| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 755943 | vjudge1 | Password (RMI18_password) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
#define vi vector<int>
#define vl vector<long long>
#define vii vector<pair<int,int>>
#define vll vector<pair<long long,long long>>
#define pb push_back
#define ll long long
#define ld long double
#define nl '\n'
#define boost ios::sync_with_stdio(false)
#define mp make_pair
#define se second
#define fi first
#define fore(i, y) for(ll i = 0; i < y; i++)
#define forr(i,x,y) for(int i = x;i<=y;i++)
#define forn(i,y,x) for(ll i = y; i >= x; i--)
#define all(v) v.begin(),v.end()
#define sz(v) v.size()
#define clr(v,k) memset(v,k,sizeof(v))
#define rall(v) v.rbegin() , v.rend()
#define pii pair<int,int>
#define pll pair<ll , ll>
const ll MOD = 1e9 + 7;
const ll INF = 1e18 + 1;
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (PGCD)
ll lcm(ll a , ll b) {return a * (b / gcd(a , b));} // least common multiple (PPCM)
// HERE IS THE SOLUTION
vector<vector<string>> possible(5);
void brute()
{
forr(i , 'a' , 'd')
{
string s = "";
s+=i;
forr(j , 'a' , 'd')
{
s+= j;
possible[2].pb(s);
forr(k , 'a' , 'd')
{
s+= k;
possible[3].pb(s);
forr(l , 'a' , 'd')
{
s+= l;
possible[4].pb(s);
s.erase(s.begin() + 3);
}
s.erase(s.begin() + 2);
}
s.erase(s.begin() + 1);
}
s.erase(s.begin());
}
}
int query(string str)
{
if(str == "abcd")return 4;
if(str == "abcdaaaa")return 8;
return 0;
}
string guess(int n,int s )
{
brute();
string t = "";
do
{
if(sz(t) + s > n)
{
if(sz(t) + s - 1 == n)
{
for(auto str : possible[s-1])
{
if(query(t + str) == sz(t) + s - 1)
{
t = t+ str;
break;
}
}
}
else if(sz(t) + s - 2 == n)
{
for(auto str : possible[s-2])
{
if(query(t + str) == sz(t) + s-2)
{
t = t + str;
break;
}
}
}
}
for(auto str : possible[s])
{
if(query(t + str) == sz(t) + s)
{
t = t+str;
break;
}
}
}
while(query(t) != n);
return t;
}