问题
I need to get the square root of a 210 digit number accurately, I thought GMP was the right tool for the job, what am I doing wrong?
#include <stdlib.h>
#include <stdio.h>
#include "gmp.h"
int
main (int argc, char *argv[])
{
mpz_t sq_me, sq_out, test;
mpz_init(sq_me);
mpz_init(sq_out);
mpz_init(test);
mpz_set_str (sq_me, argv[1], 10);
mpz_sqrt(sq_out, sq_me);
mpz_mul(test,sq_out,sq_out);
gmp_printf ("%Zd\n\n", sq_out);
gmp_printf ("%Zd\n\n", test);
return 0;
}
Input:
24524664490027821197651766357308801846702678767833275974341445171506160083003858 72169522083993320715491036268271916798640797767232430056005920356312465612184658 17904100131859299619933817012149335034875870551067
Output:
49522383313031109809242226159886283348695660460381271324714928680654813093947239 9634016783775955618921028
24524664490027821197651766357308801846702678767833275974341445171506160083003858 72169522083993320715491034366358025027526868495267716284867043049443779615862887 47102011391915422793532619329760963626718900576784
回答1:
Here's the code you need for floating point square root, you can see that the initial input and final output are identical.
#include <stdlib.h>
#include <stdio.h>
#include "gmp.h"
int main (int argc, char *argv[]) {
mpf_t sq_me, sq_out, test;
mpf_set_default_prec (10000);
mpf_init(sq_me);
mpf_init(sq_out);
mpf_init(test);
mpf_set_str (sq_me, argv[1], 10);
mpf_sqrt(sq_out, sq_me);
mpf_mul(test,sq_out,sq_out);
gmp_printf ("Input: %Ff\n\n", sq_me);
gmp_printf ("Square root: %.200Ff\n\n", sq_out);
gmp_printf ("Re-squared: %Ff\n\n", test);
return 0;
}
Here's the output with your parameter:
Input: 2452466449002782119765176635730880184670267876783327597434144
51715061600830038587216952208399332071549103626827191679864079776723243005
60059203563124656121846581790410013185929961993381701214933503487587055106
7.000000
Square root: 4952238331303110980924222615988628334869566046038127132471492
86806548130939472399634016783775955618921028.19202568258368255653837168412
92356432661548614332014106174638951390596672950394981098992388116308833260
04535647648563996144250924277757344248059826024201642748515325655438898558
17807282091590722890002
Re-squared: 2452466449002782119765176635730880184670267876783327597434144
51715061600830038587216952208399332071549103626827191679864079776723243005
60059203563124656121846581790410013185929961993381701214933503487587055106
7.000000
回答2:
You're getting the result as an integer and then squaring that. The input number must not be a perfect square, so it is truncating the decimals and decreasing the precision of the number. Look into the 'mpf' category of functions for floats rather than 'mpz' for integers.
回答3:
I am using a libgmp-3.dll
In vb.net I wrote this function for a high precision square root. I haven't tested it thoroughly but it gets me the Square Root of 3 to any precision I need
Public Shared Function SquareRoot(ByVal Value As BigInt, ByVal Precision As Integer) As String
Dim Ten As New BigInt(10)
Dim DecLen As Integer = Value.Sqrt.ToString.Length
Dim RootDigits As String = (Value * Ten.Power(Precision * 2)).Sqrt
'Add trailing zeros
RootDigits = RootDigits.PadRight((DecLen + Precision), "0")
Return RootDigits.Substring(0, DecLen) & "." & RootDigits.Substring(DecLen)
End Function
来源:https://stackoverflow.com/questions/822734/square-root-of-bignum-using-gmp