https://codeforces.com/contest/1104/problem/D
题解:倍增+二分
先特判等于1的情况
因为如果查询的两个数都在左边的话,肯定返回y,所以按倍增查找
1 2
2 4
4 8
8 16
的顺序倍增查询, 查出来目标在哪个区间
对区间【l,r】进行变形二分查找
先固定左边界 l 询问 l+l/2如果返回y代表目标不在【l,l+l/2】上,查询【l+l/2,3*l/4】
以此类推
最终可以得到答案
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
int ans,cnt,flag,temp;
int a[N];
char str[N];
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//scanf("%d",&n);
//scanf("%d",&t);
//while(t--){}
fflush(stdout);
ll x,y;
ll pos=0;
scanf("%s",str);
while(1){
if(str[0]=='s'){
x=1,y=2;
pos=0;
printf("? %I64d %I64d\n",x,y);
fflush(stdout);
scanf("%s",str);
if(str[0]=='x'){
printf("? %I64d %I64d\n",2ll,3ll);
fflush(stdout);
scanf("%s",str);
if(str[0]=='x'){
printf("! %I64d\n",1ll);
fflush(stdout);
}else{
printf("! %I64d\n",2ll);
fflush(stdout);
}
}else{
x*=2;
y*=2;
printf("? %I64d %I64d\n",x,y);
fflush(stdout);
}
}else if(str[0]=='x'){
pos=x;
x/=2;
while(x){
printf("? %I64d %I64d\n",pos,pos+x);
fflush(stdout);
scanf("%s",str);
if(str[0]=='y')
pos+=x;
x/=2;
}
printf("! %I64d\n",pos+1);
fflush(stdout);
}else if(str[0]=='y'){
x*=2;
y*=2;
printf("? %I64d %I64d\n",pos+x,pos+y);
fflush(stdout);
}else{
break;
}
scanf("%s",str);
}
//cout << "Hello world!" << endl;
return 0;
}