用PHP中的bcmath计算第N个根

Question

We are looking for the Nth root in PHP. We need to do this with a very large number, and the windows calculator returns 2. With the following code we are getting 1. Does anybody have an idea how this works?

echo bcpow(18446744073709551616, 1/64);

Answer 1

Well it seems that PHP and the BC lib has some limits, and after searching on the internet i found this interesting article/code:

So you should use this function:

<?php

function NRoot($num, $n) { 
    if ($n<1) return 0; // we want positive exponents 
    if ($num<=0) return 0; // we want positive numbers 
    if ($num<2) return 1; // n-th root of 1 or 2 give 1 

    // g is our guess number 
    $g=2; 

    // while (g^n < num) g=g*2 
    while (bccomp(bcpow($g,$n),$num)==-1) { 
        $g=bcmul($g,"2"); 
    } 
    // if (g^n==num) num is a power of 2, we're lucky, end of job 
    if (bccomp(bcpow($g,$n),$num)==0) { 
        return $g; 
    } 

    // if we're here num wasn't a power of 2 :( 
    $og=$g; // og means original guess and here is our upper bound 
    $g=bcdiv($g,"2"); // g is set to be our lower bound 
    $step=bcdiv(bcsub($og,$g),"2"); // step is the half of upper bound - lower bound 
    $g=bcadd($g,$step); // we start at lower bound + step , basically in the middle of our interval 

    // while step!=1 

    while (bccomp($step,"1")==1) { 
        $guess=bcpow($g,$n); 
        $step=bcdiv($step,"2"); 
        $comp=bccomp($guess,$num); // compare our guess with real number 
        if ($comp==-1) { // if guess is lower we add the new step 
            $g=bcadd($g,$step); 
        } else if ($comp==1) { // if guess is higher we sub the new step 
            $g=bcsub($g,$step); 
        } else { // if guess is exactly the num we're done, we return the value 
            return $g; 
        } 
    } 

    // whatever happened, g is the closest guess we can make so return it 
    return $g; 
}

echo NRoot("18446744073709551616","64");

?>

Hope this was helpful ...