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Double precision in C++ (or pow(2, 1000))

开发者 https://www.devze.com 2023-01-10 02:09 出处:网络
I\'m working on Project Euler to brush up on my C++ coding skills in preparation for the programming challenge(s) we\'ll be having this next semester (since they don\'t let us use Python, boo!).

I'm working on Project Euler to brush up on my C++ coding skills in preparation for the programming challenge(s) we'll be having this next semester (since they don't let us use Python, boo!).

I'm on #16, and I'm trying to find a way to keep real precision for 2¹°°°

For instance:

int main(){
    double num = pow(2, 1000);
    printf("%.0f", num):
    return 0;
}

prints

10715086071862673209484250490600018105614050000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Which is missing most of the numbers (from python):

>>> 2**1000

10715086071862673209484250490600开发者_开发知识库018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376L

Granted, I can write the program with a Python 1 liner

sum(int(_) for _ in str(2**1000))

that gives me the result immediately, but I'm trying to find a way to do it in C++. Any pointers? (haha...)

Edit:

Something outside the standard libs is worthless to me - only dead-tree code is allowed in those contests, and I'm probably not going to print out 10,000 lines of external code...


If you just keep track of each digit in a char array, this is easy. Doubling a digit is trivial, and if the result is greater than 10 you just subtract 10 and add a carry to the next digit. Start with a value of 1, loop over the doubling function 1000 times, and you're done. You can predict the number of digits you'll need with ceil(1000*log(2)/log(10)), or just add them dynamically.

Spoiler alert: it appears I have to show the code before anyone will believe me. This is a simple implementation of a bignum with two functions, Double and Display. I didn't make it a class in the interest of simplicity. The digits are stored in a little-endian format, with the least significant digit first.




typedef std::vector<char> bignum;

void Double(bignum & num)
{
    int carry = 0;
    for (bignum::iterator p = num.begin();  p != num.end();  ++p)
    {
        *p *= 2;
        *p += carry;
        carry = (*p >= 10);
        *p -= carry * 10;
    }
    if (carry != 0)
        num.push_back(carry);
}

void Display(bignum & num)
{
    for (bignum::reverse_iterator p = num.rbegin();  p != num.rend();  ++p)
        std::cout << static_cast<int>(*p);
}

int main(int argc, char* argv[])
{
    bignum num;
    num.push_back(1);
    for (int i = 0;  i < 1000;  ++i)
        Double(num);
    Display(num);
    std::cout << std::endl;
    return 0;
}


You need a bignum library, such as this one.


You probably need a pointer here (pun intended)

In C++ you would need to create your own bigint lib in order to do the same as in python.


C/C++ operates on fundamental data types. You are using a double which has only 64 bits to store a 1000 bit number. double uses 51 bit for the significant digits and 11 bit for the magnitude.

The only solution for you is to either use a library like bignum mentioned elsewhere or to roll out your own.


UPDATE: I just browsed to the Euler Problem site and found that Problem 13 is about summing large integers. The iterated method can become very tricky after a short while, so I'd suggest to use the code from Problem #13 you should have already to solve this, because 2**N => 2**(N-1) + 2**(N-1)

Using bignums is cheating and not a solution. Also, you don't need to compute 2**1000 or anything like that to get to the result. I'll give you a hint:

Take the first few values of 2**N:

0 1 2 4 8 16 32 64 128 256 ...

Now write down for each number the sum of its digits:

1 2 4 8 7 5 10 11 13 ...

You should notice that (x~=y means x and y have the same sum of digits)

1+1=2, 1+(1+2)=4, 1+(1+2+4)=8, 1+(1+2+4+8)=16~=7 1+(1+2+4+8+7)=23~=5

Now write a loop.

Project Euler = Think before Compute!


If you want to do this sort of thing on a practical basis, you're looking for an arbitrary precision arithmetic package. There are a number around, including NTL, lip, GMP, and MIRACL.

If you're just after something for Project Euler, you can write your own code for raising to a power. The basic idea is to store your large number in quite a few small pieces, and implement your own carries, borrows, etc., between the pieces.


Isn't pow(2, 1000) just 2 left-shifted 1000 times, essentially? It should have an exact binary representation in a double float. It shouldn't require a bignum library.

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