AtCoder Regular Contest 078 E

E - Awkward Response

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Time limit : 2sec / Memory limit : 256MB

Score : 800 points

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.

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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.

首先确定位数len，再确定前（len-1）位，再确定最后一位。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))
#define pos first
#define index second
#define mp make_pair
using namespace std;
const int inf=1e9+10;
const ll llinf=1e16+10;
const int maxn=2e5+10;
const int maxm=1e2+10;
const int mod=1e9+7;
int n;
char op[10];
int init()
{
    int k=10;
    for(int i=1;i<=9;i++)
    {
        printf("? %lld
",k);
        fflush(stdout);
        scanf("%s",op);
        if(op[0]=='N')
            return i;
        k=k*10;
    }
    k=9;
    for(int i=1;i<=9;i++)
    {
        printf("? %lld
",k);
        fflush(stdout);
        scanf("%s",op);
        if(op[0]=='Y')
            return i;
        k=k*10+9;
    }
    return 10;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int len=init();
    if(len==10)
    {
        printf("! %d
",(int)1e9);
        return 0;
    }
    int ans=0;
    for(int i=1;i<len;i++)
    {
        int l=0,r=9,t=0;
        while(l<=r)
        {
            int mid=(l+r)/2;
            printf("? %d
",ans*10+mid);
            fflush(stdout);
            scanf("%s",op);
            if(op[0]=='Y')
            {
                t=max(t,mid);
                l=mid+1;
            }
            else
                r=mid-1;
        }
        ans=ans*10+t;
    }
    for(int i=(len==1?1:0);i<=9;i++)
    {
        printf("? %lld
",((ll)ans*10+i)*10);
        fflush(stdout);
        scanf("%s",op);
        if(op[0]=='Y')
        {
            ans=ans*10+i;
            break;
        }
    }
    printf("! %d
",ans);
    return 0;
}