Submission #1236619

#	Time	Username	Problem	Language	Result	Execution time	Memory
1236619		lioow	Poklon (COCI17_poklon7)	C++20	120 / 120	181 ms	96280 KiB

poklon

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define pb push_back
#define int long long
#define repp(i,x,n) for(int i=x;i<=n;i++)
#define rep(i,x,n) for(int i=x;i>=n;i--)
#define cy cout<<"YES"<<endl
#define cn cout<<"NO"<<endl
#define ca cout<<"Alice"<<endl
#define r0 return 0
#define fi first
#define se second
#define liow ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define jelek cout<<"jelek"<<endl
#define pii pair<int,int>
#define all(v) v.begin(),v.end()
#define pi pair<pair<int,int>,pair<int,int>>
#define p3 pair<int,pair<int,int>>
#define p2
#define tp tuple<int,int,int>
#define fl fflush(stdout)
#define lb long double
#define p5 pair<int,pair<pair<int,int>,pair<int,int>>>
#pragma GCC optimize ("O2")
#pragma GCC optimize ("unroll-loops")  
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<pair<int,int>, null_type,  less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef long long bryson;
const int mod=1e9+7;
const int SQMAX=635,INF=1e18;
//const int mod=998244353;
//const int MOD=1e6+3;
mt19937_64 rng((unsigned int) chrono::steady_clock::now().time_since_epoch().count());
pii dr[8]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int dxx[4]={1,1,-1,-1};
int dyy[4]={1,-1,-1,1};
int mul(int x,int y){return (x%mod*y%mod)%mod;}
int expo(int b,int e){
	if(e==0) return 1;
	int tmp=expo(b,e/2);
	if(e%2) return mul(tmp,mul(tmp,b));
	else return mul(tmp,tmp);
}
const int maxn=1e6+5;
const int N=18;
const long double eps=1e-6;

double maks=0;
int ang=0,dep=0;
int l[maxn],r[maxn];
void dfs(int idx,int depth){
	if(l[idx]<0 && maks<log2(-l[idx])+depth){
		maks=log2(-l[idx])+depth;
		ang=-l[idx];
		dep=depth;
	}
	if(r[idx]<0 && maks<log2(-r[idx])+depth){
		maks=log2(-r[idx])+depth;
		ang=-r[idx];
		dep=depth;
	}
	if(l[idx]>0) dfs(l[idx],depth+1);
	if(r[idx]>0) dfs(r[idx],depth+1);
}

void solve(){
	int n;cin>>n;
	repp(i,1,n){
		cin>>l[i]>>r[i];
	}
	dfs(1,1);
	string s="";
	while(ang){
		s+=(ang%2==1?"1":"0");
		ang/=2;
	}
	reverse(all(s));
	while(dep>0){
		dep--;
		s+="0";
	}
	cout<<s<<endl;
}

signed main(){
	liow;
	int tc=1;
//	cin>>tc;
	while(tc--){
		solve();
	}
}

• Subtask 1: 120 / 120