user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x1,-0x20(%rbp)
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movss  D5(%rip),%xmm0
  movss  %xmm0,-0x1c(%rbp)
  jmp    L35
L22:
  movss  -0x1c(%rbp),%xmm0
  movss  D10(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x1c(%rbp),%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd -0x1c(%rbp),%xmm2
  movq   %xmm2,%rdx
  mov    -0x20(%rbp),%eax
  movapd %xmm0,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D130(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  addl   $0x1,-0x20(%rbp)
L35:
  pxor   %xmm1,%xmm1
  cvtss2sd -0x1c(%rbp),%xmm1
  movsd  D14(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  D14(%rip),%xmm1
  comisd %xmm1,%xmm0
  ja     L22
  movl   $0x1,-0x20(%rbp)
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movsd  D14(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  jmp    L23
L53:
  movsd  -0x18(%rbp),%xmm0
  movsd  D59(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%eax
  movapd %xmm0,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D130(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movsd  -0x18(%rbp),%xmm1
  movsd  D14(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
L23:
  movsd  -0x18(%rbp),%xmm1
  movsd  D14(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  D14(%rip),%xmm1
  comisd %xmm1,%xmm0
  ja     L53
  movl   $0x1,-0x20(%rbp)
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  fld1
  fstpt  -0x10(%rbp)
  jmp    L20
L34:
  fldt   -0x10(%rbp)
  fldt   0xdbf(%rip)        # 2060 <_fini+0xd74>
  fdivrp %st,%st(1)
  fstpt  -0x10(%rbp)
  mov    -0x20(%rbp),%eax
  push   -0x8(%rbp)
  push   -0x10(%rbp)
  push   -0x8(%rbp)
  push   -0x10(%rbp)
  mov    %eax,%esi
  lea    D137(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x20,%rsp
  addl   $0x1,-0x20(%rbp)
L20:
  fldt   -0x10(%rbp)
  fld1
  faddp  %st,%st(1)
  fld1
  fxch   %st(1)
  fcomip %st(1),%st
  fstp   %st(0)
  ja     L34
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D48
type_infered: string
name: retrieve_data
arguments: label: D5
type_infered: f32
name: retrieve_data
arguments: label: D10
type_infered: f32
name: retrieve_data
arguments: label: D130
type_infered: string
name: retrieve_data
arguments: label: D14
type_infered: byte[8]
name: retrieve_data
arguments: label: D16
type_infered: string
name: retrieve_data
arguments: label: D59
type_infered: byte[8]
name: retrieve_data
arguments: label: D60
type_infered: string
name: retrieve_data
arguments: label: D137
type_infered: string

tool: D48: Float:

tool: D5: 1

tool: D10: 2

tool: D130: %i. %f => %e


tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D16: Double:

tool: {"D59": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

tool: D60: Long double:

tool: D137: %i. %Lf => %Le


assistant: ```c
int main() {
  int l = 1;
  printf("Float:\n");
  float n = 1.0;
  while (1.0 + n > 1.0) {
    n = n / 2.0;
    printf("%i. %f => %e\n", l, n, n);
    l++;
  }
  l = 1;
  printf("Double:\n");
  double k = 1.0;
  while (1.0 + k > 1.0) {
    k = k / 2.0;
    printf("%i. %f => %e\n", l, k, k);
    k++;
  }
  l = 1;
  printf("Long double:\n");
  long double m = 1.0;
  while (1.0 + m > 1.0) {
    m = m / 2.0;
    printf("%i. %Lf => %Le\n", l, m, m);
    l++;
  }
}
```