user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  lea    D82(%rip),%rdi
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  lea    D17(%rip),%r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D54(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x2c(%rbp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D66(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  lea    -0x30(%rbp),%rsi
  call   <scanf@plt>
  mov    -0x30(%rbp),%r9d
  mov    -0x2c(%rbp),%edi
  cmp    %edi,%r9d
  jle    L61
  cmp    $0x1,%edi
  jle    L108
  movslq %r9d,%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  and    $0xfffffffffffffff0,%rdx
  sub    %rax,%rcx
L64:
  cmp    %rcx,%rsp
  je     L22
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L64
L22:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L24
L90:
  mov    %rsp,%rbx
  mov    %edi,%edx
  mov    %rbx,%rax
  lea    (%rbx,%rdx,4),%rdx
  nopl   0x0(%rax,%rax,1)
L72:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L72
  movslq %edi,%rax
  sub    $0x1,%r9d
  sub    $0x1,%edi
  xor    %r8d,%r8d
  lea    (%rbx,%rax,4),%rcx
  nopl   0x0(%rax,%rax,1)
L55:
  movl   $0x0,(%rcx)
  mov    %r8d,%esi
  movslq %edi,%rax
  xor    %edx,%edx
  cmp    %r8d,%edi
  jl     L131
  nopl   0x0(%rax,%rax,1)
L17:
  add    (%rbx,%rax,4),%edx
  sub    $0x1,%rax
  mov    %edx,(%rcx)
  cmp    %eax,%esi
  jle    L17
L131:
  add    $0x1,%edi
  add    $0x1,%r8
  add    $0x4,%rcx
  cmp    %r9d,%edi
  jne    L55
  lea    D80(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%eax
  test   %eax,%eax
  jle    L10
  xor    %r12d,%r12d
  lea    D67(%rip),%r13
  nopl   0x0(%rax)
L142:
  mov    (%rbx,%r12,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r12
  call   <printf@plt>
  mov    -0x30(%rbp),%eax
  cmp    %r12d,%eax
  jg     L142
L10:
  sub    $0x1,%eax
  lea    D16(%rip),%rdi
  cltq
  mov    (%rbx,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
L14:
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L112
  lea    -0x18(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
L61:
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L14
L24:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L90
L108:
  lea    D22(%rip),%rdi
  call   <puts@plt>
  mov    $0x2,%eax
  jmp    L14
L112:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D82
type: string
name: extract_symbolic_data
arguments: data_label: D17
type: string
name: extract_symbolic_data
arguments: data_label: D54
type: string
name: extract_symbolic_data
arguments: data_label: D66
type: string
name: extract_symbolic_data
arguments: data_label: D80
type: string
name: extract_symbolic_data
arguments: data_label: D67
type: string
name: extract_symbolic_data
arguments: data_label: D16
type: string
name: extract_symbolic_data
arguments: data_label: D14
type: string
name: extract_symbolic_data
arguments: data_label: D22
type: string

tool: D82: Calculo do enesimo termo de Fitromacci

tool: D17: %d

tool: D54: Digite um inteiro k: 

tool: D66: Digite um inteiro n: 

tool: D80: Segue a sequecia Fitromacci: 

tool: D67: %d 

tool: D16: 
O ultimo termo de Fitromacci eh: %d

tool: D14: 
K nao pode ser menor do que N

tool: D22: 
K precisa ser maior do que 1

assistant: ```c
int main() {
  int n, k, i, j;
  int ultimo_termo = 0;
  printf("Calculo do enesimo termo de Fitromacci\n");
  printf("Digite um inteiro k: ");
  scanf("%d", &k);
  printf("Digite um inteiro n: ");
  scanf("%d", &n);
  if (n <= k) {
    printf("\nK nao pode ser menor do que N\n");
    return 1;
  }
  if (k <= 1) {
    printf("\nK precisa ser maior do que 1\n");
    return 2;
  }
  int vetor[n];
  for (i = 0; i < k; i++) {
    vetor[i] = 1;
  }
  for (i = k; i < n; i++) {
    vetor[i] = 0;
    for (j = i - 1; j >= i - k; j--) {
      vetor[i] += vetor[j];
    }
  }
  printf("Segue a sequecia Fitromacci: ");
  for (i = 0; i < n; i++) {
    printf("%d ", vetor[i]);
  }
  printf("\nO ultimo termo de Fitromacci eh: %d", vetor[n - 1]);
  return 0;
}
```