Submission #61811

#TimeUsernameProblemLanguageResultExecution timeMemory
61811Benqpopa (BOI18_popa)C++11
100 / 100
117 ms824 KiB
#include "popa.h" #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; /*vi seq = {12, 4, 16, 2, 2, 20}; int g(int l, int r) { int ret = 0; FOR(i,l,r+1) ret = __gcd(ret,seq[i]); return ret; } int query(int a, int b, int c, int d) { return g(a,b) == g(c,d); }*/ bool divi(int a, int b) { return query(b,a,a,a); } int join(int a, int b, int* R) { // a is higher R[a] = b; return a; } int solve(int N, int* Left, int* Right) { vi st; F0R(i,N) { int cur = -1; while (sz(st) && divi(i,st.back())) { cur = join(st.back(),cur,Right); st.pop_back(); } // if (sz(st)) cout << "OOPS " << i << " " << st.back() << " " << divi(i,st.back()) << "\n"; st.pb(i); Left[i] = cur, Right[i] = -1; /*for (int i: st) cout << i << " "; cout << "\n";*/ } int cur = -1; while (sz(st)) { cur = join(st.back(),cur,Right); st.pop_back(); } return cur; } /*int main() { ios_base::sync_with_stdio(0); cin.tie(0); int* Left = new int[6]; int* Right = new int[6]; cout << solve(6,Left,Right) << "\n"; F0R(i,6) cout << Left[i] << " "; cout << "\n"; F0R(i,6) cout << Right[i] << " "; cout << "\n"; }*/ /* 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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