Submission #734688

#	Time	Username	Problem	Language	Result	Execution time	Memory
734688		rainliofficial	Combo (IOI18_combo)	C++11	5 / 100	1 ms	464 KiB

combo

#include "combo.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define uid(a, b) uniform_int_distribution<int>(a, b)(rng)
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

/* 
Date: 2023/05/02 11:56
Problem Link: 
Topic(s):
Reflection:
Solution Notes:
We can obtain the first character in 2 moves:
guess AB
    if (>=1): then A or B
    else: then X or Y


Then we want to be able to determine 3 characters for every 1 query. 
Notice we get 4N queries, so it has something to do with 4 possibilities with every query. 

I initally didn't utilize the 4N constraint, and I thought about queries 2 new characters every time. 
If we have X, we query XAB: I hoped this would amortize to N queries. However, this would still end up
with the issue of having to determine > 2 possibilities with 1 query. 

Another key constraint is that the first character in S cannot appear again. We can thus use the first character, 
(e.g. X), to split differrent strings that we want to query. 

Consider the following method: 
X___ X___ X___ X___
What can we put in each of the empty spaces to for sure know the next character?  
XA XBA XBB XBC

If A is the next character -> return 1
If B is next char -> return 2
If Y is next character -> 0
*/

const int MAXN = 2e5+5, INF = 1e9;
vector<char> c = {'A', 'B', 'X', 'Y'};


string guess_sequence(int n){
    string s; 
    int x = press(s + "AB");
    if (x > 0){
    // start with A
        if (press(s + 'A') > 0) s += 'A';
        else s += "B";
    }else{
        if (press(s + 'X') > 0) s += "X";
        else s += "Y";
    }
    vector<char> cc;
    for (char j : c) if (j != s[0]) cc.push_back(j);
    for (int i=1; i<n-1; i++){
        string q; 
        q += s + cc[0]; 
        for (int j=0; j<3; j++){
            q += s + cc[1] + cc[j];
        }
        x = press(q);
        if (x == sz(s)) s += cc[2];
        else if (x == sz(s) + 1) s += cc[0];
        else if (x == sz(s) + 2) s += cc[1];
        else{
            assert(false);
        }
    }
    assert(sz(s) == n-1);
    if (press(s + cc[0]) > sz(s)) s += cc[0];
    else if (press(s + cc[1]) > sz(s)) s += cc[1];
    else s += cc[2];
    return s;
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 95