0) return; vi v; F0R(i,l) { v.pb(i); } for (; sz(v) < K; ) { v.pb(m--); if (m < l) m = r; } for (int i: v) B.upd(i,i,-1); ins(v); } } } void print() { F0R(i,K) { vl st(N); for (auto a: pre[i]) { // cout << "AH " << i << " " << a.f << " " << a.s << "\n"; // [x,x+K-1-i] // x in [i,a.f-(K-1-i)] FOR(j,max((int)a.f-(K-1-i)+1,i),a.f+1) { vi v; F0R(z,i) v.pb(z); FOR(z,j-i,j-i+K-i) v.pb(z%(a.f-i+1)+i); ins(v,a.s); } if (i < a.f-(K-1-i)+1) st[i] += a.s, st[a.f-(K-1-i)+1] --; } F0R(j,N) { if (j) st[j] += st[j-1]; if (st[j]) { vi v; F0R(z,i) v.pb(z); FOR(z,j,j+K-i) v.pb(z); ins(v,st[j]); } } } 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 */
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