user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D22(%rip),%rdi
  lea    D19(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0xc(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L27
  lea    0x10(%rsp),%r12
  xor    %ebx,%ebx
  mov    %r12,%rbp
  cs nopw 0x0(%rax,%rax,1)
L314:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    %ebx,%eax
  jg     L314
  test   %eax,%eax
  jle    L27
  sub    $0x1,%eax
  lea    0x4(%r12),%rdx
  xor    %ebx,%ebx
  lea    (%rdx,%rax,4),%rcx
  jmp    L202
  nopw   0x0(%rax,%rax,1)
L195:
  add    $0x4,%rdx
L202:
  xor    %eax,%eax
  test   %ebx,%ebx
  jle    L175
  mov    (%r12),%eax
  cmp    0x40(%rsp),%eax
  je     L18
  cmp    $0x1,%ebx
  je     L233
  cmp    0x44(%rsp),%eax
  je     L18
  cmp    $0x2,%ebx
  je     L293
  cmp    0x48(%rsp),%eax
  je     L18
  cmp    $0x3,%ebx
  je     L225
  cmp    0x4c(%rsp),%eax
  je     L18
  cmp    $0x4,%ebx
  je     L73
  cmp    0x50(%rsp),%eax
  je     L18
  cmp    $0x5,%ebx
  je     L203
  cmp    0x54(%rsp),%eax
  je     L18
  cmp    $0x6,%ebx
  je     L385
  cmp    0x58(%rsp),%eax
  je     L18
  cmp    $0x7,%ebx
  je     L329
  cmp    0x5c(%rsp),%eax
  je     L18
  cmp    $0x8,%ebx
  je     L238
  cmp    0x60(%rsp),%eax
  je     L18
  cmp    $0x9,%ebx
  je     L174
  cmp    0x64(%rsp),%eax
  je     L18
  mov    $0xa,%eax
L175:
  cmp    %eax,%ebx
  jne    L18
  mov    (%r12),%esi
  movslq %ebx,%rax
  add    $0x1,%ebx
  mov    %esi,0x40(%rsp,%rax,4)
L18:
  mov    %rdx,%r12
  cmp    %rcx,%rdx
  jne    L195
  lea    D55(%rip),%rdi
  call   <puts@plt>
  test   %ebx,%ebx
  jle    L161
  lea    -0x1(%rbx),%eax
  lea    0x40(%rsp),%rbp
  lea    0x44(%rsp,%rax,4),%r12
  lea    D51(%rip),%rbx
  nopw   0x0(%rax,%rax,1)
L271:
  mov    0x0(%rbp),%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L271
L161:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L233:
  mov    $0x1,%eax
  jmp    L175
L293:
  mov    $0x2,%eax
  jmp    L175
L225:
  mov    $0x3,%eax
  jmp    L175
L73:
  mov    $0x4,%eax
  jmp    L175
L203:
  mov    $0x5,%eax
  jmp    L175
L385:
  mov    $0x6,%eax
  jmp    L175
L329:
  mov    $0x7,%eax
  jmp    L175
L238:
  mov    $0x8,%eax
  jmp    L175
L174:
  mov    $0x9,%eax
  jmp    L175
L27:
  lea    D55(%rip),%rdi
  call   <puts@plt>
  jmp    L161
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D22
type_infered: string
name: extract_symbolic_data
arguments: label: D19
type_infered: string
name: extract_symbolic_data
arguments: label: D9
type_infered: string
name: extract_symbolic_data
arguments: label: D55
type_infered: string
name: extract_symbolic_data
arguments: label: D51
type_infered: string

tool: D22: Enter number of elements in array

tool: D19: %d

tool: D9: Enter %d integers


tool: D55: Array obtained after removing duplicate elements

tool: D51: %d


assistant: ```c
int main() {
  int n, a[10], b[10], count = 0, c, d;
  printf("Enter number of elements in array\n");
  scanf("%d", &n);
  printf("Enter %d integers\n", n);
  for (c = 0; c < n; c++)
    scanf("%d", &a[c]);
  for (c = 0; c < n; c++) {
    for (d = 0; d < count; d++) {
      if (a[c] == b[d])
        break;
    }
    if (d == count) {
      b[count] = a[c];
      count++;
    }
  }
  printf("Array obtained after removing duplicate elements\n");
  for (c = 0; c < count; c++)
    printf("%d\n", b[c]);
  return 0;
}
```