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    $0x48,%rsp
  mov    %edi,-0x64(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D76(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x58(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D71(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x54(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x54(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x48(%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
L15:
  cmp    %rdx,%rsp
  je     L48
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L15
L48:
  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,-0x40(%rbp)
  mov    -0x58(%rbp),%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x58(%rbp),%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,0x4(%rax)
  movl   $0x2,-0x50(%rbp)
  jmp    L40
L67:
  mov    -0x50(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x40(%rbp),%rax
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%ecx
  mov    -0x50(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x40(%rbp),%rax
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  add    %eax,%ecx
  mov    -0x40(%rbp),%rax
  mov    -0x50(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x50(%rbp)
L40:
  mov    -0x54(%rbp),%eax
  cmp    %eax,-0x50(%rbp)
  jl     L67
  mov    -0x58(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4c(%rbp)
  jmp    L23
L6:
  mov    -0x40(%rbp),%rax
  mov    -0x4c(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  mov    %eax,%esi
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4c(%rbp)
L23:
  mov    -0x54(%rbp),%eax
  cmp    %eax,-0x4c(%rbp)
  jl     L6
  mov    $0x0,%eax
  mov    %rbx,%rsp
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L69
  call   <__stack_chk_fail@plt>
L69:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D13
data_type: string
name: disassemble_data
arguments: tag: D76
data_type: string
name: disassemble_data
arguments: tag: D0
data_type: string
name: disassemble_data
arguments: tag: D12
data_type: string
name: disassemble_data
arguments: tag: D71
data_type: string
name: disassemble_data
arguments: tag: D22
data_type: string
name: disassemble_data
arguments: tag: D56
data_type: string

tool: D13: ===== Exercise 8 =====

tool: D76: FIBONACCI NUMBERS!

tool: D0: Enter the first number. : 

tool: D12: %i

tool: D71: How many numbers will be shown?: 

tool: D22: --> %i First Fibonacci Numbers


tool: D56: %i


assistant: ```c
int main(int argc, const char *argv[]) {
  int number;
  printf("===== Exercise 8 =====\n");
  printf("FIBONACCI NUMBERS!\n");
  printf("Enter the first number. : ");
  scanf("%i", &number);
  printf("How many numbers will be shown?: ");
  int TheNumberOfNumber;
  scanf("%i", &TheNumberOfNumber);
  int fibonacci[TheNumberOfNumber];
  fibonacci[0] = number;
  fibonacci[1] = number;
  for (int i = 2; i < TheNumberOfNumber; i++) {
    fibonacci[i] = fibonacci[i - 2] + fibonacci[i - 1];
  }
  printf("--> %i First Fibonacci Numbers\n", number);
  for (int j = 0; j < TheNumberOfNumber; j++) {
    printf("%i\n", fibonacci[j]);
  }
  return 0;
}
```