Function main

CodeChef_problems/FINXOR/solution.cpp:26–64

The approach to solve this problem heavily relies on the property that A+B = (A^B) + 2*(A&B) The interactor gives us the value of A1^k + A2^k + A3^k ... AN^k Using the above mentioned property, we know A + k = (A^k) + 2*(A&K) Hence, (A^k) = A+k - (2*(A&K)) Which implies -> A1^k + A2^k + A3^k ... AN^k = A1 + A2 + A3 .. AN + N*k - 2*(A1&k + A2&k + A3&k .. AN&k) Lets us denote the A1 + A2 + A3 .. AN

as Ai <= 10^6, we do not need any further questions.
*/
int32_t main()
{	
	int t;
	cin>>t;
while(t--){
	int n;
	cin>>n;
	int s=0,o,i=1;
	int d[21]={0};
	cout<<1<<" "<<(1<<20)<<endl;
	cin>>s;
	s = s - n*(1<<20); //Sum for Equation 1
	if(s&1){
		d[0] = 1; //The 0th bit could simply be obtained by taking an AND with 1
	}
	else{
		d[0] = 0;
	}
	while(i!=20){//Remaining 19 bits could be obtained by the remaining 19 questions.
		cout<<1<<" "<<(1<<i)<<endl;
		cin>>o;
		d[i] = (s + n*(1<<i) - o)/(1<<(i+1));//((1<<i)*2) = (1<<(i+1))
		i++;
	}
	int x=0;
	int y=1;
	for(int i=0;i<=19;++i){
		x += (d[i]%2)*y;
		y = y*2;
	}
	cout<<2<<" "<<x<<endl;
	int ans;
	cin>>ans;
	if(ans==-1){
		return 0;
	}
}
	return 0;
}