This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define X first
#define Y second
#define y0 y12
#define y1 y22
#define INF 987654321987654321
#define PI 3.141592653589793238462643383279502884
#define fup(i,a,b,c) for(int (i)=(a);(i)<=(b);(i)+=(c))
#define fdn(i,a,b,c) for(int (i)=(a);(i)>=(b);(i)-=(c))
#define MEM0(a) memset((a),0,sizeof(a));
#define MEM_1(a) memset((a),-1,sizeof(a));
#define ALL(a) a.begin(),a.end()
#define SYNC ios_base::sync_with_stdio(false);cin.tie(0)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef double db;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> Pi;
typedef pair<ll, ll> Pll;
typedef pair<ld, ld> Pd;
typedef vector<int> Vi;
typedef vector<ll> Vll;
typedef vector<ld> Vd;
typedef vector<Pi> VPi;
typedef vector<Pll> VPll;
typedef vector<Pd> VPd;
typedef tuple<int, int, int> iii;
typedef tuple<int, int, int, int> iiii;
typedef tuple<ll, ll, ll> LLL;
typedef vector<iii> Viii;
typedef vector<LLL> VLLL;
typedef complex<double> base;
const int MOD = 42043;
ll POW(ll a, ll b, ll MMM = MOD) { ll ret = 1; for (; b; b >>= 1, a = (a*a) % MMM)if (b & 1)ret = (ret*a) % MMM; return ret; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { if (a == 0 || b == 0)return a + b; return a*(b / gcd(a, b)); }
int dx[] = { 0,1,0,-1,1,1,-1,-1 }, dy[] = { 1,0,-1,0,1,-1,1,-1 };
int ddx[] = { -1,-2,1,-2,2,-1,2,1 }, ddy[] = { -2,-1,-2,1,-1,2,1,2 };
#include "minerals.h"
bool in[100000];
int pr;
void f(Vi &A,Vi &B)
{
int n=A.size();
if(n==1)
{
Answer(A[0],B[0]);
return;
}
int m=n/2;
Vi Aa,Ab,Ba,Bb;
bool ok=in[A[0]]==in[B[0]];
fup(i,0,m-1,1)
{
int t=Query(A[i]);
Aa.pb(A[i]);
in[A[i]]^=1;
pr=t;
}
fup(i,m,n-1,1)Ab.pb(A[i]);
fup(i,0,n-1,1)
{
int t=Query(B[i]);
in[B[i]]^=1;
if(ok)
{
if(t==pr)Ba.pb(B[i]);
else Bb.pb(B[i]);
}
else
{
if(t==pr)Bb.pb(B[i]);
else Ba.pb(B[i]);
}
pr=t;
}
f(Aa,Ba);f(Ab,Bb);
}
void Solve(int n){
Vi A,B;
fup(i,1,2*n,1)
{
int t=Query(i);
in[i]=1;
if(t==pr)B.pb(i);
else A.pb(i);
pr=t;
}
f(A,B);
}
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