Restructuring

This commit is contained in:
2024-07-01 13:49:44 +00:00
parent f11b705ef0
commit 8d60e1b905
194 changed files with 1296 additions and 112 deletions
CT.rbcomposer.jsoncomposer.lockct.bat
data/primes
nbproject
phpunit.xmlsettings.yml
solutions
CodeChef/easy
HS08TEST
TEST
CodeGolf/guns-and-zips
Generic
ProjectEuler
src
template.rt
tests
vendor

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<?php
while($s=explode(' ',trim(fgets(STDIN)))) {
if($s[1]%5) {
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if($new)
echo $new;
else
echo $s[1];
}
}

@ -0,0 +1,7 @@
<?php
echo fgets(STDIN);
while($s = fgets(STDIN)) {
if(trim($s) == '42')
die;
echo $s;
}

@ -0,0 +1,53 @@
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?????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`?

@ -0,0 +1,11 @@
title: FizzBuzz
url: http://99-bottles-of-beer.net/
desc: Generate the lyrics to the song 99 Bottles of Beer as an output.
solution: See the code...
solutions:
solve.php:
desc: Basic solution
language: php

@ -0,0 +1,18 @@
<?
for ($beer=99; $beer>=0; $beer--)
{
if ($beer==0) {
echo "No more bottles of beer on the wall, no more bottles of beer.".PHP_EOL;
echo "Go to the store and buy some more, 99 bottles of beer on the wall.".PHP_EOL;
}
elseif ($beer==1) {
echo "1 bottle of beer on the wall, 1 bottle of beer.".PHP_EOL;
echo "Take one down and pass it around, no more bottles of beer on the wall.". PHP_EOL . PHP_EOL;
} elseif ($beer==2) {
echo $beer." bottles of beer on the wall, ".$beer." bottles of beer." . PHP_EOL;
echo "Take one down and pass it around, ".($beer-1)." bottle of beer on the wall.".PHP_EOL . PHP_EOL;
} else {
echo $beer." bottles of beer on the wall, ".$beer." bottles of beer." . PHP_EOL;
echo "Take one down and pass it around, ".($beer-1)." bottles of beer on the wall.".PHP_EOL . PHP_EOL;
}
}

@ -0,0 +1,17 @@
title: FizzBuzz
url: http://imranontech.com/2007/01/24/using-fizzbuzz-to-find-developers-who-grok-coding/
desc: The classical coding challenge. Write a program that prints the numbers from 1 to 100. But for multiples of three print Fizz instead of the number and for the multiples of five print Buzz. For numbers which are multiples of both three and five print FizzBuzz.
solution: See the code...
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution in Ruby
language: ruby
solve.c:
desc: ANSI C solution (Tested with TCC)
language: c

@ -0,0 +1,22 @@
#include <stdio.h>
int main( )
{
int i = 0;
for(i = 1; i <= 100; i++)
{
if(i % 3 == 0)
{
printf( "Fizz" );
}
if(i % 5 == 0)
{
printf( "Buzz" );
}
if(i % 3 && i % 5)
{
printf( "%d", i );
}
printf( "\n" );
}
}

@ -0,0 +1,13 @@
<?php
for($i=1; $i<=100; $i++) {
if($i % 3 == 0) {
print "Fizz";
}
if($i % 5 == 0) {
print "Buzz";
}
if($i % 3 AND $i % 5) {
print $i;
}
print PHP_EOL;
}

@ -0,0 +1,12 @@
(1..100).each do |i|
if(i % 3 == 0)
print "Fizz";
end
if(i % 5 == 0)
print "Buzz"
end
if(i % 3 != 0 and i % 5 != 0)
print i
end
print "\n";
end

Binary file not shown.

@ -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;
}

Binary file not shown.

@ -0,0 +1,19 @@
title: HelloWorld
url: http://en.wikipedia.org/wiki/Hello_world_program
desc: A "Hello world" program is a computer program that outputs "Hello, world" on a display device.
solution: Do i need to say more
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution in Ruby
language: ruby
solve.c:
desc: ANSI C solution (Tested with TCC)
language: c
solve.js:
desc: NodeJS solution
language: js

@ -0,0 +1,7 @@
#include <stdio.h>
int main( )
{
printf( "Hello World" );
}

@ -0,0 +1 @@
console.log("Hello World");

@ -0,0 +1,2 @@
<?php
echo 'Hello World';

@ -0,0 +1 @@
puts "Hello World"

@ -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;
}

@ -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

@ -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);
}
}
}

@ -0,0 +1,4 @@
title: Generic
url: http://google.com
desc: Various problems that doesn't fit any catagory

@ -0,0 +1,26 @@
title: Add all the natural numbers below one thousand that are multiples of 3 or 5.
url: http://projecteuler.net/problem=1
desc: |
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
solution: |
Loop numbers from 1-1000 - Use mod to find the multiples
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution in Ruby
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

@ -0,0 +1,15 @@
#include <stdio.h>
int main( )
{
int sum = 0;
int i = 0;
for(i = 1; i < 1000; i++)
{
if(i % 3 == 0 || i % 5 == 0)
{
sum += i;
}
}
printf( "%i", sum );
}

@ -0,0 +1,9 @@
sum = 0;
i = 0;
for(i=1; i<1000;i++) {
if(i % 3 == 0 || i % 5 == 0) {
sum += i;
}
}
console.log(sum);

@ -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)

@ -0,0 +1,10 @@
<?php
$sum = 0;
$i = 0;
for($i=1; $i<1000;$i++) {
if($i % 3 == 0 OR $i % 5 == 0) {
$sum += $i;
}
}
echo $sum;

@ -0,0 +1,7 @@
sum = 0
(1..1000).each do |i|
if(i % 3 == 0 or i % 5 == 0)
sum += i
end
end
puts sum

@ -0,0 +1,27 @@
title: By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
url: http://projecteuler.net/problem=2
desc: |
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
solution: |
Step through the fibonacci sequence while you sum each even entry. Quit as you reach 4million
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution in Ruby
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

@ -0,0 +1,18 @@
#include <stdio.h>
int main( )
{
int sum = 2;
int fib[3] = { 1, 2, 3 };
while(fib[2] < 4000000)
{
fib[2] = fib[0] + fib[1];
if(fib[2] % 2 == 0)
sum += fib[2];
fib[0] = fib[1];
fib[1] = fib[2];
}
printf( "%i", sum );
}

@ -0,0 +1,11 @@
sum = 2;
fib = new Array( 1, 2, 3 );
while(fib[2] < 4000000)
{
fib[2] = fib[0] + fib[1];
if(fib[2] % 2 == 0)
sum += fib[2];
fib[0] = fib[1];
fib[1] = fib[2];
}
console.log(sum );

@ -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 )

@ -0,0 +1,12 @@
<?php
$fib = array(1,2,3);
$sum = 2;
while($fib[2] < 4000000) {
$fib[2] = $fib[0] + $fib[1];
if($fib[2] % 2 == 0) {
$sum += $fib[2];
}
$fib[0] = $fib[1];
$fib[1] = $fib[2];
}
echo $sum;

@ -0,0 +1,11 @@
fib = [1, 2, 3];
sum = 2;
while(fib[2] < 4000000)
fib[2] = fib[0] + fib[1];
if(fib[2] % 2 == 0)
sum += fib[2];
end
fib[0] = fib[1];
fib[1] = fib[2];
end
puts sum;

@ -0,0 +1,15 @@
title: Find the largest prime factor of a composite number.
url: http://projecteuler.net/problem=3
desc: |
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
solution: |
Use factorization to find the largest prime factor
solutions:
solve.php:
desc: Basic solution
language: php

@ -0,0 +1,33 @@
<?php
$number = 600851475143;
echo max(factorize($number));
/**
* Returns a sorted array of the prime factorization of $num
* @staticvar array $aFactors
* @param type int Number to factorize
* @return type array Prime factors
*/
function factorize($num) {
// Returns a sorted array of the prime factorization of $num
// Caches prior results. Returns empty array for |$num|<2
// eg. factorize(360) => [5, 3, 3, 2, 2, 2]
static $aFactors = array();
if (2 > $num = abs($num))
return array(); // negatives, 1, 0
if ($aFactors[$key = "x$num"]) { // handles doubles
// Been there, done that
if (($factor = $aFactors[$key]) == $num)
return array($num);
return array_merge(factorize($num / $factor), array($factor));
}
// Find a smallest factor
for ($sqrt = sqrt($num), $factor = 2; $factor <= $sqrt; ++$factor)
if (floor($num / $factor) == $num / $factor)
return array_merge(factorize($num / $factor), array($aFactors[$key] = $factor));
return (array($aFactors[$key] = $num));
}

@ -0,0 +1,17 @@
title: Find the largest palindrome made from the product of two 3-digit numbers.
url: http://projecteuler.net/problem=4
desc: |
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.
solution: |
See code
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution in Ruby
language: ruby

@ -0,0 +1,13 @@
<?php
$max = 0;
for($num1 = 1000; $num1>100; $num1--) {
for($num2 = 1000; $num2>100; $num2--) {
$sum = $num1 * $num2;
//Check if palindrome
if($sum > $max AND strrev($sum) == $sum)
$max = $sum;
}
}
echo $max;

@ -0,0 +1,10 @@
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
end
end
end
print max

@ -0,0 +1,20 @@
title: What is the smallest number divisible by each of the numbers 1 to 20?
url: http://projecteuler.net/problem=5
desc: |
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
solution: |
See code
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution in Ruby
language: ruby
solve.js:
desc: Basic solution for NodeJS
language: js

@ -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);
}
}
}

@ -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

@ -0,0 +1,12 @@
<?php
for($i=20;true;$i+=20) {
$div = 19;
while(!($i % $div)) {
$div--;
if($div == 0) {
echo $i;
die;
}
}
}

@ -0,0 +1,13 @@
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

@ -0,0 +1,27 @@
title: What is the smallest number divisible by each of the numbers 1 to 20?
url: http://projecteuler.net/problem=6
desc: |
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 55^2 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
solution: |
See code
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic Ruby solution
language: ruby
solve.c:
desc: ANSI C solution (Tested with TCC)
language: c
solve.js:
desc: Javascript solution for NodeJS
language: javascript

@ -0,0 +1,19 @@
#include <stdio.h>
#include <math.h>
int main( )
{
int num;
int result;
int square = 0;
int sum = 0;
for(num=1; num<=100; num++) {
square += pow(num,2);
sum += num;
}
result = pow(sum,2) - square;
printf("%d", result);
}

@ -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);

@ -0,0 +1,9 @@
<?php
$square =0;
$sum =0;
for($num=1;$num<101;$num++) {
$square += pow($num,2);
$sum += $num;
}
echo pow($sum,2) - $square;

@ -0,0 +1,7 @@
square = 0
sum = 0
(1..100).each do |num|
square += num**2;
sum += num;
end
puts sum**2 - square;

@ -0,0 +1,20 @@
title: Find the 10001st prime.
url: http://projecteuler.net/problem=7
desc: |
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
solution: |
Move up the among the prime numbers till you reach 10001
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution
language: ruby
solve.c:
desc: ANSI C (Compiled with TCC) solution
language: c

@ -0,0 +1,30 @@
#include <stdio.h>
#include <math.h>
int is_prime( int prime) {
int max_test = sqrt( prime );
int i;
for (i = 3; i <= max_test; i+=2){
if (prime % i == 0){
return 0;
}
}
return 1;
}
int main( )
{
//Use start val
int i=13;
int primes = 6;
do{
i+= 2;
if(is_prime(i)){
primes++;
}
} while(primes < 10001);
printf("%d", i);
}

@ -0,0 +1,18 @@
<?php
function is_prime($prime) {
$sqrt = sqrt($prime);
for ($i = 3; $i <= $sqrt; $i+=2){
if ($prime%$i == 0) return false;
}
return true;
}
$i=13;
$primes = 6;
do{
$i+= 2;
if(is_prime($i)){
$primes++;
}
}while($primes < 10001);
echo $i;

@ -0,0 +1,21 @@
def is_prime(prime)
sqrt = Math.sqrt(prime)
i=3
while(i <= sqrt)
if (prime % i == 0)
return false
end
i+= 2
end
return true;
end
i=13
primes = 6
until (primes == 10001)
i+= 2
if(is_prime(i))
primes+=1
end
end
puts i

@ -0,0 +1,18 @@
title: Discover the largest product of five consecutive digits in the 1000-digit number.
url: http://projecteuler.net/problem=8
desc: |
Find the greatest product of five consecutive digits in the 1000-digit number.
solution: |
Bruteforce (Unless someone found a smarter way)
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

@ -0,0 +1,20 @@
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

@ -0,0 +1,8 @@
<?php
$haystack = str_replace(array('\n','\r\n'),'',file_get_contents('php://stdin'));
for($i=0;$i<1000-4;$i++) {
$cmp = $haystack[$i] * $haystack[$i+1] * $haystack[$i+2] * $haystack[$i+3] * $haystack[$i+4];
$max = ($cmp > $max) ? $cmp : $max;
}
echo $max;

@ -0,0 +1,13 @@
haystack = ''
ARGF.lines("\n") do |line|
haystack << line.strip
end
max = 0
(1..(1000-4)).each do |index|
compare = haystack[index].to_i * haystack[index+1].to_i * haystack[index+2].to_i * haystack[index+3].to_i * haystack[index+4].to_i
if(compare > max)
max = compare
end
end
puts max

@ -0,0 +1,23 @@
title: Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000.
url: http://projecteuler.net/problem=9
desc: |
A Pythagorean triplet is a set of three natural numbers, a b c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
solution: |
Make a nested forloop for a and b in the range 1-1000 - Then c = 1000-a-b - Test if solution is valid.
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution
language: ruby
solve.c:
desc: ANSI C solution compiled with TCC
language: c

@ -0,0 +1,24 @@
#include "stdio.h"
#include "math.h"
int main( )
{
int a;
int b;
int c;
int cmp;
for(a = 1; a < 1000; a++)
{
for(b = 1; b < 1000; b++)
{
//Calculate the only valid value for c
c = 1000 - a - b;
if( pow(c,2) == (pow(a, 2 ) + pow( b, 2 )))
{
int result = a * b * c;
printf("%i", result);
return 0;
}
}
}
}

@ -0,0 +1,11 @@
<?php
for ($a = 1; $a < 1000; $a++) {
for ($b = 1; $b < 1000; $b++) {
//Make it run reverse in order to find solution quickly
$c = 1000 - $a - $b;
if (pow($a, 2) + pow($b, 2) == pow($c, 2)) {
echo $a * $b * $c;
die;
}
}
}

@ -0,0 +1,9 @@
(1..1000).each do |a|
(1..1000).each do |b|
c = 1000 - a - b
if( a ** 2 + b ** 2 == c ** 2)
puts a * b *c
exit
end
end
end

@ -0,0 +1,20 @@
title: Calculate the sum of all the primes below two million.
url: http://projecteuler.net/problem=10
desc: |
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
solution: |
Use a primenumber tester
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution
language: ruby
solve.c:
desc: C solution compiled with gcc-4.3.4 (Support for long long needed)
language: c

@ -0,0 +1,28 @@
#include <stdio.h>
#include <math.h>
int is_prime( int prime) {
int max_test = sqrt( prime );
int i;
for (i = 3; i <= max_test; i+=2){
if (prime % i == 0){
return 0;
}
}
return 1;
}
int main( )
{
long long sum = 2+3;
long number;
for(number = 5; number < 2000000; number+=2) {
if (is_prime(number))
{
sum += number;
}
}
printf("%llu",sum);
}

@ -0,0 +1,18 @@
<?php
function is_prime($prime) {
$sqrt = sqrt($prime);
for ($i = 3; $i <= $sqrt; $i+=2){
if ($prime%$i == 0) return false;
}
return true;
}
$sum = 2+3;
for($number = 5; $number < 2000000; $number+=2) {
if (is_prime($number))
{
$sum += $number;
}
}
echo $sum;

@ -0,0 +1,21 @@
def is_prime(prime)
sqrt = Math.sqrt(prime)
i=3
while(i <= sqrt)
if (prime % i == 0)
return false
end
i+= 2
end
return true
end
number = 3
sum = 5
until(number >= 2000000)
number += 2
if(is_prime(number))
sum += number
end
end
puts sum;

@ -0,0 +1,20 @@
title: What is the greatest product of four adjacent numbers on the same straight line in the 20 by 20 grid?
url: http://projecteuler.net/problem=11
desc: |
In the 2020 grid below, four numbers along a diagonal line have been marked in red.
What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid?
(See file input)
solution: |
Bruteforce (Unless someone found a smarter way)
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

@ -0,0 +1,20 @@
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

@ -0,0 +1,39 @@
<?php
foreach(file('php://stdin') as $line) {
$matrix[] = explode(' ', $line);
}
$max = 0;
//Left and right
for ($y = 0; $y < 19; $y++) {
for ($x = 0; $x < 16; $x++) {
$cmp = $matrix[$y][$x] * $matrix[$y][$x + 1] * $matrix[$y][$x + 2] * $matrix[$y][$x + 3];
$max = ($cmp > $max) ? $cmp : $max;
}
}
//Vertical
for($x = 0; $x < 19; $x++) {
for ($y = 0; $y < 16; $y++) {
$cmp = $matrix[$y][$x] * $matrix[$y + 1][$x] * $matrix[$y + 2][$x] * $matrix[$y + 3][$x];
$max = ($cmp > $max) ? $cmp : $max;
}
}
//Diagonally left->down
for($x = 0; $x < 16; $x++) {
for ($y = 0; $y < 16; $y++) {
$cmp = $matrix[$y][$x] * $matrix[$y + 1][$x + 1] * $matrix[$y + 2][$x + 2] * $matrix[$y + 3][$x + 3];
$max = ($cmp > $max) ? $cmp : $max;
}
}
//Diagonally right->down
for($x = 4; $x < 19; $x++) {
for ($y = 0; $y < 16; $y++) {
$cmp = $matrix[$y][$x] * $matrix[$y + 1][$x - 1] * $matrix[$y + 2][$x - 2] * $matrix[$y + 3][$x - 3];
$max = ($cmp > $max) ? $cmp : $max;
}
}
echo $max;

@ -0,0 +1,53 @@
matrix = []
linenum = 0
ARGF.lines("\n") do |line|
matrix[linenum] = Array.new
line.split.each do |str|
matrix[linenum] << str.to_i
end
linenum += 1
end
max = 0
max = 0;
#Left and right
(0..19).each do |y|
(0..16).each do |x|
cmp = matrix[y][x] * matrix[y][x + 1] * matrix[y][x + 2] * matrix[y][x + 3];
if (cmp > max)
max = cmp
end
end
end
#Vertical
(0..16).each do |y|
(0..19).each do |x|
cmp = matrix[y][x] * matrix[y + 1][x] * matrix[y + 2][x] * matrix[y + 3][x];
if (cmp > max)
max = cmp
end
end
end
#Diagonally left->down
(0..16).each do |y|
(0..16).each do |x|
cmp = matrix[y][x] * matrix[y + 1][x + 1] * matrix[y + 2][x + 2] * matrix[y + 3][x + 3];
if (cmp > max)
max = cmp
end
end
end
#Diagonally right->down
(0..16).each do |y|
(3..19).each do |x|
cmp = matrix[y][x] * matrix[y + 1][x - 1] * matrix[y + 2][x - 2] * matrix[y + 3][x - 3];
if (cmp > max)
max = cmp
end
end
end
puts max

@ -0,0 +1,30 @@
title: What is the value of the first triangle number to have over five hundred divisors?
url: http://projecteuler.net/problem=12
desc: |
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
solution: |
Bruteforce - Still kinda slow - Guessing it could be improved
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution
language: ruby
solve.c:
desc: ANSI C solution compiled with TCC
language: c

@ -0,0 +1,23 @@
#include <stdio.h>
#include <math.h>
int main( )
{
long num = 3;
long tri_add = 2;
long div;
int divisors;
int max = 0;
do
{
num += ++tri_add;
divisors = 2;
int square_root = sqrt(num);
for(div = 2; div <= square_root; div++)
{
if(num % div == 0)
divisors+=2;
}
} while(divisors <= 500);
printf( "%i", num );
}

@ -0,0 +1,13 @@
<?php
$num = 3;
$tri_add = 2;
do {
$num += ++$tri_add;
$divisors = 2;
$square_root = sqrt($num);
for($div = 2; $div <= $square_root; $div++) {
if($num % $div == 0)
$divisors+= 2;
}
} while ($divisors <= 500);
echo $num;

@ -0,0 +1,17 @@
num = 3;
tri_add = 2;
begin
tri_add += 1
num += tri_add
divisors = 2;
square_root = Math.sqrt(num);
div = 2
while(div <= square_root)
if(num % div == 0)
divisors += 2;
end
div += 1
end
end until (divisors > 500);
puts num;

@ -0,0 +1,19 @@
title: Find the first ten digits of the sum of one-hundred 50-digit numbers.
url: http://projecteuler.net/problem=13
desc: |
Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
(See file input)
solution: |
Use a full precission math lib
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

@ -0,0 +1,100 @@
37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690

@ -0,0 +1,7 @@
<?php
$numbers = file('php://stdin');
$total = 0;
foreach($numbers as $number) {
$total = bcadd($total,trim($number));
}
echo substr($total,0,10);

@ -0,0 +1,13 @@
numbers = []
linenum = 0
ARGF.lines("\n") do |line|
numbers[linenum] = Array.new
numbers[linenum] = line.to_i
linenum += 1
end
sum=0
numbers.each do |num|
sum += num
end
puts sum.to_s[0..9]

@ -0,0 +1,27 @@
title: Find the longest sequence using a starting number under one million.
url: http://projecteuler.net/problem=14
desc: |
The following iterative sequence is defined for the set of positive integers:
n -> n/2 (n is even)
n -> 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
solution: |
Bruteforce
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution
language: ruby
solve.c:
desc: ANSI C compiled with TCC
language: c

@ -0,0 +1,34 @@
#define MAX 1000000-1
#include <stdio.h>
#include <math.h>
int main( )
{
unsigned int max_steps = 0;
unsigned int max_start;
unsigned int test;
unsigned long ctest;
unsigned int steps;
for(test = MAX; test > 1 ; test--)
{
ctest = test;
steps = 1;
while(ctest != 1)
{
ctest = ( ctest % 2 ) ? ctest * 3 + 1 : ctest / 2;
steps++;
}
if(steps > max_steps)
{
max_start = test;
max_steps = steps;
}
}
printf( "%lu", max_start );
}

@ -0,0 +1,13 @@
<?php
$max_steps = 0;
for($test=999999;$test>1;$test--) {
$ctest = $test;
$steps = 1;
while($ctest != 1) {
$ctest = ($ctest % 2) ? $ctest*3+1 : $ctest/2;
$steps++;
}
if($steps > $max_steps) { $max_start = $test; $max_steps = $steps;}
}
echo $max_start;

@ -0,0 +1,18 @@
max_start = 0
max_steps = 0
test = 1
while(test < 1000000)
ctest = test
steps = 1
while(ctest != 1)
ctest = ( ctest % 2 != 0) ? ctest * 3 + 1 : ctest / 2
steps += 1
end
if(steps > max_steps)
max_start = test
max_steps = steps
end
test += 1
end
puts max_start

@ -0,0 +1,19 @@
title: Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?
url: http://projecteuler.net/problem=15
desc: |
Starting in the top left corner of a 22 grid, there are 6 routes (without backtracking) to the bottom right corner.
How many routes are there through a 2020 grid?
solution: |
(2n)!/n!^2 - Where n is the size of the grid -
todo: Find a more "programmable way" instead of this cheating mathsolution
solutions:
solve.php:
desc: Using BCMath to cope with large numbers
language: php
solve.rb:
desc: Basic solution
language: ruby

@ -0,0 +1,11 @@
<?php
define('GRID_SIZE',20);
function factorial($num) {
if($num == 0)
return 1;
else
return bcmul($num,factorial(bcsub($num,1)));
}
echo bcdiv(factorial(2* GRID_SIZE),bcpow(factorial( GRID_SIZE),2));

@ -0,0 +1,11 @@
GRID_SIZE = 20
def factorial(num)
if(num == 0)
return 1
else
return num * factorial(num - 1)
end
end
puts factorial(2 * GRID_SIZE) / factorial(GRID_SIZE) ** 2

@ -0,0 +1,17 @@
title: What is the sum of the digits of the number 2^1000?
url: http://projecteuler.net/problem=16
desc: |
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
solution: |
Bruteforce
solutions:
solve.php:
desc: Using BCMath to cope with large numbers
language: php
solve.rb:
desc: Basic solution
language: ruby

@ -0,0 +1,7 @@
<?php
$num = bcpow(2,1000);
$sum = 0;
for($c =0; $c < strlen($num);$c++) {
$sum += $num[$c];
}
echo $sum;

@ -0,0 +1,6 @@
num = 2 ** 1000
sum = 0
num.to_s.chars do |digit|
sum += digit.to_i
end
puts sum

@ -0,0 +1,15 @@
title: How many letters would be needed to write all the numbers in words from 1 to 1000?
url: http://projecteuler.net/problem=17
desc: |
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
solution: Just a matter of having a correct number-to-word function
todo: Make a clean implementation
solutions:
solve.php:
desc: Urggh - Solve using a lib from bas@startpunt.cc and corrected a small error so 210 whould include an "and" correctly
language: php

@ -0,0 +1,79 @@
<?php
function n2w_hunderds($number)
{
$test=$number*1;
if (empty($test))return;
$lasts=array('one','two','three','four','five','six','seven','eight','nine');
$teens=array('eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen');
$teen=array('ten','twenty','thirty','forty','fifty','sixty','seventy','eighty','ninety');
/* written by bas@startpunt.cc */
$string='';
$j=strlen($number);
$done=false;
for($i=0; $i<strlen($number); $i++)
{
if($j==2)
{
if(strlen($number)>2)
{
if($number[0]!=0)$string.= ' hundred ';
if($number % 100 != 0)$string.= 'and ';
}
if ($number[$i]==1)
{
if($number[$i+1]==0) $string.=$teen[$number[$i]-1];
else
{
$string.=$teens[$number[$i+1]-1];
$done=true;
}
}
else
{
if(!empty($teen[$number[$i]-1]))$string.=$teen[$number[$i]-1].' ';
}
}
elseif($number[$i]!=0 && !$done) $string.=$lasts[$number[$i]-1];
$j--;
}
return $string;
}
function n2w($number,$uk=0)
{
if(!is_string($number))$number.="";
if(!$uk)$many=array('', ' thousand ',' million ',' billion ',' trillion ');
else $many=array('', ' thousand ',' million ',' milliard ',' billion ');
$string='';
if(strlen($number)%3!=0)
{
$string.=n2w_hunderds(substr($number,0, strlen($number)%3 ));
$string.=$many[floor(strlen($number)/3)];
}
for($i=0; $i<floor(strlen($number)/3); $i++)
{
$string.=n2w_hunderds(substr($number,strlen($number)%3+($i*3),3));
if($number[strlen($number)%3+($i*3)]!=0)$string.=$many[floor(strlen($number)/3)-1-$i];
}
return $string;
}
$len = 0;
for($i = 1; $i <= 1000; $i++) {
$len += strlen(str_replace(' ','',n2w($i)));
}
echo $len;

@ -0,0 +1,25 @@
title: Find the maximum sum travelling from the top of the triangle to the base.
url: http://projecteuler.net/problem=18
desc: |
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below
(See file input)
solution: Instead of bruteforcing from top to bottom simply add the numbers from the bottom up and let the "highest" number win in a shootout
solutions:
solve.php:
desc: Basic solution
language: php
parameters: < ProjectEuler\018\input
solve.rb:
desc: Basic solution
language: ruby
parameters: < ProjectEuler\018\input

@ -0,0 +1,15 @@
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

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<?php
foreach(file('php://stdin') as $line) {
$tri[] = explode(' ',trim($line));
}
for($y=count($tri);$y>=0;$y--) {
for($x=0;$x<count($tri[$y]);$x++) {
$cur = $tri[$y][$x];
if($cur + $tri[$y+1][$x] > $cur + $tri[$y+1][$x+1]) {
$tri[$y][$x] = $cur + $tri[$y+1][$x];
} else {
$tri[$y][$x] = $cur + $tri[$y+1][$x+1];
}
}
}
echo $tri[0][0];

@ -0,0 +1,20 @@
tri = []
linenum = 0
ARGF.lines("\n") do |line|
tri[linenum] = Array.new
line.split.each do |str|
tri[linenum] << str.to_i
end
linenum += 1
end
(0..13).reverse_each do |y|
0.upto(tri[y].size()-1) do |x|
if( tri[y+1][x] > tri[y+1][x+1])
tri[y][x] = tri[y][x] + tri[y+1][x]
else
tri[y][x] = tri[y][x] + tri[y+1][x+1]
end
end
end
puts tri[0][0]

@ -0,0 +1,25 @@
title: How many Sundays fell on the first of the month during the twentieth century?
url: http://projecteuler.net/problem=19
desc: |
You are given the following information, but you may prefer to do some research for yourself.
1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
solution: Instead of using buildtin DateTime language features this use a homemade algorithm
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution
language: ruby

@ -0,0 +1,38 @@
<?php
$days_in_month = array(
1 => 31,
2 => 28,
3 => 31,
4 => 30,
5 => 31,
6 => 30,
7 => 31,
8 => 31,
9 => 30,
10 => 31,
11 => 30,
12 => 31
);
$day_cycle = 0;
for($year=1900; $year <= 2000; $year++) {
for($month=1; $month <= 12; $month++) {
$cdates = $days_in_month[$month];
if($month == 2) {
if($year % 4 == 0) {
$cdates = 29;
}
if($years % 100 == 0) {
$cdates = 28;
}
if($years % 400 == 0) {
$cdates = 29;
}
}
$day_cycle += $cdates;
if($day_cycle % 7 == 0) {
if($year >= 1901) {$sum++; }
}
}
}
echo $sum;

@ -0,0 +1,9 @@
require "date"
sundays = 0
(1901..2000).each do |year|
(1..12).each do |month|
dt = Date.new(year,month,1)
sundays += (dt.wday == 0) ? 1 : 0
end
end
puts sundays

@ -0,0 +1,18 @@
title: Find the sum of digits in 100!
url: http://projecteuler.net/problem=20
desc: |
n! means n * (n - 1) * ... 3 * 2 * 1
For example, 10! = 10 * 9 * ... * 3 * 2 * 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
solution: Bruteforce
solutions:
solve.php:
desc: Basic solution
language: php
solve.rb:
desc: Basic solution
language: ruby

@ -0,0 +1,15 @@
<?php
define('FACTORIAL',100);
function factorial($num) {
if($num == 0)
return 1;
else
return bcmul($num,factorial(bcsub($num,1)));
}
$sum = 0;
$fac = factorial(FACTORIAL);
for($c = 0; $c < strlen($fac); $c++) {
$sum += $fac[$c];
}
echo $sum;

@ -0,0 +1,16 @@
FACTORIAL = 100
def factorial(num)
if(num == 0)
return 1
else
return num * factorial(num - 1)
end
end
number = factorial(FACTORIAL)
sum = 0
number.to_s.each_char do |c|
sum += c.to_i
end
puts sum

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