"1 million factorial"의 두 판 사이의 차이

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== just by using [http://www.manpagez.com/man/1/bc/ bc] ==
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rewrote a factorial function not to use recursive calling.
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<syntaxhighlight lang="c">
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define g(x) {
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    answer=1;
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    for(i=1;i<x+1;i++) {
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        answer *= i;
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    }
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    return answer;
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}
  
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g(1000000);
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</syntaxhighlight>
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real    578m28.905s<br>
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user    450m56.192s<br>
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sys    12m33.006s<br>
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(with poor computing power (centOS virtual machine on windows7. i5 processor))
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5,565,709 digits starting <div style='word-wrap: break-word;'>82639316883312400623766461031726662911353479789638730451677758855633796110356450844465305113114639733516068042108785885414647469506478361823012109754232995901156417462491737988838926919341417654578323931987280247219893964365444552161533920583519938798941774206240841593987701818807223169252057737128436859815222389311521255279546829742282164292748493887784712443572285950934362117645254493052265841197629905619012120241419002534128319433065076207004051595915117186613844750900755834037427137686877042093751023502633401248341314910217684549431273636399066971952961345733318557782792616690299056202054369409707066647851950401003675381978549679950259346666425613978573559764142083506&hellip; </div>ending with 249,998 zeros(about 4.49% of the whole digits)
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* more easy way to get the number of trailing zeros(from [http://answers.yahoo.com/question/index?qid=20071017145425AAaGkd2 here])
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<blockquote>
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Every multiple of 5 contributes a zero.<br>
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Every multiple of 25 contributes a second zero<br>
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Every multiple of 125 contributes a third zero<br>
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Every multiple of 625 contributes a fourth zero<br>
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etc.<br>
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<br>
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floor(1000000/5) + floor(1000000/25) + ... floor(1000000/5^8)<br>
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= 200,000 + 40,000 + 8,000 +1,600 +320 + 64 +12 + 2 <br>
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= 249,998 trailing zeros
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</blockquote>
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* someone can get this in about only 10sec!! [http://www.makewebgames.com/showthread.php/24362-Factorial-Program-in-C-1-Million-Factorial here]<br>He's saying he used FFT
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== ref ==
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http://www.luschny.de/math/factorial/FastFactorialFunctions.htm
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http://www.luschny.de/math/factorial/csharp/FactorialPrimeSwing.cs.html
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http://answers.google.com/answers/threadview/id/509662.html
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http://en.wikipedia.org/wiki/Elliptic_curve_factorization

2017년 6월 19일 (월) 12:05 판

just by using bc

rewrote a factorial function not to use recursive calling.

define g(x) {
    answer=1;
    for(i=1;i<x+1;i++) {
        answer *= i;
    }
    return answer;
}

g(1000000);

real 578m28.905s
user 450m56.192s
sys 12m33.006s
(with poor computing power (centOS virtual machine on windows7. i5 processor))

5,565,709 digits starting

82639316883312400623766461031726662911353479789638730451677758855633796110356450844465305113114639733516068042108785885414647469506478361823012109754232995901156417462491737988838926919341417654578323931987280247219893964365444552161533920583519938798941774206240841593987701818807223169252057737128436859815222389311521255279546829742282164292748493887784712443572285950934362117645254493052265841197629905619012120241419002534128319433065076207004051595915117186613844750900755834037427137686877042093751023502633401248341314910217684549431273636399066971952961345733318557782792616690299056202054369409707066647851950401003675381978549679950259346666425613978573559764142083506…

ending with 249,998 zeros(about 4.49% of the whole digits)

  • more easy way to get the number of trailing zeros(from here)

Every multiple of 5 contributes a zero.
Every multiple of 25 contributes a second zero
Every multiple of 125 contributes a third zero
Every multiple of 625 contributes a fourth zero
etc.

floor(1000000/5) + floor(1000000/25) + ... floor(1000000/5^8)
= 200,000 + 40,000 + 8,000 +1,600 +320 + 64 +12 + 2
= 249,998 trailing zeros

  • someone can get this in about only 10sec!! here
    He's saying he used FFT

ref

http://www.luschny.de/math/factorial/FastFactorialFunctions.htm

http://www.luschny.de/math/factorial/csharp/FactorialPrimeSwing.cs.html

http://answers.google.com/answers/threadview/id/509662.html

http://en.wikipedia.org/wiki/Elliptic_curve_factorization