这里的模式是什么？ 我怎样才能在 c 中编写该代码？

Question

But I have issues. I don't understand the pattern very well. I don't know how to make an even number part. I have some guess. This pattern has 2 parts and I don't know how to make this in c. What I've been able to do so far:

#include <stdio.h>
#include <stdlib.h>
    
int main(int argc, char *argv[])
{
    int a,n;

    printf("Enter a number:");
    scanf("%d",&a),

    n=0;
    while(n<=a)    
    {
        printf("%d|\n",n);
        n++;
    }
    
    return 0;
}

Answer 1

You need two loops for each row
One will print the even numbers (0,2,4,6..). The number of even numbers (if you count zero) is just the row number. Don't print a new line yet.

Then, at the end of the row, print numbers starting at the row number, and ending with (the last even number you printed)+1

At the end of each row print a newline.

I hope this will give you a hint as to how to continue

Answer 2

Pseudocode of the algorithm:

n = user_input()
for i in 0 .. n:
  for j in 0 .. i:
      print(2 * j)
  for j in 0 .. i:
      print(i + j)
  print newline

Converting that to C, we get something pretty similar.

int main() {
  int i, j, n;
  printf("Enter a number:");
  scanf("%d",&n);

  for (i=0; i<n; i++) {
    for (j = 0; j < i; j++) {
      printf("%2d ", 2 * j);
    }
    for (j = 0; j < i; j++) {
      printf("%2d ", i + j);
    }
    printf("\n");
  }
}