Find square difference: Difference between revisions
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n is 501 |
n is 501 |
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</pre> |
</pre> |
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=={{header|C}}== |
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<lang c>#include<stdio.h> |
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#include<stdlib.h> |
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int f(int n) { |
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int i, i1; |
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for(i=1;i*i-i1*i1<n;i1=i, i++); |
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return i; |
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} |
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int main(void) { |
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printf( "%d\n", f(1000) ); |
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return 0; |
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}</lang> |
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{{out}} |
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<pre>501</pre> |
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=={{header|Dart}}== |
=={{header|Dart}}== |
Revision as of 18:19, 19 November 2021
- Task
Find and show on this page the least positive integer number n, where difference of n*n and (n-1)*(n-1) greater than 1000.
The result is 501 because 501*501 - 500*500 = 251001 - 250000 = 1001 > 1000.
ALGOL 68
Using the same school maths ( or math for those in the US ) in the Wren version but using a calculation... <lang algol68>BEGIN # find the lowest positive n where the difference between n^2 and (n-1)^2 is > 1000 #
INT rqd diff = 1000; # n^2 - ( n - 1 )^2 is n^2 - n^2 + 2n - 1, i.e. 2n - 1 # # so 2n - 1 > 1000 or n > 1001/2 # print( ( "n is ", whole( ( ( rqd diff + 1 ) OVER 2 ) + 1, 0 ) ) )
END</lang>
- Output:
n is 501
C
<lang c>#include<stdio.h>
- include<stdlib.h>
int f(int n) {
int i, i1; for(i=1;i*i-i1*i1<n;i1=i, i++); return i;
}
int main(void) {
printf( "%d\n", f(1000) ); return 0;
}</lang>
- Output:
501
Dart
<lang dart>import 'dart:math';
int leastSquare(int gap) {
for (int n = 1;; n++) { if (pow(n, 2) - pow((n - 1), 2) > gap) { return n; } }
}
void main() {
print(leastSquare(1000));
}</lang>
- Output:
501
Fermat
<lang>Func F(n) =
i:=0; while i^2-(i-1)^2<n do i:=i+1 od; i.;
!!F(1000);</lang>
- Output:
501
FreeBASIC
<lang freebasic>function fpow(n as uinteger) as uinteger
dim as uinteger i while i^2-(i-1)^2 < n i+=1 wend return i
end function
print fpow(1001)</lang>
- Output:
501
Perl
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Least_square use warnings;
my $n = 1; $n++ until $n ** 2 - ($n-1) ** 2 > 1000; print "$n\n";</lang>
- Output:
501
Phix
Essentially Wren equivalent, but explained in excruciating detail especially for enyone that evidently needs elp, said Eeyore.
with javascript_semantics printf(1,""" n*n - (n - 1)*(n - 1) > 1000 n*n - (n*n - 2*n + 1) > 1000 n*n - n*n + 2*n - 1 > 1000 2*n - 1 > 1000 2*n > 1001 n > 500.5 n = %d """,ceil(500.5))
- Output:
n*n - (n - 1)*(n - 1) > 1000 n*n - (n*n - 2*n + 1) > 1000 n*n - n*n + 2*n - 1 > 1000 2*n - 1 > 1000 2*n > 1001 n > 500.5 n = 501
Python
<lang python> import math print("working...") limit1 = 6000 limit2 = 1000 oldSquare = 0 newSquare = 0
for n in range(limit1):
newSquare = n*n if (newSquare - oldSquare > limit2): print("Least number is = ", end = ""); print(int(math.sqrt(newSquare))) break oldSquare = n*n
print("done...") print() </lang>
- Output:
working... Least number is = 501 done...
Raku
<lang perl6>say first { $_² - ($_-1)² > 1000 }, ^Inf;</lang>
- Output:
501
Ring
<lang ring> load "stdlib.ring" see "working..." + nl limit1 = 6000 limit2 = 1000 oldPrime = 0 newPrime = 0
for n = 1 to limit1
newPrime = n*n if newPrime - oldPrime > limit2 see "Latest number is = " + sqrt(newPrime) + nl exit ok oldPrime = n*n
next
see "done..." + nl </lang>
- Output:
working... Latest number is = 501 done...
Wren
n needs or be such that n² - (n² - 2n + 1) > 1000 or n > 500.5. <lang ecmascript>System.print(500.5.ceil)</lang>
- Output:
501