Click here to Skip to main content
14,237,722 members
Rate this:
Please Sign up or sign in to vote.
See more:
i need a combination like this
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 
1 2 3 4 5 6   8 
1 2 3 4 5   7 8 
1 2 3 4   6 7 8 
1 2 3   5 6 7 8 
1 2   4 5 6 7 8 
1   3 4 5 6 7 8 

For next iteration 
1 2 3 4 5 6 7 
1 2 3 4 5 6     9
1 2 3 4 5   7   9
1 2 3 4   6 7   9
1 2 3   5 6 7   9
1 2   4 5 6 7   9
1   3 4 5 6 7   9

like wise till
1   3 4 5 6 7 8 
1   3 4 5 6 7   9
1   3 4 5 6   8 9
1   3 4 5   7 8 9
1   3 4   6 7 8 9
1   3   5 6 7 8 9
1     4 5 6 7 8 9


What I have tried:

I have tired but not getting the logic. Please help me.
Posted
Updated 7-Feb-17 21:59pm
Comments
Peter Leow 8-Feb-17 1:09am
   
"I have tired but not getting the logic" then show what you have tried.
OriginalGriff 8-Feb-17 1:59am
   
What have you tried?
Where are you stuck?
What help do you need?
Rate this:
Please Sign up or sign in to vote.

Solution 1

Look at the digits arranged as a square
(1,1)                      (1,9)
  \                         /
   1  2  3  4  5  6  7  8  9
   1  2  3  4  5  6  7  8  9
   1  2  3  4  5  6  7  8  9
   1  2  3  4  5  6  7  8  9
   1  2  3  4  5  6  7  8  9
   1  2  3  4  5  6  7  8  9
   1  2  3  4  5  6  7  8  9
   1  2  3  4  5  6  7  8  9
   1  2  3  4  5  6  7  8  9
  /                         \ 
(9,1)                      (9,9)


Then at iteration k you must show all the square items, but:
  • The ones of column k, that is (i,k).
  • The ones on the diagonal, that is (k, 10-k).
   
Rate this:
Please Sign up or sign in to vote.

Solution 2

Quote:
I have tired but not getting the logic. Please help me.
First help yourself with method.
You have noted that
You don't understand the logic:
- try to draw by hand the result wanted for each iteration in order.
- try to analyze the differences between each iteration an note what t(he changes are.
By this point, you should have got the logic, otherwise, talk to your teacher.

are you sure about the 9 in first iteration ?
   
Rate this:
Please Sign up or sign in to vote.

Solution 3

#include<stdio.h>

int main()
{
	int a[9]={1,2,3,4,5,6,7,8,9};
	int i,j=9,k;
	
	while(j>0)
	{
		for(i=1;i<8;i++)
		{
			for(k=0;k<9;k++)
			{
				if(k!=i&&k!=j)
			{
				printf("%d",a[k]);
				
			}
			else
			{
				printf(" ");
				continue;
			}
			}
			printf("\n");	
		}
		printf("\n");
		j--;
	}
	return 0;
}
   
Comments
Patrice T 8-Feb-17 4:08am
   
A question and a solution without a single sentence!
Is it your solution or code that don't work ?
R!sh! 8-Feb-17 4:14am
   
my solution. :)
Patrice T 8-Feb-17 4:17am
   
So, use Accept answer (near "Solution 3") to signal it's solved and to close the question.
R!sh! 17-Feb-17 0:18am
   
can you give your answer please??
R!sh! 8-Feb-17 4:14am
   
finally i did it. :D
Patrice T 8-Feb-17 11:35am
   
I see that your program do not match the example of the question, first iteration is not the same (the 9 on first line).
R!sh! 16-Feb-17 23:55pm
   
it is coming in reverse order.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)




CodeProject, 503-250 Ferrand Drive Toronto Ontario, M3C 3G8 Canada +1 416-849-8900 x 100