Odtnhj

Let n an integer and 0<=b<=63, b natural number. Find the b-th bit for the number n on it's 64 bit representation with sign.

and T be the number of test cases.

This is my attempt:

#include <iostream>
#define f cin
#define g cout
using namespace std;

int T;
long long n;
int b;

int main()

 f >> T;
 for(int i = 1; i <= T; ++i)
 
 f >> n >> b;
 int ans = 0;
 bool ok = true;
 while(n)
 
 if(b == ans)
 
 g << n % 2;
 ok = false;
 break;
 
 n /= 2;
 ++ans;
 
 if(ok) g << 0;
 
 return 0;

but it does not work on all test cases... also is there another way to do this? or is there another way to store the bits? is there some special libraries? can you do this more efficiently with other tools? can you give me some information to read about bitmasks? and where and when you should use them and how are they usefull?

Computers already store the integers in its bitwise representation. All you need are bitwise operators to know a particular bit.

int bthbit(long long n, int b) 
 if (n & (1ULL << b)) return 1;
 return 0;

The solution uses bitwise & operator after left-shifting 1 by b bits. You may want to read about bitwise operators and bitmasks .