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I need to determine x from the expression root(y).

"Y" has a range to its separated value with maximum 10^1000. I solved it in the normal way and I saw the right result. But when Y is very large, the program outputs the wrong answer.

#include<stdio.h>
#include<math.h>
int main()
{
    long long int x,y;
    scanf("%lld",&y);
    x=sqrt(y);
    printf("%lld",x);
    return 0;
}
ManoDestra
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mhhabib
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    the idea of this problem, that you are to implement the sqrt yourself for big numbers. – Eugene Sh. Apr 12 '16 at 17:56
  • 676=4*169=(2^2)*(13^13)=22^132 , Therefore √676=√(4⋅169)=√2^2 * √13^2 = 26 – Khalil Khalaf Apr 12 '16 at 17:58
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    If you really want to do numbers as large as `10^1000` then `long long` is no where near large enough. – lurker Apr 12 '16 at 17:59
  • @FirstStep and suppose √Y is prime? – Weather Vane Apr 12 '16 at 18:17
  • `sqrt` is a floating point function, not an integer. – too honest for this site Apr 12 '16 at 18:18
  • [Unums](http://www.johngustafson.net/unums.html) bear some interesting implications for this problem, but the implementation would be far too complex (and possibly not worth the time for this specific case?) to warrant an official answer to that effect. – CodeMouse92 Apr 12 '16 at 18:32
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    Related: [square root of bignum using GMP](http://stackoverflow.com/questions/822734/square-root-of-bignum-using-gmp) – Mark Plotnick Apr 12 '16 at 18:54
  • There are roughly 10 binary digits for every 3 decimal digits (as 2^10 is about the same as 10^3). So 10^1000 is roughly 2^3000. You would need more than 45 64-bit words to store a 10^1000. You need to learn to walk before you can run. I would start off implementing sqrt of two 64-bit words (128-bit integer) first and then generalize to multi-word sqrt. – Z boson Apr 13 '16 at 06:15

1 Answers1

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Use some bignum library, for example GMP

pelya
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