user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    D30(%rip),%rdi
  call   <puts@plt>
  lea    -0x3c(%rbp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L28:
  cmp    %rdx,%rsp
  je     L18
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L28
L18:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L13
  orq    $0x0,-0x8(%rsp,%rax,1)
L13:
  mov    %rsp,%r13
  test   %esi,%esi
  jle    L148
  mov    %r13,%rax
  mov    %esi,%edx
  lea    0x0(%r13,%rdx,4),%rdx
L118:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L118
L148:
  cmp    $0x1,%esi
  jle    L5
  mov    $0x1,%ebx
  mov    $0x0,%r14d
  jmp    L39
L0:
  mov    -0x3c(%rbp),%esi
  add    $0x1,%rbx
  cmp    %ebx,%esi
  jle    L11
L39:
  mov    %ebx,%r15d
  mov    %ebx,%edi
  call   <vsotaDeljiteljev@plt>
  mov    %eax,%r12d
  cmp    %ebx,%eax
  je     L0
  mov    %eax,%edi
  call   <vsotaDeljiteljev@plt>
  cmp    %ebx,%eax
  jne    L0
  cmpl   $0x1,0x0(%r13,%rbx,4)
  je     L0
  add    %r12d,%r15d
  add    %r15d,%r14d
  movl   $0x1,0x0(%r13,%rbx,4)
  movslq %r12d,%r12
  movl   $0x1,0x0(%r13,%r12,4)
  jmp    L0
L11:
  test   %r14d,%r14d
  je     L5
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r14d,%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L48:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L5:
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L48
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D30
type_infered: string
name: retrieve_data
arguments: label: D7
type_infered: string
name: retrieve_data
arguments: label: D17
type_infered: string
name: retrieve_data
arguments: label: D9
type_infered: string
name: retrieve_data
arguments: label: D28
type_infered: string

tool: D30: Please insert upper bound:

tool: D7: %d

tool: D17: Sum of the amicable numbers below %d is:


tool: D9: %d


tool: D28: There are no amicable numbers below %d


assistant: ```c
int main() {
  int meja;
  printf("Please insert upper bound:\n");
  scanf("%d", &meja);
  int tabela[meja];
  int n;
  for (n = 0; n < meja; n++) {
    tabela[n] = 0;
  }
  int i;
  int vsota = 0;
  for (i = 1; i < meja; i++) {
    int delitelji = vsotaDeljiteljev(i);
    if (i != delitelji && i == vsotaDeljiteljev(delitelji) && tabela[i] != 1) {
      vsota += i + delitelji;
      tabela[i] = 1;
      tabela[delitelji] = 1;
    }
  }
  if (vsota == 0) {
    printf("There are no amicable numbers below %d\n", meja);
    return 0;
  }
  printf("Sum of the amicable numbers below %d is:\n", meja);
  printf("%d\n", vsota);
  return 0;
}
```