Submission #1124916

#	Time	Username	Problem	Language	Result	Execution time	Memory
1124916		pan	Stone Arranging 2 (JOI23_ho_t1)	C++20	100 / 100	142 ms	16260 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
using namespace std;
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
//typedef __int128 ull;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;

int main()
{
	ll n;
	cin >> n;
	vector<ll> arr(n);
	for (ll i=0; i<n; ++i) cin >> arr[i];
	deque<pi> dq;
	unordered_map<ll, ll> cnt;
	for (ll i=0; i<n; ++i) 
	{
		if (cnt[arr[i]])
		{
			ll ps = 1;
			while (dq.size() && dq.back().f != arr[i]) {cnt[dq.back().f]--; ps += dq.back().s; dq.pop_back();}
			dq.back().s += ps;
		}
		else
		{
			dq.pb(mp(arr[i], 1));
			cnt[arr[i]]++;
		}
	}
	for (pi u: dq) for (ll i=0; i<u.s; ++i) cout << u.f << endl;
	return 0;
}

• Subtask 1: 25 / 25
• Subtask 2: 35 / 35
• Subtask 3: 40 / 40