Added a few generic problems\nSolved Euler027
This commit is contained in:
FuryFire
53
CodeGolf/guns-and-zips/input.1
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53
CodeGolf/guns-and-zips/input.1
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CodeGolf/guns-and-zips/output.1
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CodeGolf/guns-and-zips/output.1
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?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
0
CodeGolf/guns-and-zips/solve.php
Normal file
BIN
Generic/Fibonacci/a.exe
Normal file
BIN
Generic/Fibonacci/a.exe
Normal file
Binary file not shown.
51
Generic/Fibonacci/solve.c
Normal file
51
Generic/Fibonacci/solve.c
Normal file
@ -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;
|
||||
}
|
||||
BIN
Generic/Fibonacci/solve.exe
Normal file
BIN
Generic/Fibonacci/solve.exe
Normal file
Binary file not shown.
1
Generic/HelloWorld/solve.js
Normal file
1
Generic/HelloWorld/solve.js
Normal file
@ -0,0 +1 @@
|
||||
console.log("Hello World");
|
||||
20
Generic/MultiTable/solve.c
Normal file
20
Generic/MultiTable/solve.c
Normal file
@ -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
10
Generic/Seive/desc.yml
Normal file
@ -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
22
Generic/Seive/solve.c
Normal file
@ -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);
|
||||
}
|
||||
}
|
||||
}
|
||||
23
ProjectEuler/026/desc.yml
Normal file
23
ProjectEuler/026/desc.yml
Normal file
@ -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
20
ProjectEuler/026/solve.php
Normal file
@ -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 $max_div;
|
||||
13
ProjectEuler/026/solve.rb
Normal file
13
ProjectEuler/026/solve.rb
Normal file
@ -0,0 +1,13 @@
|
||||
DIGITS = 1000
|
||||
|
||||
max = 10 ** (DIGITS-1)
|
||||
fcurrent = 1
|
||||
fprev = 1
|
||||
term = 2
|
||||
while fcurrent < max do
|
||||
term = term + 1
|
||||
fnext = fcurrent + fprev
|
||||
fprev = fcurrent;
|
||||
fcurrent = fnext;
|
||||
end
|
||||
puts term
|
||||
20
ProjectEuler/027/desc.yml
Normal file
20
ProjectEuler/027/desc.yml
Normal file
@ -0,0 +1,20 @@
|
||||
title: Quadratic primes
|
||||
url: http://projecteuler.net/problem=27
|
||||
|
||||
desc: |
|
||||
Euler discovered the remarkable quadratic formula:
|
||||
n² + n + 41
|
||||
It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.
|
||||
The incredible formula n² - 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.
|
||||
Considering quadratics of the form:
|
||||
n² + an + b, where |a| < 1000 and |b| < 1000
|
||||
where |n| is the modulus/absolute value of n
|
||||
e.g. |11| = 11 and |-4| = 4
|
||||
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0
|
||||
|
||||
solution: Bruteforce
|
||||
|
||||
solutions:
|
||||
solve.php:
|
||||
desc: Basic Solution - needs BCMath
|
||||
language: php
|
||||
32
ProjectEuler/027/solve.php
Normal file
32
ProjectEuler/027/solve.php
Normal file
@ -0,0 +1,32 @@
|
||||
<?php
|
||||
function is_prime($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;
|
||||
}
|
||||
$n++;
|
||||
}
|
||||
|
||||
}
|
||||
}
|
||||
13
ProjectEuler/027/solve.rb
Normal file
13
ProjectEuler/027/solve.rb
Normal file
@ -0,0 +1,13 @@
|
||||
DIGITS = 1000
|
||||
|
||||
max = 10 ** (DIGITS-1)
|
||||
fcurrent = 1
|
||||
fprev = 1
|
||||
term = 2
|
||||
while fcurrent < max do
|
||||
term = term + 1
|
||||
fnext = fcurrent + fprev
|
||||
fprev = fcurrent;
|
||||
fcurrent = fnext;
|
||||
end
|
||||
puts term
|
||||
5
nbproject/private/private.properties
Normal file
5
nbproject/private/private.properties
Normal file
@ -0,0 +1,5 @@
|
||||
copy.src.files=false
|
||||
copy.src.target=
|
||||
index.file=
|
||||
run.as=SCRIPT
|
||||
url=http://localhost/codingtests/
|
||||
7
nbproject/project.properties
Normal file
7
nbproject/project.properties
Normal file
@ -0,0 +1,7 @@
|
||||
include.path=${php.global.include.path}
|
||||
php.version=PHP_54
|
||||
source.encoding=UTF-8
|
||||
src.dir=.
|
||||
tags.asp=false
|
||||
tags.short=true
|
||||
web.root=.
|
||||
9
nbproject/project.xml
Normal file
9
nbproject/project.xml
Normal file
@ -0,0 +1,9 @@
|
||||
<?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>
|
||||
Reference in New Issue
Block a user