Row/Col :S

BaldPython · 2005-09-03T12:38:04

I actually wanted to make this shape in C++: 00*00 0***0 ***** 0***0 00*00 For this purpose I first defined the number of rows and number of cols above. Then, I made it with five loops. I know it's not a good approach. I tried hard to find a connection between row and column to confine it in two loops but I couldnt make one succesful thing. Can anyone help? I also want to practice on such patterns, can anyone suggest a site where I can find tutorials on it.

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Started by BaldPython September 03, 2005 06:23 AM

11 comments, last by BaldPython 18 years, 7 months ago

BaldPython

100

Author

September 03, 2005 12:19 PM

Quote:Original post by xMcBaiNx
I' sorry, you need to start counting the rows from 0 to n-1. So if there are 5 rows, they will be labeled 0 to 4. The absolute values takes care of the problem, see the example below for 5 rows.

spacesBeforeStarsRow1 = |(0 - (5-1)/2)| = 2
spacesBeforeStarsRow2 = |(1 - (5-1)/2)| = 1
spacesBeforeStarsRow3 = |(2 - (5-1)/2)| = 0
spacesBeforeStarsRow4 = |(3 - (5-1)/2)| = 1
spacesBeforeStarsRow5 = |(4 - (5-1)/2)| = 2

(Note that the loops have to start from 0 now.)

Sir, I've been saying that it would require four loops, one to move from 0 to N-1. Second to print spaces before Stars, third to print asteriks and fourth to print spaces after stars.

I was looking for an algorithm/pseudocode/hint-program to solve it in two loops.

jfclavette

1,058

September 03, 2005 12:26 PM

Quote:Original post by BaldPython
Quote:Original post by xMcBaiNx
I' sorry, you need to start counting the rows from 0 to n-1. So if there are 5 rows, they will be labeled 0 to 4. The absolute values takes care of the problem, see the example below for 5 rows.

spacesBeforeStarsRow1 = |(0 - (5-1)/2)| = 2
spacesBeforeStarsRow2 = |(1 - (5-1)/2)| = 1
spacesBeforeStarsRow3 = |(2 - (5-1)/2)| = 0
spacesBeforeStarsRow4 = |(3 - (5-1)/2)| = 1
spacesBeforeStarsRow5 = |(4 - (5-1)/2)| = 2

(Note that the loops have to start from 0 now.)

Sir, I've been saying that it would require four loops, one to move from 0 to N-1. Second to print spaces before Stars, third to print asteriks and fourth to print spaces after stars.

I was looking for an algorithm/pseudocode/hint-program to solve it in two loops.

And I've been trying to help you see the light. The "hint" program is ready to be compiled in the post before your last one.

I teleported home one night; With Ron and Sid and Meg; Ron stole Meggie's heart away; And I got Sydney's leg. <> I'm blogging, emo style

BaldPython

100

Author

September 03, 2005 12:38 PM

Quote:Original post by xMcBaiNx
*** Source Snippet Removed ***

Thank you sir, this is what I came up with:

#include <iostream>#include <cmath>int main() {	      int spacesBeforeStars=0;      int spacesAfterStars=0;      int numberOfStars=0;      int sizeOfDiamond=0;        std::cout << "Enter the size of the diamond: ";    std::cin  >> sizeOfDiamond;	      for(int x=0; x<sizeOfDiamond; x++) {       spacesBeforeStars = abs((x - (sizeOfDiamond-1)/2));       spacesAfterStars  = spacesBeforeStars;       numberOfStars	 = sizeOfDiamond - (spacesBeforeStars*2);            for(int y=0; y<sizeOfDiamond; y++) {        (y>=spacesBeforeStars && y<=sizeOfDiamond-spacesAfterStars-1) ? std::cout << "X" : std::cout << " ";      }      std::cout << std::endl;    }    return 1;}

Row/Col :S

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Row/Col :S

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