https://codeforces.com/contest/1407/problem/C
Because it is a 1-n arrangement, then if pi%pj<pj%pi, then it must be pj=pj% pi, and compare it with two uncertain numbers each time, a total of 2*(n-1) queries You can find 1-n-1.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxl=3e5+10;
int n,m,cnt,tot,cas,ans;
int a[maxl];
bool vis[maxl],numin[maxl];
char s[maxl];
inline void prework()
{
scanf("%d",&n);
}
inline int print(int u,int v)
{
int x;
printf("? %d %d\n",u,v);
fflush(stdout);
scanf("%d",&x);
return x;
}
inline void mainwork()
{
if(n==1)
{
printf("! 1");
fflush(stdout);
return;
}
int t1,t2,last=1,now;
for(int i=1;i<=n-1;i++)
{
for(int j=1;j<=n;j++)
if(!vis[j] && last!=j)
{
now=j;
break;
}
t1=print(last,now);
t2=print(now,last);
if(t1>t2)
{
vis[last]=true;numin[t1]=true;
a[last]=t1;last=now;
}
else
{
vis[now]=true;numin[t2]=true;
a[now]=t2;last=last;
}
}
for(int i=1;i<=n;i++)
if(!vis[i])
a[i]=n;
printf("!");
for(int i=1;i<=n;i++)
printf(" %d",a[i]);
fflush(stdout);
}
int main()
{
int t=1;
//scanf("%d",&t);
for(cas=1;cas<=t;cas++)
{
prework();
mainwork();
}
return 0;
}