user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D101(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D78(%rip),%rdi
  call   <scanf@plt>
  mov    $0x1,%r15d
  jmp    L14
L15:
  add    $0x1,%r15d
L14:
  mov    0x4(%rsp),%ebx
  mov    %r15d,%esi
  mov    $0xa,%edi
  call   <power@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
  test   %eax,%eax
  jne    L15
  mov    %r15d,%esi
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %r15d,%r15d
  jle    L74
  mov    %r13d,%r14d
  jmp    L36
L50:
  add    %ebp,%r14d
  lea    D30(%rip),%rdi
  call   <puts@plt>
  jmp    L54
L77:
  mov    %r14d,%esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r15d,%r13d
  je     L23
L36:
  mov    %r13d,%ebp
  add    $0x1,%r13d
  mov    %r13d,%esi
  mov    $0xa,%edi
  call   <power@plt>
  mov    %eax,%r12d
  mov    %eax,%esi
  lea    D104(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  cltd
  idiv   %r12d
  mov    %eax,%ebx
  mov    %eax,%esi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  mov    0x4(%rsp),%edi
  call   <fmod@plt>
  mov    %eax,%r12d
  mov    %eax,%esi
  lea    D94(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    $0xa,%edi
  call   <power@plt>
  mov    %eax,%ebp
  mov    %eax,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  imul   %ebp,%ebx
  add    %ebx,%r14d
  lea    0x0(%rbp,%rbp,1),%ebx
  cmp    %r12d,%ebx
  jle    L50
L54:
  cmp    %r12d,%ebx
  setg   %al
  movzbl %al,%eax
  cmp    %ebp,%eax
  jl     L77
  sub    %ebp,%r12d
  lea    0x1(%r14,%r12,1),%r14d
  lea    D96(%rip),%rdi
  call   <puts@plt>
  jmp    L77
L74:
  mov    %r13d,%r14d
L23:
  mov    %r14d,%esi
  lea    D64(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L91
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L91:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D101
type_infered: string
name: read_data
arguments: tag: D78
type_infered: string
name: read_data
arguments: tag: D40
type_infered: string
name: read_data
arguments: tag: D30
type_infered: string
name: read_data
arguments: tag: D23
type_infered: string
name: read_data
arguments: tag: D104
type_infered: string
name: read_data
arguments: tag: D11
type_infered: string
name: read_data
arguments: tag: D94
type_infered: string
name: read_data
arguments: tag: D6
type_infered: string
name: read_data
arguments: tag: D96
type_infered: string
name: read_data
arguments: tag: D64
type_infered: string

tool: D101: Enter a number: 

tool: D78: %d

tool: D40: the digit is: %d



tool: D30: A

tool: D23: The n1 is: %d



tool: D104: The base is: %d


tool: D11: The fra is: %d


tool: D94: The mod is: %d


tool: D6: The quan is: %d


tool: D96: B

tool: D64: 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;
}
```