user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D145(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D207(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L143
L220:
  movl   $0x0,-0x14(%rbp)
  jmp    L32
L99:
  lea    D209(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L32:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L99
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L143:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L220
  lea    D123(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L212
L41:
  movl   $0x0,-0x14(%rbp)
  jmp    L285
L310:
  lea    D209(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L285:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L310
  addl   $0x1,-0x10(%rbp)
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L212:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L41
  lea    D110(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L164
L217:
  movl   $0x0,-0x14(%rbp)
  jmp    L33
L216:
  lea    D209(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L33:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L216
  subl   $0x1,-0x10(%rbp)
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L164:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L217
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L119
L50:
  movl   $0x0,-0x14(%rbp)
  jmp    L253
L94:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L282
  lea    D209(%rip),%rax
  jmp    L206
L282:
  lea    D195(%rip),%rax
L206:
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L253:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L94
  addl   $0x1,-0x10(%rbp)
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L119:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L50
  lea    D80(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L291
L102:
  movl   $0x0,-0x14(%rbp)
  jmp    L211
L62:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jge    L46
  lea    D195(%rip),%rax
  jmp    L265
L46:
  lea    D209(%rip),%rax
L265:
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L211:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L62
  subl   $0x1,-0x10(%rbp)
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L291:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L102
  lea    D98(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L198
L104:
  movl   $0x0,-0x14(%rbp)
  jmp    L114
L295:
  lea    D209(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L114:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L295
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jne    L236
  movl   $0x1,-0xc(%rbp)
L236:
  cmpl   $0x0,-0xc(%rbp)
  jne    L118
  addl   $0x1,-0x10(%rbp)
  jmp    L150
L118:
  subl   $0x1,-0x10(%rbp)
L150:
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L198:
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x18(%rbp)
  jl     L104
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L225
  call   <__stack_chk_fail@plt>
L225:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D145
type_guessed: string
name: read_data
arguments: data_tag: D207
type_guessed: string
name: read_data
arguments: data_tag: D209
type_guessed: string
name: read_data
arguments: data_tag: D37
type_guessed: string
name: read_data
arguments: data_tag: D123
type_guessed: string
name: read_data
arguments: data_tag: D110
type_guessed: string
name: read_data
arguments: data_tag: D32
type_guessed: string
name: read_data
arguments: data_tag: D195
type_guessed: string
name: read_data
arguments: data_tag: D80
type_guessed: string
name: read_data
arguments: data_tag: D98
type_guessed: string

tool: D145: %d

tool: D207: 
------------ Exemplo 1 ------------



tool: D209: * 

tool: D37: 


tool: D123: 
------------ Exemplo 2 ------------



tool: D110: 
------------ Exemplo 3 ------------



tool: D32: 
------------ Exemplo 4 ------------



tool: D195:   

tool: D80: 
------------ Exemplo 5 ------------



tool: D98: 
------------ Exemplo 6 ------------



assistant: ```c
int main() {
  int i, j, x;
  scanf("%d", &x);
  printf("\n------------ Exemplo 1 ------------\n\n");
  for (i = 0; i < x; i++) {
    for (j = 0; j < x; j++) {
      printf("* ");
    }
    printf("\n");
  }
  printf("\n------------ Exemplo 2 ------------\n\n");
  int k = 1;
  for (i = 0; i < x; i++) {
    for (j = 0; j < k; j++) {
      printf("* ");
    }
    k++;
    printf("\n");
  }
  printf("\n------------ Exemplo 3 ------------\n\n");
  k = x;
  for (i = 0; i < x; i++) {
    for (j = 0; j < k; j++) {
      printf("* ");
    }
    k--;
    printf("\n");
  }
  printf("\n------------ Exemplo 4 ------------\n\n");
  k = 0;
  for (i = 0; i < x; i++) {
    for (j = 0; j < x; j++) {
      printf(j >= k ? "* " : "  ");
    }
    k++;
    printf("\n");
  }
  printf("\n------------ Exemplo 5 ------------\n\n");
  k = x - 1;
  for (i = 0; i < x; i++) {
    for (j = 0; j < x; j++) {
      printf(j < k ? "  " : "* ");
    }
    k--;
    printf("\n");
  }
  printf("\n------------ Exemplo 6 ------------\n\n");
  k = 1;
  int r = 0;
  for (i = 0; i < x + (x - 1); i++) {
    for (j = 0; j < k; j++) {
      printf("* ");
    }
    if (k == x) {
      r = 1;
    }
    if (r == 0) {
      k++;
    } else {
      k--;
    }
    printf("\n");
  }
  return 0;
}
```