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

```asm
<main>:
  endbr64
  push   %r15
  lea    D109(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D87(%rip),%rdi
  call   <scanf@plt>
  jmp    L31
  cs nopw 0x0(%rax,%rax,1)
L41:
  add    $0x1,%ebx
L31:
  mov    0x14(%rsp),%r13d
  mov    %ebx,%esi
  mov    $0xa,%edi
  call   <power@plt>
  mov    %eax,%r8d
  mov    %r13d,%eax
  cltd
  idiv   %r8d
  mov    %eax,%r13d
  test   %eax,%eax
  jne    L41
  mov    %ebx,%esi
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <printf@plt>
  lea    D108(%rip),%r15
  mov    %ebx,0xc(%rsp)
  mov    %r13d,%ebx
  jmp    L72
  nopw   0x0(%rax,%rax,1)
L75:
  xor    %eax,%eax
  cmp    %r13d,%ebx
  setg   %al
  cmp    %r12d,%eax
  jge    L69
L59:
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D41(%rip),%rdi
  call   <printf@plt>
  cmp    0xc(%rsp),%r14d
  je     L19
  mov    %r14d,%ebx
L72:
  lea    0x1(%rbx),%r14d
  mov    $0xa,%edi
  mov    %r14d,%esi
  call   <power@plt>
  mov    %r15,%rdi
  mov    %eax,%r12d
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%eax
  lea    D35(%rip),%rdi
  cltd
  idiv   %r12d
  mov    %eax,%esi
  mov    %eax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%edi
  mov    %r12d,%esi
  call   <fmod@plt>
  lea    D96(%rip),%rdi
  mov    %eax,%esi
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    $0xa,%edi
  call   <power@plt>
  lea    D17(%rip),%rdi
  mov    %eax,%r12d
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%eax
  lea    (%r12,%r12,1),%ebx
  imul   %r12d,%eax
  add    %eax,%ebp
  cmp    %r13d,%ebx
  jg     L75
  lea    D34(%rip),%rdi
  add    %r12d,%ebp
  call   <puts@plt>
  xor    %eax,%eax
  cmp    %r13d,%ebx
  setg   %al
  cmp    %r12d,%eax
  jl     L59
L69:
  sub    %r12d,%r13d
  lea    D44(%rip),%rdi
  lea    0x1(%rbp,%r13,1),%ebp
  call   <puts@plt>
  jmp    L59
  nopl   0x0(%rax)
L19:
  mov    %ebp,%esi
  lea    D85(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D109
type_infered: string
name: disassemble_data
arguments: data_label: D87
type_infered: string
name: disassemble_data
arguments: data_label: D30
type_infered: string
name: disassemble_data
arguments: data_label: D108
type_infered: string
name: disassemble_data
arguments: data_label: D41
type_infered: string
name: disassemble_data
arguments: data_label: D35
type_infered: string
name: disassemble_data
arguments: data_label: D96
type_infered: string
name: disassemble_data
arguments: data_label: D17
type_infered: string
name: disassemble_data
arguments: data_label: D34
type_infered: string
name: disassemble_data
arguments: data_label: D44
type_infered: string
name: disassemble_data
arguments: data_label: D85
type_infered: string

tool: D109: Enter a number: 

tool: D87: %d

tool: D30: the digit is: %d



tool: D108: The base is: %d


tool: D41: The n1 is: %d



tool: D35: The fra is: %d


tool: D96: The mod is: %d


tool: D17: The quan is: %d


tool: D34: A

tool: D44: B

tool: D85: The number of ones is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int num;
  int frac = 1;
  int digit = 1;
  printf("Enter a number: ");
  scanf("%d", &num);
  while (1) {
    frac = num / power(10, digit);
    if (frac == 0)
      break;
    digit += 1;
  }
  printf("the digit is: %d\n\n", digit);
  int j;
  int n1 = 0;
  for (j = 0; j < digit; j++) {
    int base = power(10, j + 1);
    printf("The base is: %d\n", base);
    int fra = num / base;
    printf("The fra is: %d\n", fra);
    int mod = fmod(num, base);
    printf("The mod is: %d\n", mod);
    int quan = power(10, j);
    printf("The quan is: %d\n", quan);
    n1 += fra * quan;
    if (mod > 2 * quan - 1) {
      n1 += quan;
      printf("A\n");
    }
    if (2 * quan > mod > quan - 1) {
      n1 += mod - quan + 1;
      printf("B\n");
    }
    printf("The n1 is: %d\n\n", n1);
  }
  printf("The number of ones is: %d\n", n1);
  getchar();
  getchar();
  return 0;
}
```