Submission #760216

#TimeUsernameProblemLanguageResultExecution timeMemory
760216OrazBMoney (IZhO17_money)C++14
0 / 100
1 ms2644 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <functional> using namespace __gnu_pbds; using namespace std; typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; //Dijkstra->set //set.find_by_order(x) x-position value //set.order_of_key(x) number of strictly less elements don't need *set.?? #define N 100005 #define wr cout << "Continue debugging\n"; #define all(x) (x).begin(), (x).end() #define ll long long int #define pii pair <int, int> #define pb push_back #define ff first #define ss second int n, mn, a[N], c[N]; vector<int> E[N]; void bit(int x){ if (x == n){ set<int> s; vector<int> pos(n+1, 0); for (int i = 1; i <= n; i++){s.insert(a[i]);E[i].clear();} int idx = 0; for (auto i : s) pos[i] = idx++; idx = 0; for (int i = 1; i <= n; i++){ E[idx].pb(a[i]); if (c[i]) idx++; } vector<int> vec(n+1, 0); for (int i = 0; i <= idx; i++){ int ind = pos[E[i][0]]; if (vec[ind]) ind++; for (int j = 0; j < (int)E[i].size(); j++){ vec.insert(vec.begin()+ind+j, E[i][j]); } } for (int i = 2; i <= n; i++){ if (vec[i] < vec[i-1]) return; } mn = min(mn, idx); return; } for (int i = 0; i < 2; i++){ c[x] = i; bit(x+1); } } int chk(){ mn = 1e9; bit(1); return mn; } int main () { ios::sync_with_stdio(false); cin.tie(0); cin >> n; for (int i = 1; i <= n; i++){ cin >> a[i]; } cout << chk(); }
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