Submission #1104371

#	Time	Username	Problem	Language	Result	Execution time	Memory
1104371		EkinOnal	Mobile (BOI12_mobile)	C++17	0 / 100	1063 ms	67396 KiB

mobile

//#pragma GCC optimize("O3,unroll-loops,Ofast")
//#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")
#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
using namespace std;
// using namespace __gnu_pbds;
 
#define MAX 200005
#define pb push_back
#define mp make_pair 
#define int long long
#define f first
#define s second
#define vi vector<int>
#define pii pair<int,int>
#define si set<int>
#define vpii vector<pair<int,int>> 
const int mod = 1e9+7;
const int INF = 1e18;
// myMap.begin()->first :  key
// myMap.begin()->second : value
//#define double long double

int epow(int a,int b){int ans=1;while(b){if(b&1) ans*=a;a*=a;b>>=1;a%=mod;ans%=mod;}return ans%=mod;}
int gcd(int a,int b) {if(a<b)swap(a,b);while(b){int tmp=b;b=a%b;a=tmp;}return a;}
int mul(long long a,long long b){return ((a%mod)*(b%mod))%mod;}
int sum(long long a,long long b){return ((a%mod)+(b%mod))%mod;}
 // typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
// typedef
// tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_multiset;


void solve(){
	int n,L; cin>>n>>L;
	vpii v(n+2); for(int i=1;i<=n;i++) cin>>v[i].f>>v[i].s;

	long double l=0,r=2e9;
	while(r-l>1e-4){
		vector<pair<long double,long double>> arr;
		long double mid=(l+r)/2.0l;

		for(int i=1;i<=n;i++){
			long double a = sqrt(mid*mid-v[i].s*v[i].s);
			arr.pb({v[i].f-a,v[i].f+a});
		}

		sort(arr.begin(),arr.end());
		long double mx=0.0;
		for(auto u : arr){
			if(u.f<=mx) mx=max(mx,u.s);
		}		

		if(mx<(long double)L) l=mid;
		else r=mid;
	}
	cout << fixed << setprecision(6);
	cout<<l<<endl;

  	

	return;
}
 
int32_t main(/*int32_t argc, char* argv[]*/){
	ios_base::sync_with_stdio(0); cin.tie(0);
	// freopen("dining.in", "r", stdin);
	// freopen("dining.out", "w", stdout);		
 
	int t=1;
//	cin >> t;
	while (t--) solve();
 
	return 0;	 			
}

• Subtask 1: 0 / 4
• Subtask 2: 0 / 4
• Subtask 3: 0 / 4
• Subtask 4: 0 / 4
• Subtask 5: 0 / 4
• Subtask 6: 0 / 5
• Subtask 7: 0 / 5
• Subtask 8: 0 / 5
• Subtask 9: 0 / 5
• Subtask 10: 0 / 5
• Subtask 11: 0 / 5
• Subtask 12: 0 / 5
• Subtask 13: 0 / 5
• Subtask 14: 0 / 5
• Subtask 15: 0 / 5
• Subtask 16: 0 / 5
• Subtask 17: 0 / 5
• Subtask 18: 0 / 5
• Subtask 19: 0 / 5
• Subtask 20: 0 / 5
• Subtask 21: 0 / 5