Submission #339308

#	Time	Username	Problem	Language	Result	Execution time	Memory
339308		amunduzbaev	Experiments with Gorlum (IZhO13_expgorl)	C++14	0 / 100	406 ms	492 KiB

expgorl

/** made by amunduzbaev **/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
#include <ext/rope>

using namespace __gnu_pbds;
using namespace __gnu_cxx;
using namespace std;

#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define ub upper_bound
#define lb lower_bound
#define ll long long
#define ld long double
#define pii pair<int, int>
#define pll pair<ll, ll>
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(),x.rend()
#define fastios ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define vll vector<ll>
#define vii vector<int>
#define vpii vector<pii>
#define vpll vector<pll>
#define cnt(a)__builtin_popcount(a)
template<class T> bool umin(T& a, const T& b) {return a > b? a = b, true:false;}
template<class T> bool umax(T& a, const T& b) {return a < b? a = b, true:false;}

typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;

const int N = 1e7+5;
const int mod = 1e9+7;
const ll inf = 1e18;
const ld Pi = acos(-1);
ll n, m, k, ans, a[N];

void solve(){
	string s;
	int gx, gy, lx, ly;
	cin>>k>>s>>gx>>gy>>lx>>ly;
	int endx = gx, endy = gy;
	ld maxdis = (ld)sqrt((endx - lx) * (endx - lx) + (endy - ly) * (endy - ly));
	ld mindis = (ld)sqrt((endx - lx) * (endx - lx) + (endy - ly) * (endy - ly));
	while(k--){
		for(int i=0;i<sz(s);i++){
			if(s[i] == 'L') endx--;
			else if(s[i] == 'R') endx++;
			else if(s[i] == 'F') endy++;
			else if(s[i] == 'B') endy--;
			umax(maxdis, (ld)sqrt((endx - lx) * (endx - lx) + (endy - ly) * (endy - ly)));
			umin(mindis, (ld)sqrt((endx - lx) * (endx - lx) + (endy - ly) * (endy - ly)));
		}
	}
	
	cout<<fixed<<setprecision(12)<<mindis<<" ";
	cout<<fixed<<setprecision(12)<<maxdis<<"\n";
	return;
}

int main(){
	fastios
	int t = 1;
	if(t) solve();
	else {
		cin>>t;
		while (t--) solve();
	}
	return 0;
}

• Subtask 1: 0 / 100