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this is about as easy as it gets......I guess :)Quote:
Originally posted by Sepero
Take the sentence:
"She sells sea shells by the sea shore"
Reorder the words alphabetically, and do it in your favorite language.
;)
Code:#!/usr/bin/perl
@array = ("she", "sells", "sea", "shells", "by", "the", "sea", "shore");
@array2 = sort (@array);
print ("@array2\n");
Just a comment, if I might. Being new to Linux and all that, I am offering a perspective consistent with the burning desire to learn anything!
This stuff is incredible! I notice that the thread originally started almost two years ago, and yet I've just now seen it, only because someone chanced to resurrect it, and back it came to the front.
My humble wish...that this concept rates its' own stable in the barn. A unique category. It's beautiful stuff. It's like having a window and a seat next to a bunch of ace coders, and being able to look over their shoulders and get a sense of what coding can be. It's freewheeling, and raucus, and just one heck of an education.
Please...give this concept real status!
Are there more than two of these? Just checking to see if i'm close...Quote:
Originally posted by janet loves bill
I found this thread again, thought it was lost..........
here is a challenge I did a few years ago while taking a C/C++ class and figured I'd put it out and see how many different ways it could be done (perl, VB, Bash script??, etc.).........have fun
Oh, and can't we make this sticky? I'm sure it's one of everyone's favorites in this forum. :)
Quote:
Originally posted by Sepero
Take the sentence:
"She sells sea shells by the sea shore"
Reorder the words alphabetically, and do it in your favorite language.
;)
How about:
echo 'She sells sea shells by the sea shore' | sed -e 's/[[:space:]]/\n/g' | sort
:)
or:
Bearing in mind that it's a case-sensitive sort....Code:Python 2.3.5 (#2, May 4 2005, 08:51:39)
[GCC 3.3.5 (Debian 1:3.3.5-12)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> mystring = "She sells seas shells on the sea shore"
>>> list = mystring.split(' ')
>>> list
['She', 'sells', 'seas', 'shells', 'on', 'the', 'sea', 'shore']
>>> list.sort()
>>> print list
['She', 'on', 'sea', 'seas', 'sells', 'shells', 'shore', 'the']
>>>
Haha or ruby
irb(main):001:0> "She sells seas shells on the sea shore".split.sort
=> ["She", "on", "sea", "seas", "sells", "shells", "shore", "the"]
Quote:
Find a six digit number that gives its digits reversed when multiplied by an integer between 2 and 9 inclusive.
Output:Code:(100000..999999).each { | x |
(2..9).each { |y |
if x.to_s == (x*y).to_s.reverse then
print "#{x} * #{y} = #{x*y}\n"
end
}
}
109989 * 9 = 989901
219978 * 4 = 879912
This is fun :)
Now might be a good time to mention http://www.pythonchallenge.com
Actually, you chose a wonderful time to mention this link. Thank you much for doing so.Quote:
Perl6Quote:
Originally posted by Sepero
Take the sentence:
"She sells sea shells by the sea shore"
Reorder the words alphabetically, and do it in your favorite language.
;)
Code:"She sells sea shells by the sea shore".split.sort.say;
## disregard case
"She sells sea shells by the sea shore".split.map:{ lc $_ }.sort.say;
# or
"She sells sea shells by the sea shore".split.sort:{ lc $^a cmp lc $^b }.say;
Find a six digit number that gives its digits reversed when multiplied by an integer between 2 and 9 inclusive.
Gotta keep the C++ people alive:
Code:int main() {
for (int i = 100000; i < 1000000; ++i) {
for (int j = 2; j < 10; ++j) {
int reverse((i % 10) * 100000 + ((i / 10) % 10) * 10000
+ ((i / 100) % 10) * 1000 + ((i / 1000) % 10) * 100
+ ((i / 10000) % 10) * 10 + i / 100000);
if (i * j == reverse) {
cout << i << " + " << j << " = " << reverse << endl;
}
}
}
}
Here's one I stumbled across:
Consider a character set consisting of letters, a space, and a point. Words consist of one or more, but at most 20 letters. An input text consists of one or more words separated from each other by one or more spaces and terminated by 0 or more spaces followed by a point. Input should be read from, and including, the first letter of the first word, up to and including the terminating point. The output text is to be produced such that successive words are separated by a single space with the last word being terminated by a single point. Odd words are copied in reverse order while even words are merely echoed. For example, the input string
: whats the matter with kansas.
becomes
: whats eht matter htiw kansas.
Easy, but easy ones are fun once in a while too.
You forgot to reverse 'whats' in your example....
Here is a Python solution - it's a bit clunky, but it works, and it only took a short while to code.
And in action:Code:#!/usr/bin/python
def isEven(number):
if divmod(number, 2)[1] == 0:
return True
else:
return False
def reverse(mystring):
output = ""
for loop in range(len(mystring)-1, -1, -1):
output += mystring[loop]
return output
input = "whats the matter with kansas. I like pies. Chips are good too."
data = input.split(' ')
output = ""
for word in data:
point = False
if word[-1] == ".":
point = False
if word[-1] == ".":
word = word[:-1]
point = True
if isEven(len(word)):
output += word
else:
output += reverse(word)
if point:
output += ". "
else:
output += " "
print output
Code:mrben@hobbes:~/temp/script$ python words.py
stahw eht matter with kansas. I like pies. spihC era good oot.
mrben@hobbes:~/temp/script$
The problem is to reverse every other word. like
whats eht matter htiw kansas. I like seip. Chips era good oot.
Doh! I read it as reversing every word that had an odd number of letters......
Amended script:
Output, using 2 sentences with odd and even numbers of words:Code:#!/usr/bin/python
def isEven(number):
if divmod(number, 2)[1] == 0:
return True
else:
return False
def reverse(mystring):
output = ""
for loop in range(len(mystring)-1, -1, -1):
output += mystring[loop]
return output
input = "whats the matter with kansas. I like pies. Chips are good too."
data = input.split(' ')
output = ""
count = len(data)
if not isEven(count):
count -= 1
for index in range(0, count, 2):
output += data[index]+" "
if not data[index+1][-1] == ".":
output += reverse(data[index+1])+" "
else:
output += reverse(data[index+1][:-1])+". "
if not isEven(len(data)):
output += data[-1]
print output
Code:mrben@hobbes:~/temp/script$ python words2.py
whats eht matter htiw kansas. I like seip. Chips era good oot.
mrben@hobbes:~/temp/script$
mrben@hobbes:~/temp/script$ python words2.py
whats eht matter htiw kansas. I like seip. Chips era really doog too.
mrben@hobbes:~/temp/script$
there are at least 3 ways this can be written in perl....Quote:
Originally posted by janet loves bill
Find a six digit number that gives its digits reversed when multiplied by an integer between 2 and 9 inclusive.
here is one............:)Code:#!/usr/bin/perl -w
for($number=1e5; $number <= 5e5; $number++){
for($num=2; $num <= 9; $num++){
if($number*$num==reverse($number)){
print("$number times $num\n");}}}
exit;
A slightly cheating version in ruby :D . Doesn't take into account that there might be more then one space between words.Quote:
Consider a character set consisting of letters, a space, and a point. Words consist of one or more, but at most 20 letters. An input text consists of one or more words separated from each other by one or more spaces and terminated by 0 or more spaces followed by a point. Input should be read from, and including, the first letter of the first word, up to and including the terminating point. The output text is to be produced such that successive words are separated by a single space with the last word being terminated by a single point. Odd words are copied in reverse order while even words are merely echoed. For example, the input string
Output:Code:#!/usr/bin/ruby
sentences = 'whats the matter with kansas. I like pies. Chips are good too. '.split('. ');
reverse = true
sentences.each { | sentence |
words = sentence.split(' ')
(0..words.length).each { | i |
word = words[i]
next if word == nil
word.reverse! if reverse = !reverse
print word + (i == words.length - 1 ? '. ' : ' ')
}
}
print "\n"
whats eht matter htiw kansas. I like seip. Chips era good oot.
How about having a function which takes two vars, an unsigned char pointer and an integer. The integer tells you what bit to flip on the unsigned char the pointer points to. Not a hard one, but a little fun challenge non-the-less.
Here's my little function :), assuming the integer 0 is the least significant bit in the unsigned char.Quote:
How about having a function which takes two vars, an unsigned char pointer and an integer. The integer tells you what bit to flip on the unsigned char the pointer points to. Not a hard one, but a little fun challenge non-the-less.
Code:void flip (unsigned char *c, int i)
{
unsigned char p = (unsigned char)pow(2.0, (double)i);
*c = *c ^ p;
}
*involuntary shudder*Quote:
Originally Posted by tony_yum
:p
Instead of using a floating point math function, here's what I'd do:
;)Code:unsigned char p = (unsigned char)(1 << i);
Oh yeah :) that's for the correctionQuote:
unsigned char p = (unsigned char)(1 << i);
write a program that can access protected memory without help from the linux kernel. :-)
Alternately:Code:fd = open("/dev/mem", O_RDWR);
/* do stuff with fd */
since all memory is "protected" as far as the Intel docs go. :pCode:char *ptr = malloc(however_much_you_need);
/* do stuff with ptr */
But to be less pedantic, there's no other way to do it that I know of. Because while you do know where the paging tables are held in physical memory (you can get this from the PDBR, which is register CR3 on x86, IIRC, which user-mode code does have read access to), you don't know where in linear memory this physical address is mapped.
If you did know that piece of information, you have access to the entire 4GB linear address space, so you could maybe write to those addresses. (The page table must be mapped in, otherwise the processor wouldn't be able to do paging. So it's available somewhere, you just don't know where.) But those pages may also be marked supervisor-only (the U/S bit in the page table), which would mean writing to them would cause a segfault.
I tried my whack at a memory-editor for linux in C once, only to find out what "Protected Memory" meant :-pQuote:
Originally Posted by bwkaz
SegFaults are a pain in my
It's easy !Quote:
Originally Posted by grady
Code:void swap(int *_a, int *_b)
{
*_a=*_a+*_b;
*_b=*_a-*_b;
*_a=*_a-*_b;
}
As long as *_a + *_b <= INT_MAX, anyway... ;)
Er, well, wait, maybe it would still work.
...
...
Yeah, it would, as long as *_a and *_b are both positive. If they're negative, you can underflow and lose a bit off the top, but otherwise I think you're safe.
Using a single recursive loop, Print this:
vroooom racecar vroooom
You can use a string or array, but it should be printed one character at a time.
:)
is there any ideea in inverse(rac)=car and vroooom appearing in both ends ?
Whew! I followed all the rules, right? A single recursive loop and print a single char at a time? :D
Code:#include <stdio.h>
#include <string.h>
char str[] = "vroooom racecar vroooom";
void printStr(int pos) {
if (pos == strlen(str)) {
return;
}
printf("%c", str[pos]);
printStr(pos + 1);
}
int main() {
printStr(0);
}
LOL, yeah. Now this time, only use the string "vroooom race".Quote:
Originally Posted by tecknophreak
EDIT:
Good job technophreak. I guess I didn't make myself understood as well as I thought.
I'll try again. :)
Using:
One recursive function(/method), and the string(/array) "vroooom race"
Print:
"vroooom racecar vroooom"
this took about 10minutes
i hope you like it
Code:#include <stdio.h>
#include <string.h>
char s[] = "vroooom race";
int t=0;
int h=0;
void f(char t)
{printf("%c", t);}
void H(int p) {
if(s[p]!=0)
{
t++;
f(s[p]);
H(p+1);
h++;
if(t==1){
if(h<=5)
f(s[p]);
else
f(s[strlen(s)-p-6]);
}
else t=1;
}
}
int main() {
H(0);
scanf("%d",1);
}
Nice spx2. I hadn't expected a loop like that. There is definitely more than one way to do it. :)
well,im really good with algorithms...atleast i was
i'm retired now from contests and such.
if you like it thank you.
is there any better way to do it ?
If I may take that as a personal challenge... ;)Quote:
Originally Posted by spx2
Though, it can still be done differently. The original way that I was thinking, was to somehow switch the "vr" and "m" during the second half of recursion.Code:#include <stdio.h>
#include <string.h>
char s[] = "vroooom race";
void f(char t)
{ printf("%c", t); }
void H(int p)
{
if(s[p]!=0)
{
f(s[p]); // print "vroooom race"
H(p+1);
if(p<11 && p>6)
f(s[p]); // print "car "
else if(p<=6)
f(s[6-p]); // print "vroooom"
}
}
int main()
{
H(0);
printf("\n");
}
we shall leave that to other people(the optimization you talk about)
Yeah, I kinda wanna do it myself, but at the same time, I don't wanna spoil the fun for anyone else. ;)
Here is my solution with reversing vrooom.
Code:#include<stdio.h>
#include<string.h>
char str[] = "vrooom race";
void invVrooom () {
char str2 [] = "mooorv";
for(int x = 0; x < 6; x++) { str[x] = str2[x]; }
return;
}
void printString(int pos) {
printf("%c",str[pos]);
if(str[pos] == 'e') {
invVrooom();
return;
} else {
printString(pos+1);
}
printf("%c",str[pos]);
return;
}
int main (void) {
printString(0);
printf("\n");
return 0;
}
Is anybody else reminded of the Ralph S. Mouse books by this challenge? Teachers used to read them to us in like 2nd grade and there was a mouse who had a toy motorcycle that he could drive by saying "vrooom", and then one day he got stuck in a corner and couldn't turn around, so to back up he said "mooorv".
Funny that I remember that this many years later...:)
Ok, here's one. You know a quote has a ceaser(sp?) cipher on it, you just don't know what is is. You have to run a program/script and have the program figure out what the correct cipher is for the quote. The program will have two outputs the cipher, i.e. Shift +24, and the answer.
Here's a fun quote: zypr: "ufyr'q qylry'q jgrrjc fcjncp bmgle rm rfyr bme? jmmiq jgic fc'q rpwgle rm hskn mtcp, zsr fc ayl'r osgrc kyic gr."
I'd really like to see the script for this one since I'm sure it's quicker/easier than c/c++.