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    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0xa,-0xa4(%rbp)
  movl   $0x0,-0xac(%rbp)
  jmp    L70
L79:
  mov    -0xac(%rbp),%eax
  movslq %eax,%rdx
  mov    -0xac(%rbp),%eax
  cltq
  mov    %rdx,-0x90(%rbp,%rax,8)
  addl   $0x1,-0xac(%rbp)
L70:
  mov    -0xac(%rbp),%eax
  cmp    -0xa4(%rbp),%eax
  jl     L79
  mov    -0xa4(%rbp),%edx
  lea    -0x90(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_longs@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xa4(%rbp),%edx
  lea    -0x90(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <shuffle@plt>
  mov    -0xa4(%rbp),%edx
  lea    -0x90(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_longs@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xa4(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0xa0(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L28:
  cmp    %rdx,%rsp
  je     L23
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L28
L23:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L44
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L44:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x98(%rbp)
  movl   $0x0,-0xa8(%rbp)
  jmp    L19
L30:
  mov    -0xa8(%rbp),%eax
  cltq
  mov    -0x90(%rbp,%rax,8),%rax
  mov    %eax,%ecx
  mov    -0x98(%rbp),%rax
  mov    -0xa8(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0xa8(%rbp)
L19:
  mov    -0xa8(%rbp),%eax
  cmp    -0xa4(%rbp),%eax
  jl     L30
  mov    -0xa4(%rbp),%edx
  mov    -0x98(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_ints@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x98(%rbp),%rax
  mov    $0x9,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <quicksort_ints@plt>
  mov    -0xa4(%rbp),%edx
  mov    -0x98(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_ints@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    %rbx,%rsp
  mov    $0x0,%eax
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L78
  call   <__stack_chk_fail@plt>
L78:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_infered: string
name: parse_data
arguments: data_label: D9
type_infered: string
name: parse_data
arguments: data_label: D19
type_infered: string
name: parse_data
arguments: data_label: D39
type_infered: string

tool: D1: 


tool: D9: 



tool: D19: ints:


tool: D39: DONE


assistant: ```c
int main() {
  int n = 10;
  long A[10];
  for (int i = 0; i < n; i++)
    A[i] = i;
  print_longs(A, n);
  printf("\n");
  shuffle(A, n);
  print_longs(A, n);
  printf("\n\n");
  printf("ints:\n");
  int B[n];
  for (int i = 0; i < n; i++)
    B[i] = A[i];
  print_ints(B, n);
  printf("\n");
  quicksort_ints(B, 0, 9);
  print_ints(B, n);
  printf("\n");
  printf("DONE\n");
}
```