user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D39(%rip),%r13
  lea    D79(%rip),%r12
  lea    D78(%rip),%rbp
  lea    D85(%rip),%rbx
  jmp    L93
L109:
  cmp    $0x3,%eax
  jne    L93
  mov    $0x0,%eax
  call   <main>
  jmp    L70
L40:
  lea    D75(%rip),%rdi
  call   <printf@plt>
  lea    0x20(%rsp),%rsi
  lea    D117(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D95(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D124(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%edi
  movsd  0x20(%rsp),%xmm0
  call   <Euller@plt>
L70:
  cmpl   $0x0,0xc(%rsp)
  je     L67
L93:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D145(%rip),%rdi
  call   <printf@plt>
  lea    D135(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D124(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    $0x2,%eax
  je     L40
  jg     L109
  test   %eax,%eax
  je     L16
  cmp    $0x1,%eax
  jne    L93
  lea    D101(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    D117(%rip),%r14
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D95(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D124(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%edi
  movsd  0x18(%rsp),%xmm0
  subsd  0x10(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %edi,%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,0x20(%rsp)
  call   <Euller@plt>
  jmp    L70
L16:
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L70
L67:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L71
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L71:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D39
datatype: string
name: unpack_data
arguments: tag: D79
datatype: string
name: unpack_data
arguments: tag: D78
datatype: string
name: unpack_data
arguments: tag: D85
datatype: string
name: unpack_data
arguments: tag: D75
datatype: string
name: unpack_data
arguments: tag: D117
datatype: string
name: unpack_data
arguments: tag: D95
datatype: string
name: unpack_data
arguments: tag: D124
datatype: string
name: unpack_data
arguments: tag: D145
datatype: string
name: unpack_data
arguments: tag: D135
datatype: string
name: unpack_data
arguments: tag: D101
datatype: string
name: unpack_data
arguments: tag: D6
datatype: string

tool: D39:             Resolucao de Equacoes Diferenciais Pelo Metodo de Euller


tool: D79: 


	(1)..........................Entrar com a, b e n


tool: D78: 	(2)..........................Entrar com h e n


tool: D85: 	(3)..........................Entrar Com Novos Valores de a e b


tool: D75: Entre com h: 

tool: D117: %lf

tool: D95: Entre com o numero de intervalos: 

tool: D124: %d

tool: D145: 	(0)..........................Sair


tool: D135: 
Opcao Desejada: 

tool: D101: Entre com a: 

tool: D6: Entre com b: 

assistant: ```c
int main() {
  double a, b, h;
  int n, opcao;
  do {
    printf("            Resolucao de Equacoes Diferenciais Pelo Metodo de Euller\n");
    printf("\n\n\n\t(1)..........................Entrar com a, b e n\n");
    printf("\t(2)..........................Entrar com h e n\n");
    printf("\t(3)..........................Entrar Com Novos Valores de a e b\n");
    printf("\t(0)..........................Sair\n");
    printf("\nOpcao Desejada: ");
    scanf("%d", &opcao);
    switch (opcao) {
    case 1:
      printf("Entre com a: ");
      scanf("%lf", &a);
      printf("Entre com b: ");
      scanf("%lf", &b);
      printf("Entre com o numero de intervalos: ");
      scanf("%d", &n);
      h = (b - a) / n;
      Euller(h, n);
      break;
    case 2:
      printf("Entre com h: ");
      scanf("%lf", &h);
      printf("Entre com o numero de intervalos: ");
      scanf("%d", &n);
      Euller(h, n);
      break;
    case 3:
      main();
      break;
    case 0:
      exit(0);
      break;
    }
  } while (opcao);
  return 0;
}
```