Double precision in C++ (or pow(2, 1000))
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.
精彩评论