sqrt在c语言中什么意思 sqrt

sqrt在c语言中什么意思 sqrt

PHP 3, PHP 4, PHP 5)

sqrt -- 平方根
说明
float sqrt ( float arg )

返回 arg 的平方根。

例 1. sqrt()

<?php
// Precision depends on your precision directive
echo sqrt(9); // 3
echo sqrt(10); // 3.16227766 ...
?>

29-Mar-2006 08:03
A much easier way to find perfect squares:
This method uses integerrootsto find perfect squares instead of integer squares to find integer roots.

<?

# initial value
$i = 0;

#set any limit you want
$setSomeLimit = 10000;

# do it! .... do it! ... do it!
for ($n = 1; $n < sqrt($setSomeLimit; $n++)
{
// Find the square
$s= $n * $n;

print "
perfect_square_root is of $s is $n";
}

?>
php NOSPAM at primezero dot com
13-Jul-2005 02:15
creating perfect squares... (nodecimals)

<?

# initial value
$i = 0;

#set any limit you want
$setSomeLimit = 10000;

# do it! .... do it! ... do it!
do {

# find the square root
$sqRoot = sqrt($i);

# not a decimal? that it is a perfect square! ^_^
if (ereg("^[0-9]{1,12}$", $sqRoot)) {

# suppose that the_square_root = teh_square_root
print "
perfect_square_root is of $i is $sqRoot";
}
$i++; # next!!

# keeps running until it reaches the limit
} while ($i < $setSomeLimit);

?>
btharper1221 at gmail dot com
15-May-2005 03:48
this is something you can use if you need an exact answer to a root problem, this includes an echo outside of the function to show how it works, it will return it such as
$ret[0] is the number still part of the root (this is the exactness part of it)
$ret[1] is the number of roots and
$ret[2] is 1 or 0, 1 being that there should be an i following the number of roots (imaginary)

<?php
functionnewrt($val,$rt){
$i = 0;
if(($rt % 2 == 0) && ($val < 1)){
$i = 1;
$val = (-$val);
}
$c = 1;
if(($rt % 2 != 0) && ($val < 0)){
$c = -$c;
$val = -$val;
}
for($d = 2; pow($d,$rt) <= abs($val); $d++){
if($val % (pow($d,$rt)) == 0){
$val = ($val/(pow($d,$rt)));
$c = ($c*$d);
$d--;
}
}
$ret[0] = $val;
$ret[1] = $c;
$ret[2] = $i;
return $ret;
}//end function newrt

$ret = newrt($num,$num2);
if($ret[0] != 1){
if($ret[2] == 0) echo $ret[1]." roots of ".$ret[0];
if($ret[2] == 1) echo $ret[1]."iroots of ".$ret[0];
}elseif($ret[0] == 1){
if($ret[2] == 0) echo $ret[1];
if($ret[2] == 1) echo "$ret[1]i";
}
?>
chris DOT rutledge AT gmail DOT com
31-Mar-2005 10:24
Just a note to say you can take the square root of a negative number - it returns an imaginary number.

One might do it like this

//take initial value of $x

$abs_x = abs($x);
$answer = sqrt($abs_x);

echo $answer;

if ($x < 0) echo"i";
jouhni at web dot de
17-Feb-2005 07:46
To get any root of a number your can use the pow() function:

pow(8, 1/3)

which gives you the third root of eight.

Jouhni
php user
13-May-2004 07:25
How to calculate triangle's area if you haven't height measure.
You need only sides' measure.
it would have to work.

(formula of erone)

$s1 = side 1
$s2 = side 2
$s3 = side 3

<?
function area($s1,$s2,$s3)
{
$calc1 = ($s1+$s2+$s3)/2;
$calc2 = $p*($p-$s1)*($p-$s2)*($p-$s3);
$result = round(sqrt($p),2);
return $result;
}
?>

example of I use:

<?
//This will make to return the area of a triangle with sides of 3,4,5
$area = area(3,4,5);
echo $area;
?>
bishop
18-Jul-2003 05:09
Compute a triangle's area:

function triangleArea($a, $b /* ... */) {
if (func_num_args() === 3) {
$s = .5 * ($a + $b + ($c = func_get_arg(2)));
return sqrt($s * ($s-$a) * ($s-$b) * ($s-$c));
} else {
return (.5 * $a * $b);
}
}

With 2 args, parameters are triangle's base and height. With 3 args, parameters are the lengths of each side. Argument order is not important.

Either way, you get the area.
tcn at pandora dot be
13-Nov-2002 08:37
Checking if the square you want to take from a number is valid or not (you can't take the root of a negative number)

if ($i<0)
{
echo "Error! The number you specified can't berooted! It's negative!
}
else
{
echo "The square root of your number is ".sqrt($i);
}
morvandiumNOSPAM at yahoo dot com
05-Nov-2001 12:44
I find this useful for any grid system where distances need to be computed (maps and graphs,mostly).

Just use the distance between points formula (pseudo-code): sqrt((x2-x1)^2+(y2-y1)^2) to find the distance between two points.

  

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