Submission #447382

#	Time	Username	Problem	Language	Result	Execution time	Memory
447382		Khizri	Sure Bet (CEOI17_sure)	C++17	0 / 100	0 ms	332 KiB

sure

#include <bits/stdc++.h>
using namespace std;
//------------------------------DEFINE------------------------------
//******************************************************************
#define IOS ios_base::sync_with_stdio(false); cin.tie(0),cout.tie(0)
#define ll long long
#define pb push_back		 
#define F first																 
#define S second 															 
#define INF 1e18
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define pii pair<int,int>
#define pll pair<ll,ll>
#define OK cout<<"Ok"<<endl;
#define MOD (ll)(1e9+7)
#define endl "\n"
//******************************************************************
//----------------------------FUNCTION------------------------------
//******************************************************************
ll gcd(ll a,ll b){
	if(a>b) swap(a,b);
	if(a==0) return a+b;
	return gcd(b%a,a);
}
ll lcm(ll a,ll b){
	return a/gcd(a,b)*b;
}
bool is_prime(ll n){
	ll k=sqrt(n);
	if(n==2) return true;
	if(n<2||n%2==0||k*k==n) return false;
	for(int i=3;i<=k;i+=2){
		if(n%i==0){
			return false;
		}
	}
	return true;
}
//*****************************************************************
//--------------------------MAIN-CODE------------------------------
const int mxn=1000+5;
int t=1,n;
double a[mxn],b[mxn],dp[mxn][mxn][3],m;
vector<pair<double,pair<double,double>>>vt;
void solve(){
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i]>>b[i];
	}
	double ans=0;
	vt.pb({{0},{0,0}});
	for(int j=1;j<=n;j++){
		int q=vt.size();
		for(int i=0;i<q;i++){
			vt.pb({vt[i].F+1,{vt[i].S.F+a[j],vt[i].S.S}});
			vt.pb({vt[i].F+1,{vt[i].S.F,vt[i].S.S+b[j]}});
			vt.pb({{1},{a[j],0}});
			vt.pb({{1},{0,b[j]}});
			ans=max(ans,min(vt[i].S.F+a[j]-(vt[i].F+1),vt[i].S.S-(vt[i].F+1)));
			ans=max(ans,min(vt[i].S.F-(vt[i].F+1),vt[i].S.S+b[j]-(vt[i].F+1)));
		}
	}
	printf("%.4lf",(double)ans);
}
int main(){
	IOS;
	//cin>>t;
	while(t--){
		solve();
	}
	return 0;
}

• Subtask 1: 0 / 20
• Subtask 2: 0 / 40
• Subtask 3: 0 / 40