Codeforces-1088D：Ehab and another another xor problem（思维）

D. Ehab and another another xor problem
time limit per test 1 second
memory limit per test 256 megabytes
inputstandard input
outputstandard output
This is an interactive problem!

Ehab plays a game with Laggy. Ehab has 2 hidden integers (a,b). Laggy can ask a pair of integers (c,d) and Ehab will reply with:

$1$ if $a⊕c>b⊕d.$
$0$ if $a⊕c=b⊕d.$
$-1$ if $a⊕c<b⊕d.$
Operation a⊕b is the bitwise-xor operation of two numbers a and b.

Laggy should guess (a,b) with at most 62 questions. You’ll play this game. You’re Laggy and the interactor is Ehab.

It’s guaranteed that $0≤a,b<2^{30}$ .

Input
See the interaction section.

Output
To print the answer, print “! a b” (without quotes). Don’t forget to flush the output after printing the answer.

Interaction
To ask a question, print “? c d” (without quotes). Both c and d must be non-negative integers less than $2^{30}$ . Don’t forget to flush the output after printing any question.

After each question, you should read the answer as mentioned in the legend. If the interactor replies with -2, that means you asked more than 62 queries and your program should terminate.

To flush the output, you can use:-

fflush(stdout) in C++.
System.out.flush() in Java.
stdout.flush() in Python.
flush(output) in Pascal.
See the documentation for other languages.
Hacking:

To hack someone, print the 2 space-separated integers a and b $(0≤a,b<2^{30})$ .

Example
inputCopy
1
-1
0
outputCopy
? 2 1
? 1 2
? 2 0
! 3 1

思路：很明显我们需要一位一位的来确定a，b。从高位向低位来确定会简单点，假设a，b的高前几位已经确定好，当前正在确定第k位。有四种情况：
$a[k]=0，b[k]=0；$
$a[k]=0，b[k]=1；$
$a[k]=1，b[k]=0；$
$a[k]=1，b[k]=1；$
然后询问 $L=ask(a|(1<<k),b);R=ask(a,b|(1<<k));$
若 $a[k]=0，b[k]=0，$ 则 $L=1，R=-1；$
若 $a[k]=0，b[k]=1，$ 则 $L=R$ ，此时要根据a，b低 $k$ 位大小的比较才能确定；
若 $a[k]=1，b[k]=0，$ 则 $L=R，$ 此时要根据a，b低 $k$ 位大小的比较才能确定；
若 $a[k]=1，b[k]=1，$ 则 $L=-1，R=1；$

所以当 $L=R$ 时，若 $a_{低k位}>b_{低k位}，a[k]=1，b[k]=0$ ；否则 $a[k]=0，b[k]=1.$

如何确定低 $k$ 位的大小呢，我们可以一开始就比较出a，b的大小，然后逐步向下推。

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e6+10;
typedef long long ll;
int ask(ll x,ll y)
{
    printf("? %lld %lld\n",x,y);
    fflush(stdout);
    int ans;
    scanf("%d",&ans);
    return ans;
}
int main()
{
    ll a=0,b=0;
    int tag=ask(0,0);       //确定a,b大小
    for(int i=29;i>=0;i--)
    {
        int L=ask(a^(1ll<<i),b);
        int R=ask(a,b^(1ll<<i));
        if(L==R)
        {
            if(tag==-1)b^=1ll<<i;//低i位 a<b
            if(tag==1)a^=1ll<<i; //低i位 a>b
            tag=L;      //递推tag
        }
        else if(L==-1)
        {
            a^=1ll<<i;
            b^=1ll<<i;
        }
    }
    printf("! %lld %lld\n",a,b);
    return 0;
}

Codeforces-1088D：Ehab and another another xor problem（思维）

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