Babbage problem: Difference between revisions
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<p>Comentarios "in spanish":</p>
<p>
Documentación:</p>
<p>Establece que se trabajará con números sin decimales:</p>
"decimales(0)" (A)
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La variable que controlará 11 iteraciones del proceso
principal. Se inicializa con 10, y se decrementará hasta
llegar a 0, de acuerdo a lo descrito en 1.4, 1.5 y 1.6:
"i" (
Esto permitirá encontrar 11 números que cumplan con lo
-------------------------------------------------------------
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Comentarios:
JNZ
#() : macro que permite resolver expresiones infijas en
tiempo de compilación.
</p>
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=={{header|Elena}}==
{{trans|Smalltalk}}
ELENA
<syntaxhighlight lang="elena">import extensions;
import system'math;
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var n := 1;
until(n.sqr().mod
{
n += 1
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{{out}}
<pre>{{x->25264},{x->99736}}</pre>
clear all;close all;clc;
BabbageProblem();
function BabbageProblem
% Initialize x to 524, as the square root of 269696 is approximately 519.something
x = 524;
% Loop until the square of x modulo 1000000 equals 269696
while mod(x^2, 1000000) ~= 269696
% If the last digit of x is 4, increment x by 2
% Otherwise, increment x by 8
if mod(x, 10) == 4
x = x + 2;
else
x = x + 8;
end
% Display the result
end
</syntaxhighlight>
{{out}}
<pre>
The smallest positive integer whose square ends in 269696 = 25264
</pre>
=={{header|Maxima}}==
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Which outputs the same thing as above.
=={{header|S-BASIC}}==
Because we have to use double-precision floating point to represent the required number of digits, and because the approach to calculating a double-precision n mod 1000000 to isolate the right-most six digits of the square is particularly inefficient, the program will take a long time to find the solution (but it will, eventually!)
<syntaxhighlight lang="BASIC">
$lines
$constant true = FFFFH
$constant false = 0
var n, sq, r = real.double
var done = integer
print "Finding smallest number whose square ends in 269696"
n = 520 rem - no smaller number has a square that large
done = false
rem - no need to search beyond the number Babbage already knew
while not done and n <= 99736.0 do
begin
sq = n * n
rem - compute sq mod 1000000 by repeated subtraction
r = sq
while r >= 1000000.0 do
r = r - 1000000.0
begin
print using "The smallest number is ######"; n
print using "and its square is ##,###,###,###"; sq
done = true
end
rem - only even numbers can have a square ending in 6
end
end
</syntaxhighlight>
{{out}}
<pre>
Finding smallest number whose square ends in 269696
The smallest number is 25264
and its square is 638,269,696
</pre>
=={{header|Scala}}==
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638269696
</pre>
▲=={{header|Tiny Craft Basic}}==
▲<syntaxhighlight lang="basic">10 print "calculating..."
▲ 40 let n = n + 2
▲50 if (n ^ 2) % 1000000 <> 269696 then 30
▲60 print "The smallest number whose square ends in 269696 is: ", n
=={{header|Transd}}==
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=={{header|Wren}}==
{{libheader|Wren-fmt}}
<syntaxhighlight lang="
The answer must be an even number and it can't be less than the square root of 269,696.
So, if we start from that, keep on adding 2 and squaring it we'll eventually find the answer.
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*/
import "./fmt" for Fmt // this enables us to format numbers with thousand separators
var start = 269696.sqrt.ceil // get the next integer higher than (or equal to) the square root
start = (start/2).ceil * 2 // if it's odd, use the next even integer
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