/* Base Converter Base Converter takes input a number in base 10 and the base it has to be converted in and then outputs results according to following characterset rule:- 0 to 9 : [0-9] 10 to 35 : [A-Z] 36 to 61 : [a-z] Input: 456 // Input number in base 10 30 // Base in which you want to convert that number in Output: 456 30 // Inputs separated by space F6 // Output in the required base Detailed explaination at end */ #include #include using namespace std; string base = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"; string rvs(const string&); int main() { long long int i; string o=""; unsigned long int b; cin>>i>>b; if(b>base.length()) { cout<<" ERROR : Max base limit "<=b) { o+=base[i%b]; i/=b; } o+=base[i]; cout< Code 0 -> 0 9 -> 9 10 -> A 15 -> F User inputs number i and base b. String o is output string = "" A while loop is run until i >=b In each while loop o += base[i%b], this means, i%b gives a digit in base b let's assume it's n. Then a character corresponding to n in global string base is added in o. And in same loop i/=b, so it keeps decreasing. And after the loop a left over number which is smaller than b is added to o Finally o has string which has number n in base b But there's a problem, as you can observe i%b always sends character from end and is added in beginning. So last character is first in o and first character is at end in o. In simple words o is reversed number so, it has to be reversed again in order to achieve real number. O is reversed and we get final output It's same way when u convert in binary with factorization method you have to note 1 and 0 from bottom to top. */