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compare: jct:0fbf5ed2507c332de8887a38c2a39805bbb01ad6
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10 Commits
c6c387d17f ... 0fbf5ed250

Author SHA1 Message Date
Jens True
0fbf5ed250 Removing invalid previous solutions 2024-06-26 13:23:30 +00:00
Jens True
27fd0bf2ea Few more challenges on ProjectEuler 2024-06-26 13:16:36 +00:00
FuryFire
2a2623a652 Added a few generic problems\nSolved Euler027 2016-02-23 15:55:26 +01:00
FuryFire
a6355b2629 Removed incomplete solution 2012-12-17 12:20:18 +01:00
FuryFire
604de1528d Solved ProjectEuler/039,041,042 2012-12-17 12:17:09 +01:00
FuryFire
cb4a6a8081 Solved ProjectEuler/038,104 2011-04-26 16:07:20 +02:00
FuryFire
62c2c8f47b Solved ProjectEuler/035 2012-04-25 15:26:13 +02:00
FuryFire
6cc3b0ad82 Solved ProjectEuler/028,048 2011-04-24 16:22:52 +02:00
FuryFire
80dbb818a8 Begun adding support for a global settings.yml file 2011-04-12 14:12:02 +02:00
FuryFire
b25e8b7c11 Added a few lua solutions - Updated the CT script to include additional parameters 2012-04-11 14:49:45 +02:00
76 changed files with 1012215 additions and 50 deletions
Show Stats Expand all files Collapse all files

33
CT.rb
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@ -1,14 +1,36 @@
require "yaml"
require 'erb'
require 'open3'
require 'pp'
require 'benchmark'
include Benchmark
cli = { 'php' => 'php', 'ruby' => 'ruby', 'lua'=> 'lua', 'javascript'=>'node', 'js'=>'node', 'c' => 'tcc -run'}
settings = YAML::load_file("settings.yml")
problem = YAML::load_file(ARGV[0] + "/desc.yml")
output = ERB.new(IO.read('template.rt'))
puts output.result()
if(ARGV[1] == 'table')
case ARGV[1]
when 'test'
if(problem['solutions'])
problem['solutions'].each do |sol|
if(cli[sol[1]['language']])
puts "--------------\n|"+sol[0]+"\n--------------"
puts sol[1]['desc']
parameter = (sol[1]['parameters']) ? " " + sol[1]['parameters'] : ""
stdin, stdout, stderr = Open3.popen3(cli[sol[1]['language']] + " "+ARGV[0]+"/"+sol[0] + parameter)
puts "Result: " + stdout.readlines[0]
end
end
else
puts "No solutions for said problem"
end
when 'table'
require 'terminal-table'
files = Dir[ARGV[0]+"/**/*/desc.yml"]
files.sort
files.sort!
entry = []
files.each do |file|
yaml = YAML::load_file(file)
@ -30,4 +52,7 @@ if(ARGV[1] == 'table')
end
table = Terminal::Table.new(:headings => ['Code','Title','Solved'],:rows => rows)
puts table
else
output = ERB.new(IO.read('template.rt'))
puts output.result()
end

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53
CodeGolf/guns-and-zips/output.1 Normal file
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?0???????>n???h*++?Z?L&?W??H$?????????{??????~??>r?{W[?8?d?j?l4Vb t???v????-??o??`???i??N?}-???w??_??Hq??q~x?5????????[?vW??|???;]???????Zc?8c? ???^???????-?H???f?L/ ??orCy????oN?{j??????45?q?]_n/`n???y7&?????????o??"}??J?A8??????,?????\@??Z??B@-l?d?J\???CH%??+T???"T|?EX4?MJK?B?8?f??J ?>\Y%K1H?(???[?ZXt?k???????T?V?D^"1?Xy?œ??@ ??q?????i?@ 044t????)sp?nmm?{wcc?F?????~S?T?!n? >??????c#K???%O?m?Jmn?-?/_??^????????????G??G??5??+???%W?F??3W??u_?U?;w?????<???????u
A/Duc?W??o}?_B[?>?Y?w????a??? Ue[^???o???+?]???????rC?????~??-q?X????{?j?z?t: ????#?Z??m?e???^Y
1??R?B????????_?i?)?3?X?\]Z^YY?????$?
?2?ZZ??????7?,Q??m&ZP+T?f6/???n??D??b?N?O?)%DbZ^rI4???????Bp??T?*q???+5 ??@A?? I,cI???.?x?b5??z?$?p8??????4X8???i?D"?????.};?y?f??l??????W~????'S
q???????[?}K???o?????q]??'???I????u
??h???t??x#?C?_?.
???u?b?Uu??? ??Q?"???_??????6?i???/?{?Nm?k9???.????T?O????G;???O??-C??????miiI ?FH.pG??,C? ?a<?%?a(
???????A??M??????$?r?????9?|zzzjjJ"???????????g ????on???Tn?(!??????+ ?Kre?VSS[^SQ?????Uw(?KJ?????????
???????9´'H?8?I??E?M?G???
?I~1k*j?5???'????? ~`&D84M;?N????????s?????Vkg[$immmkk?=??`??\???1?? ????????A??VUU?4???q\?????????~?TVVR%|??T*moo?
??T???d%?E?E Y?"???? ??? eeei?? ?mK."_XX??|E?a???????d???(???!?xr?F?!?[???????>?o||? ?I o?#???:???'!??L2;? ?0?????x<??ES?? BQ?F???????25/8??????*??
?\.Wuu??? q/?? ?D?}9?4MQ????w
?????uuu?`?eY?????QE$?-??H$BQ?a???Q?V??O??$944? ???FQT$I?R8??aB??????||?4??2??m?9?P%I??t???T}rN?E?????f~w?qn?????????]??{??Enh???????C.???G??7??M??$???????RIEND?B`?

0
CodeGolf/guns-and-zips/solve.php Normal file
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51
Generic/Fibonacci/solve.c Normal file
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@ -0,0 +1,51 @@
#include <stdio.h>
#include <string.h>
#define SIZE 20
void add(char *a, char *b, char *result, int size) {
int i = 0;
for(i=0;i<size;i++) {
//We have overflow!
if((a[i] + b[i] + result[i] < a[i]) || (a[i] + b[i] + result[i] < b[i])) {
result[i+1]++;
}
result[i]+= a[i] + b[i];
}
}
int main( )
{
unsigned char prev[SIZE];
unsigned char current[SIZE];
unsigned char next[SIZE];
unsigned int i=0;
memset(prev,0x00,SIZE);
memset(current,0x00,SIZE);
memset(next,0x00,SIZE);
prev[0]=1;
current[0]=1;
next[0]=1;
for(i=2;i<100;i++) {
add(current,prev,next,SIZE);
memcpy(prev,current,SIZE);
memcpy(current,next,SIZE);
//prev = current;
//current = next;
int j;
for(j=0;j<SIZE;j++) {
printf("%d: %i\n",j,current[j]);
}
}
for(i=0;i<SIZE;i++) {
printf("%d: %i\n",i,current[i]);
}
return 0;
}

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3
Generic/HelloWorld/desc.yml
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@ -14,3 +14,6 @@ solutions:
solve.c:
desc: ANSI C solution (Tested with TCC)
language: c
solve.js:
desc: NodeJS solution
language: js

1
Generic/HelloWorld/solve.js Normal file
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@ -0,0 +1 @@
console.log("Hello World");

20
Generic/MultiTable/solve.c Normal file
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@ -0,0 +1,20 @@
#include <stdio.h>
#define SIZE_X 12
#define SIZE_Y 12
void main()
{
int x,y;
int MemorizeThis[SIZE_X][SIZE_Y];
for (x =1; x <= SIZE_X; x++)
{
for (y = 1; y <= SIZE_Y; y++)
{
MemorizeThis[x-1][y-1] = x * y;
printf("[%03d]", MemorizeThis[x-1][y-1]);
}
printf("\n");
}
printf("\n");
return;
}

10
Generic/Seive/desc.yml Normal file
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@ -0,0 +1,10 @@
title: Sieve of Eratosthenes
url: http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
desc:Generate primes from 2 to 100
solution: Do i need to say more
solutions:
solve.c:
desc: ANSI C solution (Tested with TCC)
language: c

22
Generic/Seive/solve.c Normal file
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@ -0,0 +1,22 @@
#include <stdio.h>
void main( )
{
unsigned char primes[100];
int p;
int j;
memset(&primes, 1, 50 );
for( p = 2; p<=10; p++) {
if (primes[p]) {
for(j = p*p; j <= 100; j+= p){
primes[j] = 0;
}
}
}
printf( "Primes:%d\n", sizeof(primes)) ;
for(p=0;p<sizeof(primes);p++) {
if(primes[p]) {
printf( "%d\n",p);
}
}
}

3
ProjectEuler/001/desc.yml
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@ -21,3 +21,6 @@ solutions:
solve.js:
desc: NodeJS solution
language: javascript
solve.lua:
desc: Basic solution
language: lua

7
ProjectEuler/001/solve.lua Normal file
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@ -0,0 +1,7 @@
sum = 0
for i=1,999 do
if i % 3 == 0 or i % 5 == 0 then
sum = sum + i
end
end
print(sum)

3
ProjectEuler/002/desc.yml
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@ -22,3 +22,6 @@ solutions:
solve.js:
desc: NodeJS solution
language: javascript
solve.lua:
desc: Basic solution
language: lua

11
ProjectEuler/002/solve.lua Normal file
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@ -0,0 +1,11 @@
sum = 2
fib = {1, 2, 3}
while fib[2] < 4000000 do
fib[3] = fib[1] + fib[2]
if(fib[3] % 2 == 0) then
sum = sum + fib[3]
end
fib[1] = fib[2]
fib[2] = fib[3]
end
print (sum )

3
ProjectEuler/005/desc.yml
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@ -15,3 +15,6 @@ solutions:
solve.rb:
desc: Basic solution in Ruby
language: ruby
solve.js:
desc: Basic solution for NodeJS
language: js

11
ProjectEuler/005/solve.js Normal file
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@ -0,0 +1,11 @@
for(i=20;true;i+=20) {
div = 19;
while(!(i % div)) {
div--;
if(div == 0) {
console.log( i );
process.exit(0);
}
}
}

12
ProjectEuler/005/solve.lua Normal file
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@ -0,0 +1,12 @@
i=0
while(true) do
i = i + 20
div = 19
while((i % div) == 0) do
div = div - 1
if(div == 0) then
print(i)
os.exit()
end
end
end

17
ProjectEuler/005/solve.rb
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@ -1,10 +1,13 @@
max = 0;
(100..1000).each do |num1|
(100..1000).each do |num2|
sum = num1 * num2
if( sum > max and sum.to_s.reverse == sum.to_s)
max = sum
i = 0
while (true)
i = i + 20
div = 19
while((i % div) == 0)
div = div - 1
if(div == 0) then
puts i
exit
end
end
end
print max

5
ProjectEuler/006/desc.yml
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@ -19,6 +19,9 @@ solutions:
solve.rb:
desc: Basic Ruby solution
language: ruby
solve.rb:
solve.c:
desc: ANSI C solution (Tested with TCC)
language: c
solve.js:
desc: Javascript solution for NodeJS
language: javascript

9
ProjectEuler/006/solve.js Normal file
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@ -0,0 +1,9 @@
square = 0;
sum = 0;
for(num=1;num<101;num++) {
square += Math.pow(num,2);
sum += num;
}
console.log( Math.pow(sum,2) - square);

2
ProjectEuler/008/desc.yml
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@ -11,6 +11,8 @@ solutions:
solve.php:
desc: Expects the haystack as stdin with optional newlines
language: php
parameters: < ProjectEuler\008\input
solve.rb:
desc: Expects the haystack as stdin with optional newlines
language: ruby
parameters: < ProjectEuler\008\input

2
ProjectEuler/010/desc.yml
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@ -16,5 +16,5 @@ solutions:
desc: Basic solution
language: ruby
solve.c:
desc: ANSI C solution compiled with gcc-4.3.4
desc: C solution compiled with gcc-4.3.4 (Support for long long needed)
language: c

2
ProjectEuler/011/desc.yml
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@ -13,6 +13,8 @@ solutions:
solve.php:
desc: Expects the matrix as stdin
language: php
parameters: < ProjectEuler\011\input
solve.rb:
desc: Expects the matrix as stdin
language: ruby
parameters: < ProjectEuler\011\input

2
ProjectEuler/013/desc.yml
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@ -12,6 +12,8 @@ solutions:
solve.php:
desc: Expects input on STDIN
language: php
parameters: < ProjectEuler\013\input
solve.rb:
desc: Expects input on STDIN
language: ruby
parameters: < ProjectEuler\013\input

2
ProjectEuler/018/desc.yml
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@ -18,6 +18,8 @@ solutions:
solve.php:
desc: Basic solution
language: php
parameters: < ProjectEuler\018\input
solve.rb:
desc: Basic solution
language: ruby
parameters: < ProjectEuler\018\input

1
ProjectEuler/022/desc.yml
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@ -12,3 +12,4 @@ solutions:
solve.php:
desc: Expects data on STDIN
language: php
parameters: < ProjectEuler\022\input

23
ProjectEuler/026/desc.yml Normal file
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@ -0,0 +1,23 @@
title: Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
url: http://projecteuler.net/problem=26
desc: |
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
solution: Bruteforce
solutions:
solve.php:
desc: Basic Solution - needs BCMath
language: php

20
ProjectEuler/026/solve.php Normal file
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@ -0,0 +1,20 @@
<?php
define('MAX',1000);
$max_len = 0;
for($div = MAX; $div > 2; $div--) {
$result = bcdiv('1',(string)$div,2000);
for($len = 2; $len < 1000; $len++) {
if(substr($result,10,$len) === substr($result,10+$len,$len)) {
if($len > $max_len) {
echo "Div: ".$div. " Len: ".$len."\n";
$max_len = $len;
$max_div = $div;
}
break;
}
}
}
echo $max_div;

13
ProjectEuler/026/solve.rb Normal file
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@ -0,0 +1,13 @@
DIGITS = 1000
max = 10 ** (DIGITS-1)
fcurrent = 1
fprev = 1
term = 2
while fcurrent < max do
term = term + 1
fnext = fcurrent + fprev
fprev = fcurrent;
fcurrent = fnext;
end
puts term

20
ProjectEuler/027/desc.yml Normal file
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@ -0,0 +1,20 @@
title: Quadratic primes
url: http://projecteuler.net/problem=27
desc: |
Euler discovered the remarkable quadratic formula:
n² + n + 41
It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.
The incredible formula n² - 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.
Considering quadratics of the form:
n² + an + b, where |a| < 1000 and |b| < 1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |-4| = 4
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0
solution: Bruteforce
solutions:
solve.php:
desc: Basic Solution - needs BCMath
language: php

30
ProjectEuler/027/problem27.php Normal file
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@ -0,0 +1,30 @@
<?php
function is_prime($prime) {
if($prime < 1)
{
return false;
}
if($prime == 1)
return true;
if($prime == 2)
return true;
$sqrt = sqrt($prime);
for ($i = 3; $i <= $sqrt; $i+=2){
if ($prime%$i == 0) return false;
}
return true;
}
$high = 0;
$max_primes = 0;
for($a=-999; $a<1000; $a++) {
for($b=-999; $b<1000; $b=$b+2) {
$n=0;
while(is_prime($n*$n + $a * $n + $b)) {
$n++;
if($n > $max_primes) {
$max_primes = $n;
echo "max: $n a: $a b: $b\n";
}
}
}
}

32
ProjectEuler/027/solve.php Normal file
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@ -0,0 +1,32 @@
<?php
function is_prime($prime) {
if($prime == 1)
return true;
if($prime == 2)
return true;
$sqrt = sqrt($prime);
for ($i = 3; $i <= $sqrt; $i+=2){
if ($prime%$i == 0) return false;
}
return true;
}
$high = 0;
$max_primes = 0;
for($a=-999; $a<1000; $a++) {
for($b=-999; $b<1000; $b=$b+2) {
$n=0;
while(true) {
echo $n*$n + $a * $n + $b . "\n";
if(!is_prime($n*$n + $a * $n + $b)) {
if($n > $max_primes) {
$max_primes = $n;
echo "max: $n a: $a b: $b\n";
}
break;
}
$n++;
}
}
}

13
ProjectEuler/027/solve.rb Normal file
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@ -0,0 +1,13 @@
DIGITS = 1000
max = 10 ** (DIGITS-1)
fcurrent = 1
fprev = 1
term = 2
while fcurrent < max do
term = term + 1
fnext = fcurrent + fprev
fprev = fcurrent;
fcurrent = fnext;
end
puts term

32
ProjectEuler/028/desc.yml Normal file
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@ -0,0 +1,32 @@
title: What is the sum of both diagonals in a 1001 by 1001 spiral?
url: http://projecteuler.net/problem=28
desc: |
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is 101.
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
solution: |
Step around in the spiral (add 2 - 3 times, then add 4 - 3times, then 6 - times etc)
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution
language: ruby
solve.c:
desc: ANSI C solution (Tested with TCC)
language: c
solve.js:
desc: NodeJS solution
language: javascript
solve.lua:
desc: Basic solution
language: lua

19
ProjectEuler/028/solve.c Normal file
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@ -0,0 +1,19 @@
#include <stdio.h>
#define SIZE 1001
#define SIDES 4
int main( )
{
int sum = 1;
int result=1;
int addition;
for(addition = 2; addition <= SIZE; addition+=2) {
int cside;
for(cside = 0; cside < SIDES; cside++) {
sum += addition;
result += sum;
}
}
printf( "%i", result );
}

13
ProjectEuler/028/solve.js Normal file
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@ -0,0 +1,13 @@
SIZE = 1001;
SIDES = 4
sum = 1;
result=1;
for(addition = 2; addition <= SIZE; addition+=2) {
for(cside = 0; cside < SIDES; cside++) {
sum += addition;
result += sum;
}
}
console.log(result);

12
ProjectEuler/028/solve.lua Normal file
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@ -0,0 +1,12 @@
SIZE = 1001;
SIDES = 4;
sum = 1;
result=1;
for addition = 2,SIZE,2 do
for cside=0,(SIDES-1) do
sum = sum + addition;
result = result + sum;
end
end
print(result)

12
ProjectEuler/028/solve.php Normal file
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@ -0,0 +1,12 @@
<?php
define('SIZE',1001);
define('SIDES',4);
$sum = 1;
$result=1;
for($addition = 2; $addition <= SIZE; $addition+=2) {
for($cside = 0; $cside < SIDES; $cside++) {
$sum += $addition;
$result += $sum;
}
}
echo $result;

12
ProjectEuler/028/solve.rb Normal file
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@ -0,0 +1,12 @@
SIZE = 1001
SIDES = 4
sum = 1;
result=1;
(2...SIZE).step(2) { |add|
SIDES.downto(1) {
sum += add;
result += sum;
}
}
puts result

34
ProjectEuler/031/problem31.php Normal file
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@ -0,0 +1,34 @@
<?php
define("TOTAL", 200);
$combo=0;
$sum = 200;
for ($a = 0; $a <= (TOTAL / 200); $a++) {
$sum_a = TOTAL - ($a * 200);
for ($b = 0; $b <= ($sum_a / 100); $b++) {
$sum_b = $sum_a - ($b * 100);
for ($c = 0; $c <= ($sum_b / 50); $c++) {
$sum_c = $sum_b - ($c * 50);
for ($d = 0; $d <= ($sum_c / 20); $d++) {
$sum_d = $sum_c - ($d * 20);
for ($e = 0; $e <= ($sum_d / 10); $e++) {
$sum_e = $sum_d - ($e * 10);
for ($f = 0; $f <= ($sum_e / 5); $f++) {
$sum_f = $sum_e - ($f * 5);
for ($g = 0; $g <= ($sum_e / 2); $g++) {
$sum_g = $sum_f - ($g * 2);
for ($h = 0; $h <= ($sum_g / 1); $h++) {
if ($a * 200 + $b * 100 + $c * 50 + $d * 20 + $e * 10 + $f * 5 + $g * 2 + $h == TOTAL) {
$combo++;
}
}
}
}
}
}
}
}
}
echo $combo;

6
ProjectEuler/032/solve.php
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@ -1,6 +0,0 @@
<?php
function pandigital($number) {
$array = count_chars($number,1);
ksort($array);
if($array == array(49=>1,50=>1,51=>1,52=>1,53=>1,54=>1,55=>1,56=>1,57=>1)) { return true;} else { return false; }
}

24
ProjectEuler/033/solve.php Normal file
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@ -0,0 +1,24 @@
<?php
$sum = [];
for($n = 10; $n <= 99; $n++)
{
for($d=$n+1;$d <= 99; $d++)
{
foreach([1,2,3,4,5,6,7,8,9] as $digit)
{
$ns = (int)str_replace($digit, '', $n);
$ds = (int)str_replace($digit, '', $d);
if($n != $ns && $d != $ds && $d != 0 && $ds != 0)
{
if($n/$d == $ns/$ds)
{
echo "$n/$d = $ns/$ds\n";
$sum[] = $ns/$ds;
}
}
}
}
}
echo array_product($sum).PHP_EOL;

2
ProjectEuler/034/desc.yml
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@ -6,7 +6,7 @@ desc: |
Find the sum of all numbers which are equal to the sum of the factorial of their digits.
Note: as 1! = 1 and 2! = 2 are not sums they are not included.
solution: Bruteforce
solutons:
solutions:
solve.php:
desc: Basic solution
language: php

11
ProjectEuler/035/desc.yml Normal file
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@ -0,0 +1,11 @@
title: How many circular primes are there below one million?
url: http://projecteuler.net/problem=35
desc: |
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
solution: Bruteforce
solutions:

61
ProjectEuler/035/solve.php Normal file
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@ -0,0 +1,61 @@
<?php
function is_prime($prime) {
if($prime < 1)
return false;
if($prime == 1)
return true;
if($prime == 2)
return true;
$sqrt = sqrt($prime);
for ($i = 3; $i <= $sqrt; $i+=2){
if ($prime%$i == 0) return false;
}
return true;
}
if (($handle = fopen("primes1000000.txt", "r")) !== FALSE) {
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
if ($data[1] >= 1000000)
break;
//If there's an even digit (or a 5) we can instantly exclude it as when rotated it will not not be a prime.
if (
strpos((string) $data[1], "0") !== false ||
strpos((string) $data[1], "2") !== false ||
strpos((string) $data[1], "4") !== false ||
strpos((string) $data[1], "5") !== false ||
strpos((string) $data[1], "6") !== false ||
strpos((string) $data[1], "8") !== false
) {
continue;
}
$primes[] = (int) $data[1];
}
fclose($handle);
}
echo "File read\n";
$count = 1+1; //2 and 5 are primes but not included
foreach($primes as $prime)
{
//echo "Now testing $prime\n";
$newprime = $prime;
for($rotate = 1; $rotate < strlen($prime); $rotate++)
{
$newprime = substr($newprime, -1, 1).substr($newprime, 0, -1);
//echo "New prime: $newprime\n";
if(!in_array($newprime, $primes))
{
continue 2;
}
}
$count++;
echo $prime.PHP_EOL;
}
echo "Total: $count\n";

18
ProjectEuler/036/desc.yml Normal file
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title: Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2.
url: http://projecteuler.net/problem=36
desc: |
The decimal number, 585 = 1001001001 (binary), is palindromic in both bases.
Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
(Please note that the palindromic number, in either base, may not include leading zeros.)
solution: We can skip all even numbers since a even number would end in binary 10 - In reverse that would be 01 and we are told to skip leading 0's
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution in Ruby
language: ruby

12
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<?php
define('MAX',1000000);
$result = 0;
for($i=1;$i<MAX;$i+=2) {
if($i == strrev($i)) {
$bin = decbin($i);
if($bin == strrev($bin)) {
$result += $i;
}
}
}
echo $result;

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MAX = 1000000
result = 0
(1..MAX).step(2) { |i|
if(i.to_s == i.to_s.reverse)
if(i.to_s(2) == i.to_s(2).reverse)
result += i
end
end
}
print result

19
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title: What is the largest 1 to 9 pandigital that can be formed by multiplying a fixed number by 1, 2, 3, ... ?
url: http://projecteuler.net/problem=38
desc: |
Take the number 192 and multiply it by each of 1, 2, and 3:
192 x 1 = 192
192 x 2 = 384
192 x 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n 1?
solution: |
Bruteforce
solutions:
solve.php:
desc: Basic solution
language: php

27
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<?php
function pandigital($number) {
$array = count_chars($number,1);
ksort($array);
if($array == array(49=>1,50=>1,51=>1,52=>1,53=>1,54=>1,55=>1,56=>1,57=>1)) { return true;} else { return false; }
}
while(true) {
$value++;
$test = array(1,2);
$key = 2;
while(true) {
$new = array();
foreach($test as $p) {
$new[] = $value * $p;
}
$var = implode('',$new);
if(strlen($var) != 9) { break; }
if(strlen($var) == 9 AND pandigital($var)) { $good[] = $var; break; }
$test[] = $key++;
}
if($value > 9999) { break; }
}
rsort($good);
echo $good[0];

18
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title: Integer right triangles
url: http://projecteuler.net/problem=39
desc: |
If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p <=1000, is the number of solutions maximised?
solution: |
Bruteforce
solutions:
solve.php:
desc: Basic solution
language: php
solve.c:
desc: Basic solution
language: c

28
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#include <stdio.h>
#define P_VALUE 1000
int p, a, b, c, max, key, result = 0;
int main( )
{
for(p=2;p<P_VALUE;p++)
{
for(a=2;a<p;a++)
{
for(b=a;b<p;b++)
{
//Find C so it runs faster...
c = p - a - b;
if(a*a+b*b == c*c) {
result++;
}
}
}
if(result > max) {
max = result;
key = p;
}
result = 0;
}
printf( "%i", key);
}

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<?php
define('P_VALUE',1000);
for($p=2;$p<P_VALUE;$p++) {
for($a=2;$a<$p;$a++) {
for($b=$a;$b<$p;$b++) {
//Find C so it runs faster...
$c = $p - $a - $b;
if($a*$a+$b*$b == $c*$c) {
$result++;
}
}
}
if($result > $max) {
$max = $result; $key = $p;
}
$result = 0;
}
echo $key;

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ProjectEuler/041/desc.yml Normal file
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title: Integer right triangles
url: http://projecteuler.net/problem=41
desc: |
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
What is the largest n-digit pandigital prime that exists?
solution: |
Bruteforce
solutions:
solve.php:
desc: Basic solution
language: php

27
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<?php
function pandigital_less($number) {
$array = count_chars($number,1);
foreach(range(1,count($array)) as $char) {
if( $array[ord($char)]!== 1) return false;
}
return true;
}
function is_prime($prime) {
//if($prime%2 == 0) return false;
$sqrt = sqrt($prime);
for ($i = 3; $i <= $sqrt; $i+=2){
if ($prime%$i == 0) return false;
}
return true;
}
//We don't need to check uneven numbers
//Has to start with a 7 and therefor be 7 digits long
for($var = 7000001; $var < 8000000; $var=$var+2) {
if(is_prime($var) AND pandigital_less($var)) { $result = $var; }
}
echo $result;

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title: Coded triangle numbers
url: http://projecteuler.net/problem=42
desc: |
The nth term of the sequence of triangle numbers is given by, tn = 0.5n(n+1); so the first ten triangle numbers are:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we shall call the word a triangle word.
Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words?
solution: Bruteforce
solutions:
solve.php:
desc: Expects data on STDIN
language: php
parameters: < ProjectEuler\042\words.txt

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ProjectEuler/042/solve.php Normal file
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<?php
$words = explode('","',substr(file_get_contents('php://STDIN'),1,-1));
//Find triangles
for($t=1;$t<40;$t++) {
$triangle_numbers[] = 0.5*$t*($t+1);
}
foreach($words as $word) {
for($c =0; $c < strlen($word); $c++) {
$char_sum += ord($word[$c]) - 64;
}
if(in_array($char_sum,$triangle_numbers)) {
$result++;
}
$char_sum = 0;
}
echo $result;

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ProjectEuler/042/words.txt Normal file
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<?php
/**
* Triangle Tn = n*(n+1)/2
* Pentagonal Pn = n*(3n-1)/2
* Hexagonal Hn = n*(2n-1)
* T(285) = P(165) = H(143) = 40755
* Find next
*/
function Tn($n)
{
return $n*($n+1)/2;
}
function Pn($n)
{
return $n*(3*$n-1)/2;
}
function Hn($n)
{
return $n*(2*$n-1);
}
echo "Tn(285)=".Tn(285).PHP_EOL;
echo "Pn(165)=".Pn(165).PHP_EOL;
echo "Hn(143)=".Hn(143).PHP_EOL;
for($t = 285; $t < 100000; $t++)
{
$triangle[$t] = Tn($t);
}
for($p = 165; $p < 100000; $p++)
{
$pentagonal[$p] = Pn($p);
}
for($h = 143; $h < 100000; $h++)
{
$hexagonal[$h] = Hn($h);
}
$result = array_intersect($triangle, $pentagonal, $hexagonal);
print_r($result);

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title: Find the last ten digits of 11 + 22 + ... + 10001000.
url: http://projecteuler.net/problem=48
desc: |
The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
solution: |
Bruteforce
solutions:
solve.php:
desc: Using BCMath to cope with large numbers
language: php
solve.rb:
desc: Basic solution
language: ruby

7
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<?php
define('MAX',1000);
$sum = 0;
for($c =1; $c <= MAX;$c++) {
$sum = bcadd($sum,bcpow($c,$c));
}
echo substr($sum,-10,10);

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MAX = 1000
sum = 0
(1..MAX).each do |digit|
sum += digit**digit
end
puts sum.to_s[-10..-1]

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title: Finding Fibonacci numbers for which the first and last nine digits are pandigital.
url: http://projecteuler.net/problem=104
desc: |
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1.
It turns out that F541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9, but not necessarily in order). And F2749, which contains 575 digits, is the first Fibonacci number for which the first nine digits are 1-9 pandigital.
Given that Fk is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.
solution: |
Bruteforce
solutions:
solve.php:
desc: Bruteforce - Sooooo slow
language: php

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<?php
function pandigital($num) {
for ($i = 1; $i <= 9; $i++) {
if(strpos($n,(string)$i) === false) {
return false;
}
}
return true;
}
$var = 1;
$k = 1;
while(true) {
$k++;
$new = bcadd($var,$prev);
$prev = $var;
$var = $new;
if(pandigital(substr($var,-9)) AND pandigital(substr($var,0,9))) { echo $k; die; }
}

1
ct.bat
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@ -1 +1,2 @@
@ECHO OFF
ruby CT.rb %1 %2

5
nbproject/private/private.properties Normal file
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copy.src.files=false
copy.src.target=
index.file=
run.as=SCRIPT
url=http://localhost/codingtests/

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include.path=${php.global.include.path}
php.version=PHP_54
source.encoding=UTF-8
src.dir=.
tags.asp=false
tags.short=true
web.root=.

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<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://www.netbeans.org/ns/project/1">
<type>org.netbeans.modules.php.project</type>
<configuration>
<data xmlns="http://www.netbeans.org/ns/php-project/1">
<name>codingtests</name>
</data>
</configuration>
</project>

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1, 2, 2
2, 3, 1
3, 5, 2
4, 7, 2
5, 11, 4
6, 13, 2
7, 17, 4
8, 19, 2
9, 23, 4
10, 29, 6
11, 31, 2
12, 37, 6
13, 41, 4
14, 43, 2
15, 47, 4
16, 53, 6
17, 59, 6
18, 61, 2
19, 67, 6
20, 71, 4
21, 73, 2
22, 79, 6
23, 83, 4
24, 89, 6
25, 97, 8
26, 101, 4
27, 103, 2
28, 107, 4
29, 109, 2
30, 113, 4
31, 127, 14
32, 131, 4
33, 137, 6
34, 139, 2
35, 149, 10
36, 151, 2
37, 157, 6
38, 163, 6
39, 167, 4
40, 173, 6
41, 179, 6
42, 181, 2
43, 191, 10
44, 193, 2
45, 197, 4
46, 199, 2
47, 211, 12
48, 223, 12
49, 227, 4
50, 229, 2
51, 233, 4
52, 239, 6
53, 241, 2
54, 251, 10
55, 257, 6
56, 263, 6
57, 269, 6
58, 271, 2
59, 277, 6
60, 281, 4
61, 283, 2
62, 293, 10
63, 307, 14
64, 311, 4
65, 313, 2
66, 317, 4
67, 331, 14
68, 337, 6
69, 347, 10
70, 349, 2
71, 353, 4
72, 359, 6
73, 367, 8
74, 373, 6
75, 379, 6
76, 383, 4
77, 389, 6
78, 397, 8
79, 401, 4
80, 409, 8
81, 419, 10
82, 421, 2
83, 431, 10
84, 433, 2
85, 439, 6
86, 443, 4
87, 449, 6
88, 457, 8
89, 461, 4
90, 463, 2
91, 467, 4
92, 479, 12
93, 487, 8
94, 491, 4
95, 499, 8
96, 503, 4
97, 509, 6
98, 521, 12
99, 523, 2
100, 541, 18

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1000000
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languages:
javascript:
command: node
ext: [js,javascript]
php:
command: php
ext: [php]
ruby:
command: ruby
ext: [ruby,rb]
tcc:
command: tcc -run
ext: [c]
lua:
command: lua
ext: [lua]