Question

As some of you may notice this question is problem 16 from Project Euler. I have solved it using the new “bigInt” feature of C# 4.0 which was fairly straightforward but which is also not really learning everything I should. I am assuming that since it is 2 ^ 1000 there would be some sort of bit shifting solutions but I can’t figure out how exactly it would work.

Does anybody know a way to calculate 2^1000 without using bigint?

Answer 1

Here is a rather naive way to do it in python just using a list(or array) of digits

digits = [1]
for n in range(1000):
    newdigits = []
    carry = 0
    for digit in digits:
        s = 2*digit+carry
        carry = s/10
        s = s%10
        newdigits.append(s)
    if carry:
        newdigits.append(carry)
    digits = newdigits
print "".join(map(str,reversed(digits)))