Submission #1146352

#	Time	Username	Problem	Language	Result	Execution time	Memory
1146352		monkey133	Bikeparking (EGOI24_bikeparking)	C++20	25 / 100	377 ms	20368 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;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//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 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int main()
{
	ll n;
	cin >> n;
	ll tot = 0;
	vector<ll> have(n), want(n);
	for (ll i=0; i<n; ++i) cin >> have[i];
	for (ll i=0; i<n; ++i) {cin >> want[i]; tot += want[i];}
	ll lo = 0, hi = tot;
	while (lo!=hi)
	{
		ll mid = (lo + hi + 1) >> 1;
		ll leftt = mid;
		vector<pi> arr;
		vector<ll> have2 = have;
		for (ll i=n-1; i>=0 && leftt; --i) 
		{
			ll take = min(want[i], leftt);
			leftt -= take;
			arr.pb(mp(take, i));
		}
		reverse(all(arr));
		bool fail = 0;
		ll idx = 0;
		for (ll i=0; i<arr.size() && !fail; ++i)
		{
			while (arr[i].f)
			{
				ll take = min(arr[i].f, have2[idx]);
				have2[idx] -= take;
				arr[i].f -= take;
				if (idx >= arr[i].s) {fail = 1; break;}
				if (have2[idx]==0) idx++;
			}
		}
		if (fail) hi = mid-1;
		else lo = mid;
	}
	
	ll leftt = lo;
	ll idx = 0;
	for (ll i=n-1; i>=0; --i)
	{
		while (min(leftt, want[i]) > 0)
		{
			ll take = min(leftt, min(want[i], have[idx]));
			have[idx] -= take;
			want[i] -= take;
			leftt -= take;
			if (have[idx]==0) idx++;
		}
	}
	// pro
	
	ll ans = lo;
	//show(ans);
	idx = 0;
	for (ll i=n-1; i>=0; --i)
	{
		//show(i);
		while (want[i]> 0)
		{
			ll take =  min(want[i], have[idx]);
			have[idx] -= take;
			want[i] -= take;
			if (idx > i) ans-= take;
			if (have[idx]==0) idx++;
		}
	}
	cout << ans << endl;
	
	return 0;
}

• Subtask 1: 16 / 16
• Subtask 2: 9 / 9
• Subtask 3: 0 / 19
• Subtask 4: 0 / 24
• Subtask 5: 0 / 32