SPOJ Solution-Gopu and Counting Bitwise Prime Numbers-Solution in C, C++, Java, Python

Input:
2
1 2
1 3

Output:
0
1 

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>
#include <limits.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
 
#define MOD (ll)1000000007
#define pb   push_back
#define EPS 1e-9
#define FOR(i,n)  for(int i = 0;i < n; i++)
#define FORE(i,a,b) for(int i = a;i <= b; i++)
#define pi(a)   printf("%d\n", a)
#define all(c)  c.begin(), c.end()
#define tr(container, it)   for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define gc getchar_unlocked
#define sdi(a, b)   si(a);si(b)
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
#define endl '\n'

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

void si(int &x){
    register int c = gc();
    x = 0;
    int neg = 0;
    for(;((c<48 || c>57) && c != '-');c = gc());
    if(c=='-') {neg=1;c=gc();}
    for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
    if(neg) x=-x;
}

const int MAXN = 65;
int isPrime[MAXN];	//isPrime[i] = 0 indicates i is prime. 
void sieve(){
	//if even check itself while calling. This function will only tells wether a number is prime or not(not for even numbers).
	isPrime[0] = isPrime[1] = 1;
	for(int i = 3; i*i <= MAXN; i+=2){
		if(isPrime[i] == 0){
			if(i*(ll)1*i <= MAXN){
				for(int j = i*i; j <= MAXN; j += (2*i)){
					isPrime[j] = 1;
				}
			}
		}
	}
}



int main(){
    io;
    sieve();
	int t;
	cin >> t;
	while(t--){
		double a, b;
		cin >> a >> b;
		int count = 0;
		int bitsA = ((double)log2(a) + 1);
		int bitsB = ((double)log2(b) + 1);
		for(int i = bitsA; i < bitsB; i++){
			if(!isPrime[i]){
				count++;
			}
		}
		//if b+1 is power of two and bitsB is prime then count+1
		ll rem = b+1;
		if(!(rem&(rem-1)) && !isPrime[bitsB]>{
			count++;
		}
		cout << count << endl;
	}    
    return 0;
}

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