Codeforces Round #610 (Div. 2) D. Enchanted Artifact Interaction + Thinking

Portal

Title:

Ideas:

First of all, we found that if we know the length of the string, we can $O(n+1)$ Solve the query in the second time. For example, the current length is$n$ , then we can construct a length of$n$ full${}^{\prime }{a}^{\text{'}}$ String, let me ask about his$cost$ , then traverse each bit and modify it${}^{\prime }{b}^{\prime \; To}$ see if the cost is reduced, if it cannot be reduced, change back to${}^{\prime }{a}^{\prime }$ , Otherwise update the cost.
In this case, we consider how to1 1Find the length in$one$ query.
First of all, it can be inserted, modified, and deleted. Modifying the length is not very realistic, we consider inserting and deleting.
First you can ask about'a''a${}^{\prime }{a}^{\text{'}}$ This character, the return value is$x$ . Now there are just a few situations:
$(1)\; The$ required string${}^{\prime }{a}^{\prime }$ , return$0$ , end directly.
$(2)\; The$ required strings are${}^{\prime }{b}^{\prime }$ , Then the length of this string must be$x$ because of this$One\; of\; the\; x$ changes is to${}^{\prime }{a}^{\text{'}}$ Change${}^{\prime }{b}^{\text{'}}$ , the rest are inserted${}^{\prime }{b}^{\prime }$。
$(3)\; The$ required string has at least${}^{\prime }{a}^{\prime }$ , Then the length of this string is$x+1$ because there is${}^{\prime }{a}^{\prime }$ , also need to insert$x$ number.
Of course, if you follow the above ideas directly, the number of times is$O(n+3)$ because we want to ask$x$ a${}^{\prime }{b}^{\prime ’S}$ cost, we also need to ask$x+.\; 1$ th${}^{\prime }{a}^{\prime }$ Cost, so we consider whether we can use the information that has been asked to solve it.
Consider if$x$ a${}^{\prime }{b}^{\prime If}$ not, suppose his return value is$y$ , then we know that the qualified length is$x+1$ , then we need to put$and--$ To increase a length, and then$y$ means that the initial state is${}^{\prime }{b}^{\prime ’S}$ cost, we can ask all${}^{\prime }{a}^{\prime }$ Operation is removed, because all${}^{\prime }{a}^{\prime }$ And all${}^{\prime }{b}^{\prime }$ Is the same, so the number of times is$O(n+2)$ .

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;

//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;

const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;

int x,y;
string s;
int a[1000];

int main()
{
    
    
//	ios::sync_with_stdio(false);
//	cin.tie(0);

    s="a";
    cout<<s<<endl; cout.flush();
    cin>>x;

    string ans;
    for(int i=1;i<=x;i++) ans+='b';
    cout<<ans<<endl; cout.flush();
    cin>>y;

    y--; s="";
    for(int i=1;i<=x+1;i++) s+='b';
    int mi=INF;
    for(int i=0;i<s.length();i++)
    {
    
    
        s[i]='a'; int now;
        cout<<s<<endl; cout.flush();
        cin>>now;
        if(now>y) s[i]='b';
        else y=now;
    }




	return 0;
}
/*

*/