Utpalendu Barman's Blog

For a^b=(a X a X a X a.....X a) eg. for a=3, b=10 This can be written as 3^10 = (3^5)^2 Using this logic we can implement long long power(long long a, long long b){     long long res=1;     while(b>0){          if(b%2==1)//Odd Number{              res*=a;          }          a*=a;          b/=2;     }     return res; }

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)

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; } }

Factors of a number n can be defined as the set of numbers which are divisible by n. eg. n=24 factors={1,2,3,4,6,8,12,24}; On native approach this can be done in time complexity O(n), but on efficient approach this can be done in O(sqrt(n)); For n=100; its factors are 1,2,4,5,10,20,25,50,100; If noticed carefully, it exists in pair, whose product form the number n; (1,100),(2,50),(4,25),(5,20),(10,10), we can see an exception for middle element 10, for odd numbers of factors. Similarly for n=24; factors: 1,2,3,4,6,8,12, 24 pairs: (1,24),(2,12),(3,8),(4,6) Further we can notice few more points: 1.  If (a,b) is a pair, 1<=a<=sqrt(n) 2. If (a,b) is a pair, b=n/a; as a*b=n Therefore progmatically, void Facotrs(int n){     for(int i=1;i<=sqrt(n);i++){         if(n/i=i)//OR i*i==n , for equal divisor element        {            print(i);         }         else{             print(i);             print(n/i);         }     } }

Instagram , a social networking site launched in  October 2010 became extensively famous after 2015. Beside social networking nowadays it became a platform to promote various kinds of business and a mode of earning and of course to become famous on social media as a public figure. But it is only possible to those, who have  extensively vast numbers of followers. After discussing to those who earns from instagram I found that there is generally 2 ways to increase/make instagram followers: Method 1: By Buying followers- you will find a large number of websites who claims to sell real followers by social marketing but 99% are fake. Method 2: By Following peoples- Maximum number of peoples uses this formula. First by following people than when these peoples follows back  in return than by unfollowing them. Recently I have developed a script based on method 2, which will enable one to mass follow and unfollow people in one click. I have transformed this script to a chrome extension,