Program to Print all subarrays of a array in efficient way
void Subarray(int arr[],int n){
for(long long i=0;i<(1<<n);i++){ for(long long j=0;j<n;j++){ if((i & 1<<j)>0) { cout<<arr[j]; } } cout<<endl; } }
Efficient method to calculate XOR series
A XOR series between integer a and b (a<=b) can be defined as the xoring result of all elements between a and b (inclusive of a and b); 1. For Series from 1 to n: f(n)=1^2^3^4^5...^n But for XOR it holds a noticeable outcome n Binary XOR from 1->n 1 0001 0001 2 0010 0011 3 0011 0000 4 0100 0100 5 0101 0001 6 0110 0111 7 0111 0000 8 1000 1000 From this result we can conclude that : 1. An element just before multiple of 4 results 0; 2. An element divisible by 4 results as the same. 3. An element just after the multiple of 4 results 1; 4. An element after 2 element from multiple of 4 results as the element is self +1 We can formulate the above as: r=n%4: case r==0: XOR=n case r==1: XOR=1 case r==2: XOR=n+1 case r==3: XOR=0 2. For Series from a to b : In this case we do, f(a) ^ f(b) = (1^2^3^4...^a) ^ (1^2^3^4...^b)=XOR(a,b)
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