The Project Euler thread

[axith]

Re: The Project Euler thread

for all it's worth, here's my friend key: 83532459359302_e704d4c6b1d37a7739410f38da156b2a

[axith]

Re: The Project Euler thread

So which keywords make python read from a text file? My eyes are buggy from looking though online reference stuff and there's several problems that require it..

[emerald000]

Re: The Project Euler thread

You will want to use:

file = open('/path/to/file')

file.read()       # Returns whole file.
file.readline()   # Will return line by line.
file.readlines()  # Returns a list of every line.

file.close()      # Close the file after you have read everything you need.

[kaiten123]

Re: The Project Euler thread

just did 96 which was cool since i never really used recursion in a way that was non-trivial before
no idea how you guys do some of the higher numbered ones, highest one i did was 243, which for some reason is many times easier than everything around it, didnt even write any code.

edit: got to lvl 3 :)

[kaiten123]

Re: The Project Euler thread

brute forced 407 lol
never saw that box at the bottom right before, is that new? or does it pop up when you solve something recent?

[leonid]

Re: The Project Euler thread

i should get back to solve rest of the problems

[Arkuski]

Re: The Project Euler thread

This is pretty awesome I'm giving some of these a go

[Reincarnate]

Re: The Project Euler thread

411 wasn't too bad -- pretty much an exercise in programming

[Reincarnate]

Re: The Project Euler thread

415... only 3 solvers after 7 and a half hours.

Yeah I can pretty much kiss my 100% goodbye

It's been fun, PE :(

[MracY]

Re: The Project Euler thread

2 mod 3 is 2

(x mod y is the remainder of x / y)

PS use Python instead, it's cooler

sumTotal = 0
num = 1
while num < 1000:
if (num % 3 == 0) or (num % 5 == 0):
sumTotal = sumTotal + num
num = num + 1
print sumTotal

Damn thing doesn't save formatting but w/e

Just as a note to people who want to join Project Euler: this solution is not necessarily representative of what you will be doing.

Here are some of my crude notes for problem 1, that used logic in stead of programming:

vertical axis is m
horizontal axis is n


Sum of the series of 5 is 0.5nm+0.5m
Sum of the series of 3 is 0.5pq+0.5q

1000/5 = 200.
The last value here is 1000, so n = 199
m = 1000-5 = 995

1000/3 = 333 with 1 remainder, so p = 333
q = 1000-1 = 999

Sum5 = 0.5*199*995+0.5*995
= 99500

Sum3 = 0.5*333*999+0.5*999
= 167166

99500+166833=266333



There is overlap between the sums of multiples of 15.
1000/15 is 66 with 10 remainder, so x = 66
y = 1000-10 = 990

Sum of this series is 0.5xy+0.5y
= 0.5*66*990+0.5*990 = 33165

So the answer is: 99500+166833-33165 = 233168

[leonid]

Re: The Project Euler thread

@Reincarnate use code tag

while 1:
  print "INDENT"
  if "INDENT":
    print "MORE INDENT"
print "NEVER REACHES HERE"

[benguino]

Re: The Project Euler thread

Maybe ill try to play around and do some more problems this weekend :) (I haven't touched PE in awhile) . gotta start on my take-home test and other school stuff first though xP

[benguino]

Re: The Project Euler thread

Started doing this in my free time again :3 Just got problem 34 this morning. Problem and code below; any tips that can make me a better programmer or mathematician is greatly appreciated :P

problem:

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.

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: (in Java)

//this will be our final answer
		int probSum = 0;
		
		//at some point, the sum of the factorials of the digits won't be large enough to add up to the number
		//so choose our digits to be 9 and find when we have too many digits
		//our upper bound for curious numbers becomes k
		int i = 1;
		int nFact = factorial(9);
		int k = 9;
		while(i*nFact > k)
		{
				k += 9*Math.pow(10, i);
				i++;
		}
		
		for(int j=10; j <=k ; j++)
		{
			String[] digits = (j+"").split("");
			
			int sum = 0;
			for(int l = 1; l < digits.length; l++)
			{
				sum += factorial(Integer.parseInt(digits[l]));
			}
			
			if(j==sum)
			{
				probSum += j;
			}
		}
		
		System.out.println(probSum);

EDIT: cool, got another :3


Problem:

Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 53 = 49714.

What is the total of all the name scores in the file?

Solution:

int totalScore = 0;
		
		File file = new File("names.txt");
		Scanner scan = null;
		try 
		{
			scan = new Scanner(file);
		} 
		catch (FileNotFoundException e) {}
		String nameString = scan.next();
		String[] names = nameString.split("\",\"");
		ArrayList<String> names2 = new ArrayList<String>();
		for(int i = 0; i < names.length; i++)
		{
			names2.add(names[i]);
		}
		
		Collections.sort(names2);
		for(int i = 0; i < names2.size(); i++)
		{
			char[] chars = names2.get(i).toCharArray();
			int charSum = 0;
			for(int j = 0; j < chars.length; j++)
			{
				charSum += chars[j] - 'A' + 1;
			}
			totalScore += charSum*(i+1);
		}
		System.out.println(totalScore);

[Arkuski]

Re: The Project Euler thread

Learned some jython, and these are all much easier

[Choofers]

Re: The Project Euler thread

I'm currently on my lunch break, so I decided that instead of studying for my ethernet and osi model exam tomorrow, I'd delve into powershell.

Had a bit of an issue getting custom scripts to work without having digital signatures, but oh well.

$count = 0
$total = 0
do{if (($count % 3 -eq 0) -or ($count % 5 -eq 0))
  {
  $total = $total + $count
  }
  else
  {
  $total = $total
  }
$count++
}
while ($count -lt 1000)
echo $total

this feels very reminiscent of java and maybe you DuDeZ will stop hating me because I'm not using BASIC

maybe ???


edit: problem two solved, I was originally incrementing my variables one number at a time but I quickly realized that this would need a variable for every number in the sequence lol

$f0 = 0
$f1 = 1
$fsum = 0
do{
  if ($f1 % 2 -eq 0)
  {
    $fsum = $fsum + $f1
  }
  if ($f0 % 2 -eq 0)
  {
    $fsum = $fsum + $f0
  }
  $f0 = $f0 + $f1
  $f1 = $f1 + $f0
  }
while ($f0 -lt 4000000)
echo $fsum