adjacency matrix!

input:
1 2
1 3

Desired O/P:
0 1 1
1 0 0
1 0 0

Here 1 and 2 are linked thus 2 and 1 are linked too, so the value of matrix at 01 and 02 is 1
Same goes for 1 and 3

Code:

#include<stdio.h>
#include<conio.h>

main()
{
    int a[2][2],i,j;
    for(i=0;i<2;i++)
    for(j=0;j<2;j++)
    scanf("%d",&a[i][j]);
}

Can you please add the code for desired output?
thanks

yea let me get right on that.

Originally Posted by enakta13

thanks

See my // comments inside your code

Code:

#include<stdio.h>
#include<conio.h> // You don't use anything declared outside of <stdio.h>
                  // Therefore you don't need to depend on a
                  // non-Standard header for your program
                  // I suggest you remove the dependency on <conio.h>

main()
// main should be defined as returning int
{
    int a[2][2],i,j;
    for(i=0;i<2;i++)
    for(j=0;j<2;j++)
    scanf("%d",&a[i][j]);
    // you REALLY, REALLY, REALLY should test the return value of scanf()
    // anyway ... scanf is a bad choice for user input. It's way better
    // to fgets() and sscanf() or strtol()
}
// Looking so so up to this point. Keep it up

Have fun.

Originally Posted by qny

See my // comments inside your code

Code:

#include<stdio.h>
#include<conio.h> // You don't use anything declared outside of <stdio.h>
                  // Therefore you don't need to depend on a
                  // non-Standard header for your program
                  // I suggest you remove the dependency on <conio.h>

main()
// main should be defined as returning int
{
    int a[2][2],i,j;
    for(i=0;i<2;i++)
    for(j=0;j<2;j++)
    scanf("%d",&a[i][j]);
    // you REALLY, REALLY, REALLY should test the return value of scanf()
    // anyway ... scanf is a bad choice for user input. It's way better
    // to fgets() and sscanf() or strtol()
}
// Looking so so up to this point. Keep it up

Have fun.

But where is the code to get the output?

Originally Posted by enakta13

But where is the code to get the output?

That's the question we should be asking you. No one will hand you a solution. You need to attempt it on your own, and we'll be here to answer questions if you get stuck along the way.