user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D20(%rip),%r14
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  mov    $0x30,%r12d
  push   %rbp
  push   %rbx
  mov    $0x20,%ebx
L40:
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  mov    %ebx,%ebp
  call   <printf@plt>
  nopl   0x0(%rax,%rax,1)
L53:
  mov    %ebp,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    %r12d,%ebp
  jne    L53
  mov    $0xa,%edi
  add    $0x10,%ebx
  lea    0x10(%rbp),%r12d
  call   <putchar@plt>
  cmp    $0x80,%ebx
  jne    L40
  mov    $0x90,%ebp
  lea    D7(%rip),%r12
  nopl   0x0(%rax)
L100:
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  mov    %ebx,%r14d
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L32:
  mov    %r14d,%esi
  mov    %r14d,%edx
  mov    %r12,%rdi
  xor    %eax,%eax
  sar    $0x6,%esi
  and    $0x3f,%edx
  add    $0x1,%r14d
  and    $0x1f,%esi
  sub    $0xffffff80,%edx
  add    $0xc0,%esi
  call   <printf@plt>
  cmp    %r14d,%ebp
  jne    L32
  mov    $0xa,%edi
  add    $0x10,%ebx
  add    $0x10,%ebp
  call   <putchar@plt>
  cmp    $0x800,%ebx
  jne    L100
  lea    D16(%rip),%r12
  jmp    L51
  nopl   0x0(%rax)
L12:
  mov    %ebp,%ebx
  cmp    $0x10000,%ebp
  je     L107
L51:
  lea    -0xd7a0(%rbx),%eax
  lea    0x10(%rbx),%ebp
  cmp    $0x85f,%eax
  jbe    L12
  lea    -0xfdd0(%rbx),%eax
  cmp    $0x1f,%eax
  jbe    L12
  lea    -0xfff0(%rbx),%eax
  cmp    $0xf,%eax
  jbe    L12
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  nopl   0x0(%rax)
L1:
  mov    %ebx,%edx
  mov    %ebx,%ecx
  mov    %ebx,%esi
  mov    %r12,%rdi
  sar    $0x6,%edx
  and    $0x3f,%ecx
  sar    $0xc,%esi
  xor    %eax,%eax
  and    $0x3f,%edx
  sub    $0xffffff80,%ecx
  add    $0xe0,%esi
  add    $0x1,%ebx
  sub    $0xffffff80,%edx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L1
  mov    $0xa,%edi
  mov    %ebp,%ebx
  call   <putchar@plt>
  cmp    $0x10000,%ebp
  jne    L51
L107:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D20
datatype: string
name: retrieve_data
arguments: data_label: D0
datatype: string
name: retrieve_data
arguments: data_label: D7
datatype: string
name: retrieve_data
arguments: data_label: D16
datatype: string

tool: D20:  %c

tool: D0: %04x:

tool: D7:  %c%c

tool: D16:  %c%c%c

assistant: ```c
int main(int argc, char **argv) {
  int k, j;
  for (k = 0x20; k < 0x80; k += 0x10) {
    printf("%04x:", k);
    for (j = 0; j < 0x10; j++)
      printf(" %c", k + j);
    printf("\n");
  }
  for (k = 0x80; k < 0x800; k += 0x10) {
    printf("%04x:", k);
    for (j = 0; j < 0x10; j++) {
      int sum = k + j;
      int b0 = 0x80 + ((sum & 0x003f));
      int b1 = 0xc0 + ((sum & 0x07c0) >> 6);
      printf(" %c%c", b1, b0);
    }
    printf("\n");
  }
  for (k = 0x800; k < 0x10000; k += 0x10) {
    if (k >= 0xd7a0 && k < 0xe000)
      continue;
    if (k >= 0xfdd0 && k < 0xfdf0)
      continue;
    if (k >= 0xfff0 && k < 0x10000)
      continue;
    printf("%04x:", k);
    for (j = 0; j < 0x10; j++) {
      int sum = k + j;
      int b0 = 0x80 + ((sum & 0x003f));
      int b1 = 0x80 + ((sum & 0x0fc0) >> 6);
      int b2 = 0xe0 + ((sum & 0xf000) >> 12);
      printf(" %c%c%c", b2, b1, b0);
    }
    printf("\n");
  }
}
```