Submission #425669

#TimeUsernameProblemLanguageResultExecution timeMemory
425669Blistering_BarnaclesToy Train (IOI17_train)C++11
5 / 100
17 ms15052 KiB
#include "train.h"
//apig's property
//Happiness can be found, even in the darkest of times, if one only remembers to turn on the light
//El Pueblo Unido Jamas Sera Vencido
//The saddest thing about betrayal? is that it never comes from your enemies
//Do or do not... there is no try
//Billions of bilious blue blistering barnacles in a thundering typhoon!
#include<bits/stdc++.h>
#define fast ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0)
#define F first
#define S second
#define pb push_back
#define vll vector< ll >
#define vi vector< int >
#define pll pair< ll , ll >
#define pi pair< int , int >
#define all(s) s.begin() , s.end()
#define sz(s) s.size()
#define str string
#define md ((s + e) / 2)
#define mid ((l + r) / 2)
#define msdp(dp) memset(dp , -1 , sizeof dp)
#define mscl(dp) memset(dp , 0 , sizeof dp)
#define C continue
#define R return
#define B break
#define lx node * 2
#define rx node * 2 + 1
#define br(o) o ; break
#define co(o) o ; continue
using namespace std;
typedef long long ll;
ll q, dp[100005], a[555555] , b[555555], k, l, m, n, o, p;
map < ll , ll > mp;
vll adj[555555];
const ll mod = 1e9+7;
str s;
vi o1 , o2 ;
ll mem(ll i){
    if(i == n)R 0 ;
    ll &r = dp[i] ;
    if(r != -1)R r ;
    if(o1[i]){
        if(o2[i]){
            if(b[i] & (1 << 1))r = 1 ;
            else r = mem(i + 1) ;
        }
        else {
            if(b[i] & (1 << 0))R r = mem(i + 1) ;
            r = 0 ;
        }
    }
    else {
        if(o2[i]){
            if(b[i] & (1 << 0))r = mem(i + 1) ;
            else r = 1 ;
        }
        else {
            if(b[i] & (1 << 1))r = 0 ;
            else r = mem(i + 1) ;
        }
    }
    R r ;
}
vector<int> who_wins(std::vector<int> a, std::vector<int> r, std::vector<int> u, std::vector<int> v) {
	o1 = a , o2 = r ;
	msdp(dp) ;
	vi ans ;
	n = sz(a) ;
	m = sz(u) ;
	for(ll i = 0 ; i < m ; i++){
        ll op = (u[i] == v[i]) ;
        b[u[i]] |= (1 << (op)) ;
	}
	for(ll i = 0 ; i < n ; i++){
        ans.pb(mem(i)) ;
	}
	R ans ;
}
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