largenumber

I need to convert a large decimal to binary how would I go about doing this? Decimal in question is this 3324679375210329505 http://www.wikihow.com/Convert-from-Decimal-to-Binary How about: String binary = Long.toString(3324679375210329505L, 2); You may want to go for BigDecimal . A BigDecimal consists of an arbitrary precision integer unscaled value and a 32-bit integer scale.The BigDecimal class provides operations for arithmetic, scale manipulation, rounding, comparison, hashing, and format conversion. The toString() method provides a canonical representation of a BigDecimal. new BigDecimal

Can you recommend good languages to do math with large numbers in? So far I've used Actionscript 2 and Objective-c and with Objective-c even using NSDecimalNumbers I was limited to 32 digits in my calculations... I would need at a minimum to be able to calculate with numbers fifty-thousand digits long. Perhaps Haskell will appeal to you. Python has arbitrary-length integers and uses them transparently, so you don't need any special code or classes for this. >>> len(str(math.factorial(123456))) 574965 Try also bc , which is probably already installed in your machine. Try java with its

X and Y are integers larger than 100 digits. Find the integer P which is within the range [ X , Y [ and that guaranties the "best" prime decomposition (i.e. the decomposition with the most unique prime factors). What I've done is just check the primality and decompose each number in the range and find the number that respects the rule. Is there any other way to do this? An example on small integers Edit: In the above example, 123456 is decomposed to 2^6 * 3^1 * 643^1 , that's 2 * 2 * 2 * 2 * 2 * 2 * 3 * 643 but only 3 unique factors. While the answer, 123690, is decomposed to 6 unique factors

This question already has an answer here: How to calculate modulus of large numbers? 10 answers I have two numbers A and B where A and B can be in the range 1<= A,B <=100^100000 How can we find the value of A^B modulo some M in C++ ?? Floris In the duplicate I pointed out, the solution I particularly like is https://stackoverflow.com/a/8972838/1967396 (see there for attribution and references) For your convenience I reproduce the code here (wrapped into an SCCE - but using C, not C++): #include <stdio.h> int modular(int base, unsigned int exp, unsigned int mod) { int x = 1; int i; int power =

This question already has answers here : Infinite loop with cin when typing string while a number is expected (4 answers) I'm learning C++ and writing little programs as I go along. The following is one such program: // This program is intended to take any integer and convert to the // corresponding signed char. #include <iostream> int main() { signed char sch = 0; int n = 0; while(true){ std::cin >> n; sch = n; std::cout << n << " --> " << sch << std::endl; } } When I run this program and keep inputs at reasonably small absolute values, it behaves as expected. But when I enter larger inputs,

I have to store the number 600851475143 in my program. I tried to store it in long long int variable and long double as well but on compiling it shows the error integer constant is too large for "long" type. I have also tried unsigned long long int too. I am using MinGW 5.1.6 for running g++ on windows. What datatype should I use to store the number? long long is fine, but you have to use a suffix on the literal. long long x = 600851475143ll; // can use LL instead if you prefer. If you leave the ll off the end of the literal, then the compiler assumes that you want it to be an int , which in