Codeforces Round #503 (by SIS, Div. 2) D. The hat -交互题，二分

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <cctype>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/
                const int maxn = 1e5+9;
                ll f[maxn];
                int n;
                void get(int m){
                    int x,y;
                    // cout<<"? "<<m<<endl;
                    printf("? %d\n",m);
                    fflush(stdout);
                    scanf("%d", &x);
                    
                    // cout<<"? "<<m+n/2<<endl;
                    // cin>>y;
                    printf("? %d\n",m + n/2);
                    fflush(stdout);
                    scanf("%d", &y);
                    f[m] = y - x;
                }
int main(){

                // cin>>n;
                scanf("%d", &n);
                int le = 1, ri = n/2;
                get(1);
                if(f[1]==0){
                    printf("! 1\n");
                    fflush(stdout);
                }
                while(le<=ri){
                    int mid = (le + ri) / 2;
                    get(mid);
                    if(f[mid]==0){
                        printf("! %d\n",mid);
                        fflush(stdout);
                        return 0;
                    }
                    if(f[1]>0){
                        if(f[mid]>0){
                            le = mid+1;
                        }
                        else ri = mid - 1;
                    }
                    else {
                        if(f[mid]<0){
                            le = mid+1;
                        }
                        else ri = mid - 1;    
                    }
                }
                printf("! -1\n");
                fflush(stdout);

                return 0;
}