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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x14,-0x20(%rbp)
  movl   $0xa,-0x1c(%rbp)
  movl   $0xf,-0x18(%rbp)
  movl   $0x5,-0x14(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  imul   -0x18(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  imul   -0x18(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x18(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  imul   %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D98(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  imul   -0x18(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1b,%edi
  call   <f@plt>
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movsd  D18(%rip),%xmm0
  mov    D77(%rip),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <g@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0xc(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D108(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    D105(%rip),%rax
  mov    $0x24,%esi
  movq   %rax,%xmm0
  mov    $0x4,%edi
  call   <h@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D54
type_guessed: string
name: disassemble_data
arguments: tag: D21
type_guessed: string
name: disassemble_data
arguments: tag: D98
type_guessed: string
name: disassemble_data
arguments: tag: D43
type_guessed: string
name: disassemble_data
arguments: tag: D70
type_guessed: string
name: disassemble_data
arguments: tag: D18
type_guessed: f64
name: disassemble_data
arguments: tag: D77
type_guessed: f64
name: disassemble_data
arguments: tag: D108
type_guessed: string
name: disassemble_data
arguments: tag: D105
type_guessed: f64

tool: D54: (a + b) * c / d = %d


tool: D21: ((a + b) * c) / d = %d


tool: D98: (a + b) * (c / d) = %d


tool: D43: a + (b * c) / d = %d


tool: D70: 


tool: D18: 6.6

tool: D77: 1.9

tool: D108: %f


tool: D105: 2.5

assistant: ```c
int main(int argc, char **argv) {
  int a = 20;
  int b = 10;
  int c = 15;
  int d = 5;
  int e;
  e = (a + b) * c / d;
  printf("(a + b) * c / d = %d\n", e);
  e = ((a + b) * c) / d;
  printf("((a + b) * c) / d = %d\n", e);
  e = (a + b) * (c / d);
  printf("(a + b) * (c / d) = %d\n", e);
  e = a + (b * c) / d;
  printf("a + (b * c) / d = %d\n", e);
  printf("\n\n");
  f(27);
  printf("\n\n");
  float x = g(1.9, 6.6);
  printf("%f\n", x);
  printf("\n\n");
  char *y = h(4, 2.5, '$');
  printf("%s\n", y);
  free(y);
  return 0;
}
```