第一天 C. Chocolate Bunny

第一天 C. Chocolate Bunny

This is an interactive problem.

We hid from you a permutation p
of length n, consisting of the elements from 1 to n. You want to guess it. To do that, you can give us 2 different indices i and j, and we will reply with pimodpj (remainder of division pi by pj

).

We have enough patience to answer at most 2⋅n

queries, so you should fit in this constraint. Can you do it?

As a reminder, a permutation of length n
is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4

in the array).
Input

The only line of the input contains a single integer n
(1≤n≤104

) — length of the permutation.
Interaction

The interaction starts with reading n

.

Then you are allowed to make at most 2⋅n

queries in the following way:

"? x y" (1≤x,y≤n,x≠y

). 

After each one, you should read an integer k
, that equals pxmodpy

.

When you have guessed the permutation, print a single line "! " (without quotes), followed by array p

and quit.

After printing a query 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 documentation for other languages.

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.

Hack format

In the first line output n
(1≤n≤104). In the second line print the permutation of n integers p1,p2,…,pn

思路如下：由于可以查询 2*n 次，所以最少两次询问要确定一个数。每次询问是做模运算，很容易想到小数模大数就是它本身。所以两次一定可以确定哪个是小数

int n;
    cin >> n;
    int tmp = 1;
    rep(i, 2, n){
        int q, p;
        cout <<"? " << tmp << " " << i << endl;
        cin >> q;
        cout << "? " << i << " " << tmp << endl;
        cin >> p;
        if(q < p) {
            a[i] = p;
        } else {
            a[tmp]=q, tmp=i;
        }
    }
    a[tmp] = n;
    cout << "!";
    rep(i, 1, n){
        cout << " " << a[i];
    }
    cout << endl;
    return 0;