user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  push   %rbx
  lea    D19(%rip),%rbx
  sub    $0x1d0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1c8(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%r14
  movaps %xmm0,(%rsp)
  lea    0x1c0(%rsp),%rbp
  movq   $0x0,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  nopl   0x0(%rax)
L235:
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%r14),%eax
  movdqa D8(%rip),%xmm3
  test   %eax,%eax
  jle    L121
L59:
  cmp    $0xa,%r13d
  je     L162
  movslq (%r14),%rax
  mov    %rax,%rsi
  imul   $0x66666667,%rax,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %esi,%eax
  add    %ecx,%ecx
  mov    %edx,(%r14)
  sub    %ecx,%eax
  test   %r13d,%r13d
  jne    L166
  mov    %eax,(%rsp)
  test   %edx,%edx
  je     L137
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  mov    %eax,%ecx
  mov    %edx,%eax
  add    %esi,%esi
  mov    %ecx,(%r14)
  mov    %ecx,%edx
  xor    %ecx,%ecx
  sub    %esi,%eax
  mov    $0x1,%esi
L238:
  movslq %r13d,%rdi
  cmp    (%rsp,%rdi,4),%eax
  lea    0x1(%r13),%edi
  je     L140
  cmp    %esi,%edi
  jge    L37
L130:
  movslq %edi,%r8
  cmp    (%rsp,%r8,4),%eax
  je     L174
  add    $0x1,%edi
  cmp    %esi,%edi
  jge    L37
  movslq %edi,%rdi
  mov    %esi,%r13d
  cmp    (%rsp,%rdi,4),%eax
  je     L206
L37:
  test   %ecx,%ecx
  jne    L174
  mov    %eax,(%rsp,%rsi,4)
  lea    0x1(%rsi),%r13d
L206:
  test   %edx,%edx
  jne    L59
  nopl   0x0(%rax,%rax,1)
L162:
  add    $0x4,%r14
  cmp    %r14,%rbp
  jne    L235
L91:
  xor    %ebx,%ebx
  mov    %rsp,%r12
  lea    D20(%rip),%rbp
  nopl   0x0(%rax)
L239:
  mov    (%r12,%rbx,4),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,%r13d
  jg     L239
L47:
  mov    0x1c8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L241
  add    $0x1d0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L140:
  mov    $0x1,%ecx
  cmp    %edi,%esi
  jg     L130
  nopl   (%rax)
L174:
  mov    %esi,%r13d
  jmp    L206
  nopl   0x0(%rax,%rax,1)
L137:
  add    $0x4,%r14
  mov    $0x1,%r13d
  cmp    %r14,%rbp
  jne    L235
  jmp    L91
  nopl   0x0(%rax,%rax,1)
L121:
  test   %r13d,%r13d
  jne    L91
  jmp    L47
L166:
  mov    %r12d,%esi
  cmovg  %r13d,%esi
  cmp    $0x3,%r13d
  jle    L134
  movd   %eax,%xmm4
  mov    %esi,%ecx
  pshufd $0x0,%xmm4,%xmm0
  shr    $0x2,%ecx
  movdqa %xmm0,%xmm1
  pcmpeqd (%rsp),%xmm1
  pand   %xmm3,%xmm1
  cmp    $0x1,%ecx
  je     L86
  pcmpeqd 0x10(%rsp),%xmm0
  movdqa %xmm3,%xmm2
  pand   %xmm0,%xmm2
  pandn  %xmm1,%xmm0
  por    %xmm2,%xmm0
  movdqa %xmm0,%xmm1
L86:
  movdqa %xmm1,%xmm2
  mov    %esi,%edi
  psrldq $0x8,%xmm2
  and    $0xfffffffc,%edi
  and    $0x3,%esi
  movslq %r13d,%rsi
  movdqa %xmm2,%xmm0
  mov    %edi,%r13d
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm2
  pandn  %xmm1,%xmm0
  por    %xmm2,%xmm0
  movdqa %xmm0,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm2
  pandn  %xmm0,%xmm1
  por    %xmm2,%xmm1
  movd   %xmm1,%ecx
  jne    L238
  jmp    L37
L134:
  movslq %r13d,%rsi
  xor    %ecx,%ecx
  xor    %r13d,%r13d
  jmp    L238
L241:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D19
type_guessed: string
name: disassemble_data
arguments: tag: D8
type_guessed: byte[16]
name: disassemble_data
arguments: tag: D20
type_guessed: string

tool: D19: %d

tool: {"D8": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D20: %d 

assistant: ```c
int main() {
  int niz[100], i, k, counter = 0, n, temp, ima;
  int niz_pomocni[10] = {};
  for (i = 0; i < 100; i++) {
    scanf("%d", &niz[i]);
    if (niz[i] <= 0)
      break;
    do {
      if (counter == 10)
        break;
      n = niz[i] % 10;
      niz[i] /= 10;
      ima = 0;
      if (counter == 0)
        niz_pomocni[counter++] = n;
      else {
        for (k = 0; k < counter; k++) {
          if (niz_pomocni[k] == n)
            ima = 1;
        }
        if (ima == 0) {
          niz_pomocni[counter++] = n;
        }
      }
    } while (niz[i] != 0);
  }
  for (i = 0; i < counter; i++)
    printf("%d ", niz_pomocni[i]);
  return 0;
}
```