Pattern problem in C programming language-CodePudding

I have the following pattern to print in the C programming language.

BBBB
BAAB
BAAB
BBBB

I have tried the following code, but it's not working.

My code :

#include <stdio.h>

int main() {
    // Write C code here
    int i,j;
    for(i=1;i<=4;i  )
        {
            for(j=1;j<=4;j  )
                {
                    if ((i==1&&j>=i)||(i==4&&j<=i)){
                        printf("%c",65 1);
                    }
                        
                
                    else{
                        printf("%c",65);
                }        
            }
        printf("\n");            
        }        
    return 0;
}

However, the pattern I am getting is the following.

BBBB
AAAA
AAAA
BBBB

CodePudding user response：

The problem with your code is your special case will only fire on the first and fourth row. We can see this a little better if we space things out.

if( 
  (i==1 && j>=i) ||  // `i==1` only on the first row
  (i==4 && j<=i)     // `i==4` only on the fourth row
) {
  printf("%c",65 1);
}

Every other iteration will use your else block that just prints A.

else {
  printf("%c",65);
}

Note: 'A' is much easier to read than 65, and 'B' much easier than 65 1.

There's plenty of ways to do this. Here's one elegant way with a single loop.

We can observe that we want to print BBBB at the start and every third row. If we start iterating at 0 we can do this when i is divisible by 3. 0/3 has a remainder of 0. 3/3 has a remainder of 0. We use the modulus operator % to get the remainder.

for(int i = 0; i < 4; i  ) {
  if( i % 3 == 0 ) {
    puts("BBBB");
  }
  else {
    puts("BAAB");
  }
}

This will continue to repeat the pattern if you extend the loop. 6/3 has a remainder of 0. 9/3 has a remainder of 0. And so on.

(You could also start with i=1 and check i%3 == 1, but get used to starting counting at 0; it makes a lot of things easier.)