X-OR
题面翻译
题目描述
本题是交互题。
有一个固定的长度为 \(n\) 的排列 \(P\)，其值域为 \([0,n-1]\)，你可以进行不超过 \(4269\) 次询问，之后你需要输出这个排列 \(P\)。
输入格式
第一行有一个正整数 \(n\)，表示排列的长度。
保证 \(3\le n\le 2048\)，\(0\le P_i\le n-1\)。
交互格式
你可以按照 ? a b 的格式进行询问，之后你会得到 \(P_a\) 与 \(P_b\) 的按位或。
当你需要输出 \(P\) 时，首先输出一个 !，之后输出 \(n\) 个整数 \(P_i\)。
题目描述
This is an interactive problem!
Ehab has a hidden permutation $ p $ of length $ n $ consisting of the elements from $ 0 $ to $ n-1 $ . You, for some reason, want to figure out the permutation. To do that, you can give Ehab $ 2 $ different indices $ i $ and $ j $ , and he'll reply with $ (p_i|p_j) $ where $ | $ is the bitwise-or operation.
Ehab has just enough free time to answer $ 4269 $ questions, and while he's OK with answering that many questions, he's too lazy to play your silly games, so he'll fix the permutation beforehand and will not change it depending on your queries. Can you guess the permutation?
输入格式
The only line contains the integer $ n $ $ (3 \le n \le 2048) $ — the length of the permutation.
输出格式
To ask a question, print "? $ i $ $ j $ " (without quotes, $ i \neq j $ ) Then, you should read the answer, which will be $ (p_i|p_j) $ .
If we answer with $ -1 $ instead of a valid answer, that means you exceeded the number of queries or made an invalid query.
Exit immediately after receiving $ -1 $ and you will see wrong answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.
To print the answer, print "! $ p_1 $ $ p_2 $ $ \ldots $ $ p_n $ " (without quotes). Note that answering doesn't count as one of the $ 4269 $ queries.
After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get idleness limit exceeded. To do this, use:
- fflush(stdout) or cout.flush() in C++;
- System.out.flush() in Java;
- flush(output) in Pascal;
- stdout.flush() in Python;
- See the documentation for other languages.
Hacks:
The first line should contain the integer $ n $ ( $ 3 \le n \le 2^{11} $ ) — the length of the permutation $ p $ .
The second line should contain $ n $ space-separated integers $ p_1 $ , $ p_2 $ , $ \ldots $ , $ p_n $ ( $ 0 \le p_i < n $ ) — the elements of the permutation $ p $ .
样例 #1
样例输入 #1
3
1
3
2
样例输出 #1
? 1 2
? 1 3
? 2 3
! 1 0 2
提示
In the first sample, the permutation is $ [1,0,2] $ . You start by asking about $ p_1|p_2 $ and Ehab replies with $ 1 $ . You then ask about $ p_1|p_3 $ and Ehab replies with $ 3 $ . Finally, you ask about $ p_2|p_3 $ and Ehab replies with $ 2 $ . You then guess the permutation.
只要找到 \(0\) 所在的位置，我们就可以轻松求出其他地方的值了。
关键在于如何快速找到 \(0\) 所在的位置。
先考虑如何快速求出某一位 \(i\) 的值。暴力的话枚举所有其他位置，询问 \(i,j\)，然后所有得到的结果取个与就行了。
发现不用枚举所有位置，期望情况下，只要随机 \(log logn\) 个位置询问取与就足够了。具体可以随机十个数。
求出某个数 \(p_i\)，后，如果询问 \(i,j\ne p_i\)，那么 \(j\) 不可能是 \(0\)，否则求出 \(p_j\) 的答案。
发现询问时 \(p_i\) 的个数会不断减少，所以只会询问 \(log\) 次可以得到 \(0\) 的位置，然后就好做了。总共询问次数 \(2n+lognlog logn\)
#include<bits/stdc++.h>
using namespace std;
const int N=3005;
mt19937 gen(time(0));
int n,p[N],lst,cnt,pt[N];
int ask(int x,int y)
{
printf("? %d %d\n",x,y);
fflush(stdout);
int a;
scanf("%d",&a);
return a;
// ++cnt;
// return pt[x]|pt[y];
}
void answer(int x)
{
// printf("%d\n",x);
for(int i=1;i<=n;i++)
if(x^i)
p[i]=ask(x,i);
p[x]=0;
putchar('!'),putchar(' ');
for(int i=1;i<=n;i++)
printf("%d ",p[i]);
fflush(stdout);
// for(int i=1;i<=n;i++)
// if(p[i]^pt[i])
// return puts("failed"),void();
// printf("succes %d",cnt);
}
int calc(int x)
{
int mn=32767;
for(int T=1;T<=10;T++)
{
int k=gen()%n+1;
if(k^x)
mn&=ask(x,k);
}
return mn;
}
int main()
{
scanf("%d",&n);
lst=1,p[1]=calc(1);
for(int i=2;i<=n;i++)
{
if(ask(i,lst)==p[lst])
{
p[i]=calc(i);
lst=i;
}
}
answer(lst);
}