Awkward Response AtCoder

Problem Statement

 

This is an interactive task.

Snuke has a favorite positive integer, N. You can ask him the following type of question at most 64 times: "Is n your favorite integer?" Identify N.

Snuke is twisted, and when asked "Is n your favorite integer?", he answers "Yes" if one of the two conditions below is satisfied, and answers "No" otherwise:

• Both n≤N and str(n)≤str(N) hold.
• Both n>N and str(n)>str(N) hold.

Here, str(x) is the decimal representation of x (without leading zeros) as a string. For example, str(123)= 123 and str(2000) = 2000. Strings are compared lexicographically. For example, 11111 < 123 and 123456789 < 9.

Constraints

 

• 1≤N≤10⁹

Input and Output

 

Write your question to Standard Output in the following format:

? n

Here, n must be an integer between 1 and 10¹⁸ (inclusive).

Then, the response to the question shall be given from Standard Input in the following format:

ans

Here, ans is either Y or N. Y represents "Yes"; N represents "No".

Finally, write your answer in the following format:

! n

Here, n=N must hold.

Judging

 

• After each output, you must flush Standard Output. Otherwise you may get TLE.
• After you print the answer, the program must be terminated immediately. Otherwise, the behavior of the judge is undefined.
• When your output is invalid or incorrect, the behavior of the judge is undefined (it does not necessarily give WA).

Sample

 

Below is a sample communication for the case N=123:

Input	Output
	? 1
Y
	? 32
N
	? 1010
N
	? 999
Y
	! 123

• Since 1≤123 and str(1)≤str(123), the first response is "Yes".
• Since 32≤123 but str(32)>str(123), the second response is "No".
• Since 1010>123 but str(1010)≤str(123), the third response is "No".
• Since 999≥123 and str(999)>str(123), the fourth response is "Yes".
• The program successfully identifies N=123 in four questions, and thus passes the case.

题意：有一个1~1e9的数字，让你去猜。

你可以做最多64个询问，每一个询问评测机会根据这个规定来返回信息。

Snuke is twisted, and when asked "Is n your favorite integer?", he answers "Yes" if one of the two conditions below is satisfied, and answers "No" otherwise:

• Both n≤N and str(n)≤str(N) hold.
• Both n>N and str(n)>str(N) hold.

思路：

先确定这个数多少位，然后每一位用二分去得到具体这一位的数字。

我们从1到10，再100 ，1000， 每一次*10的去询问，就可以得到这个数字的位数。

如果我们问10，返回Y，问100，返回N，那么这一位是2位数，可以对照规定自己琢磨为什么。

知道多少位只有，我们每一个位置

int mid;
int l=0;
int r=9;

这样二分。

这里我利用了多一位的数一定n>N来一直进入第二个条件来询问的，

比如 是二位数，我问的时候问三位数，已知位放再数字中，未知位放0，询问位通过二分进行变化，

那么一定进入第二个条件，根据字典序的关系，来确定这一位的数字是几。

对于1和100这种1和1后面只有0的数，我们通过询问全是9的数来特判。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <strstream>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d
",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int query(string ans)
{
    char x;
    cout<<"? "<<ans<<endl;
    cin>>x;
    if(x=='Y')
    {
        return 1;
    }else
    {
        return 0;
    }
}
void solve()
{
    string temp="9";
    string ans="1";
    while(!query(temp))
    {
        temp+="9";
        ans+="0";
    }
    cout<<"! "<<ans<<endl;
}
int main()
{
    //freopen("D:\common_text\code_stream\in.txt","r",stdin);
    //freopen("D:\common_text\code_stream\out.txt","w",stdout);
    string ans="1";
    string res;
    while(1)
    {
        cout<<"? "<<ans<<endl;
         cin>>res;
        if(res[0]=='Y')
        {
            ans+="0";
        }else
        {
            break;
        }
        if(ans.length()>11)
        {
            solve();
            return 0;
        }
    }
    int len=ans.length();
    string temp;
    rep(i,0,len-1)
    {
        int mid;
        int l=0;
        int r=9;
        while(l<=r)
        {
            mid=(l+r)>>1;
            ans[i]='0'+mid;
            if(query(ans))
            {
                r=mid-1;
            }else
            {
                l=mid+1;
                temp=ans;
            }
        }
        ans=temp;
    }
    temp.pop_back();
    int num=temp.length();
    stringstream ss;
    ss.clear();
    ss<<temp;
    ll fans;
    ss>>fans;
    fans++;
    cout<<"! "<<fans<<endl;

    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '
');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}