Submission #1146165

#	Time	Username	Problem	Language	Result	Execution time	Memory
1146165		monkey133	Circle Passing (EGOI24_circlepassing)	C++20	100 / 100	155 ms	8252 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());

ll n, m, q;
ll dist(ll x, ll y)
{
	return min((y - x  + 2*n)%(2*n), (x - y  + 2*n)%(2*n));
}

int main()
{
	cin >> n >> m >> q;
	vector<ll> pos(2*m);
	for (ll i=0; i<m; ++i) cin >> pos[i];
	for (ll i=0; i<m; ++i) pos[m + i] = pos[i] + n;
	sort(all(pos));
	while (q--)
	{
		ll l, r;
		cin >> l >> r;
		ll ans = dist(r, l);
		ll idx1 = lb(all(pos), l)-pos.begin()-1, idx2 = lb(all(pos), l)-pos.begin();
		if (idx1 == -1) idx1 = 2*m-1;
		if( idx2==2*m) idx2 = 0;
		//show2(pos[idx1], pos[idx2]);
		//  go to left
		ll res = dist(pos[idx1], l) + dist((pos[idx1] + n)%(2*n), r)  + 1;
		// go to right
		ll res2 = dist(pos[idx2], l) + dist((pos[idx2] + n)%(2*n), r) + 1;
		ans = min(ans, res); ans = min(ans, res2);
		cout << ans << endl;
	}
	
	
	return 0;
}

• Subtask 1: 14 / 14
• Subtask 2: 20 / 20
• Subtask 3: 22 / 22
• Subtask 4: 17 / 17
• Subtask 5: 27 / 27