How to find kth least significant bit?

[SparkXV]

This code is not passing all test cases?
Can anybody tell me what is wrong with this code?

#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
int main()
{
	//code
	ll t;
	cin>>t;
	while(t--)
	{
	    ll n,k;
	    cin>>n>>k;
	    cout<<(1&(n>>(k-1)))<<"\n"; 
	}
	return 0;
}

[jonnin]

what are you really wanting to know?

there are 2 ways to look at it..
1) you have n bits, and bits 1 & 2 (for example) are not trusted). The LSB is &4.
2) you have n bits, and only the highest z are good. find the first set bit, and count towards the 1's position until you find z or run out (if you run out all are good).

[dutch]

You should use unsigned long long.

[jonnin]

example brute force / crude / fun solution prints the number in binary and again with only N significant bits. You can do it with less code, but it would quickly become hard to follow if you get cute at it.

 int main()
 {
    int soi = 8*sizeof (unsigned long long); //in bits	
	unsigned long long  testnum = 31415926539;
	unsigned long long u = 9223372036854775808ull; //2^63
	int j;
	for(j = 0; j++ < soi; u/= 2)
	{
	   cout << (int)((testnum&u) > 0);
	}
	cout << endl; //written in binary for fun.
	
	
	u = 9223372036854775808ull; //2^63 
	int sigb = 5;
	int s = 0;
	bool b = false;
	for(j = 0; j++ < soi; u/= 2)
	{
	   if(testnum&u)
	    b = true;
	  if(b && s < sigb)
	  {
		  s++;
		  cout << (int)((testnum&u) > 0); //the last one this line prints is what you asked.
                //its location is &u, or you can convert that to the nth bit easily. 
	  }
	  else
	  {
		cout << '0';  
	  }	  
	}  
    cout << endl; 
 }

which is kind of nuts. some messing with this would get you there, just handle negatives and roundoff carefully to be sure its always correct:

cout << (int)(log2(testnum)+0.999999) << endl; //the most sig bit of the number, from the 1's bit. so that minus the number of bits you want is the least... etc...