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wu :: forums    riddles    cs (Moderators: SMQ, Icarus, Eigenray, ThudnBlunder, Grimbal, william wu, towr)    In three ways « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: In three ways  (Read 7891 times) Misha Kruk Guest In three ways   « on: Aug 1st, 2002, 8:50pm » Quote Modify Remove How well do you know C? "for" loops?   Find three ways to make the program below to print 20 copies of the dash character '-' by changing/adding one character:   int i, n=20; for (i=0; i < n; i--) {printf("-");};   Wrong solutions:       * Changing "n=20" into "n=-20" won't work.     * Changing "i--" into "i++" doesn't satisfy the one character condition.     * Changing "i=0" into "i=40" won't work.     * Changing "i < n" into "i < -n" wont' work.   IP Logged -D- Guest Re: In three ways   « Reply #1 on: Aug 1st, 2002, 10:24pm » Quote Modify Remove Well... since the initial code is obviously wrong.. the first answer would be to "fix" the code...     for( i = 0; -i < 20; i-- ) ...   IP Logged -D- Guest Re: In three ways   « Reply #2 on: Aug 1st, 2002, 10:26pm » Quote Modify Remove Answer #2... change    i--   to    n--   IP Logged AJ Guest Re: In three ways   « Reply #3 on: Aug 4th, 2002, 7:49am » Quote Modify Remove Third one:  for(i = 0 ; i + n ; i--)   Tricy one IP Logged -D- Guest Re: In three ways   « Reply #4 on: Aug 4th, 2002, 11:51am » Quote Modify Remove good one AJ.... that was tricky. -D- IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: In three ways   « Reply #5 on: Sep 22nd, 2003, 2:20am » Quote Modify This nice puzzle has a variation:   Find a way to make this program print 21 copies of '-'. The requirements are the same. IP Logged Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: In three ways   « Reply #6 on: Sep 22nd, 2003, 5:54am » Quote Modify Are we supposed to find 3 ways?   Here's one: int i, n=20;   for (i=0; i < -n; i--) {printf("-");}; IP Logged mathschallenge.net / projecteuler.net wowbagger Uberpuzzler     Gender: Posts: 727 Re: In three ways   « Reply #7 on: Sep 22nd, 2003, 6:20am » Quote Modify Sir Col, your solution doesn't work (as already noted in Misha Kruk's original post), because already the first test of i < -n is false, so there will be no "-" characters at all. IP Logged "You're a jerk, <your surname>!" Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: In three ways   « Reply #8 on: Sep 22nd, 2003, 8:53am » Quote Modify Duh, silly me. I was thinking that it looped until, rather than while.   In which case... hmm?   IP Logged mathschallenge.net / projecteuler.net James Fingas Uberpuzzler     Gender: Posts: 949 Re: In three ways   « Reply #9 on: Sep 22nd, 2003, 10:19am » Quote Modify Haha! Very very sneaky! The answer is too beautiful to give away, but here's what I was thinking about when I happened to stumble on the answer: could we use a bitwise operator instead of '+'? IP Logged Doc, I'm addicted to advice! What should I do? Sir Col Uberpuzzler impudens simia et macrologus profundus fabulae     Gender: Posts: 1825 Re: In three ways   « Reply #10 on: Sep 22nd, 2003, 2:05pm » Quote Modify Ooh, that is very sneaky. I'd better not give too much away.   Nice puzzle, Barukh. IP Logged mathschallenge.net / projecteuler.net Dudidu Full Member     Posts: 227 Re: In three ways   « Reply #11 on: Oct 8th, 2003, 11:00am » Quote Modify I'm sorry but I think that the answer should be posted so everybody could learn from it (if they want to). So...   for (i = 0 ; ~i < n ; i--)       IP Logged Niharika Jeena Guest Re: In three ways   « Reply #12 on: Apr 25th, 2005, 2:47am » Quote Modify Remove U can print it 21 times by adding "~" before i in condition..   for(i=0;~i<n;i--)   this will give the solution IP Logged d_ram22 Newbie Coder     Gender: Posts: 2 Re: In three ways   « Reply #13 on: Apr 7th, 2009, 8:36am » Quote Modify Good puzzle there the fixed questions is : Make following program print - 20 times by making one character change (add/modify).   #include<stdio.h> main() {   int i, n=20;     for (i=0; i < n; i++) {printf("-");}; }   Third Answer: hidden: #include<stdio.h> main() {   int i, n=20;     for (i=0; i ^ n; i++) {printf("-");}; } IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: In three ways   « Reply #14 on: Apr 7th, 2009, 12:17pm » Quote Modify on Apr 7th, 2009, 8:36am, d_ram22 wrote: Good puzzle there the fixed questions is : Make following program print - 20 times by making one character change (add/modify).   #include<stdio.h> main() {   int i, n=20;     for (i=0; i < n; i++) {printf("-");}; } Well that's silly.  For example, you could replace any of the spaces with a \t or a \n, or most of the \ns with spaces.  Or add/remove a space.  Or remove the last semicolon.  Or add a semicolon.  And of course you can also change i < n to i - n.   With one character change, what possible numbers of '-'s can you print? 0, 16, 20, 21, 40, infinity, ...? Though technically undefined, with gcc<3 or gcc-4 -O(0,1) (or gcc-4 -(fwrapv, fno-strict-overflow), etc.), there are also 2147483648 and 2147483649, if sizeof(int)==4.  And then there is undefined (but probably positive and finite on a 32-bit machine). IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft => cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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