user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D48(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D81(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  js     L100
  mov    $0x2,%ebx
  cmp    $0x3,%eax
  jg     L136
  lea    D68(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L44
L100:
  lea    D84(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L44
L3:
  add    $0x1,%ebx
L136:
  mov    0x4(%rsp),%edi
  call   <sqrt@plt>
  cmp    %eax,%ebx
  jg     L32
  mov    0x4(%rsp),%esi
  mov    %esi,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L3
  mov    %ebx,%edx
  lea    D47(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L49
L32:
  mov    0x4(%rsp),%esi
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L49:
  mov    $0x2,%ebx
L90:
  mov    0x4(%rsp),%edi
  call   <sqrt@plt>
  cmp    %eax,%ebx
  jg     L118
  mov    0x4(%rsp),%esi
  mov    %esi,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L128
  add    $0x1,%ebx
  jmp    L90
L128:
  mov    %ebx,%edx
  lea    D47(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L148
L118:
  mov    0x4(%rsp),%esi
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L148:
  mov    $0x2,%ebx
L133:
  mov    0x4(%rsp),%edi
  mov    %edi,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L70
  add    $0x1,%ebx
  call   <sqrt@plt>
  cmp    %eax,%ebx
  jle    L133
  mov    0x4(%rsp),%esi
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L44
L70:
  mov    %ebx,%edx
  mov    %edi,%esi
  lea    D47(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L44:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D48
type: string
name: unpack_data
arguments: tag: D81
type: string
name: unpack_data
arguments: tag: D68
type: string
name: unpack_data
arguments: tag: D84
type: string
name: unpack_data
arguments: tag: D47
type: string
name: unpack_data
arguments: tag: D36
type: string

tool: D48: Enter the number: 

tool: D81: %d

tool: D68: Number is prime

tool: D84: Number must be positive

tool: D47: Number %d is not prime, denominator is %d



tool: D36: Number %d is prime



assistant: ```c
int main() {
  int number, denominator, is_prime = 1;
  printf("Enter the number: ");
  scanf("%d", &number);
  if (number < 0) {
    printf("Number must be positive");
    return 0;
  }
  if (number <= 3) {
    printf("Number is prime");
    return 0;
  }
  for (denominator = 2; denominator <= (int)sqrt(number); denominator++) {
    if ((number % denominator) == 0) {
      printf("Number %d is not prime, denominator is %d\n\n", number, denominator);
      is_prime = 0;
      break;
    }
  }
  if (is_prime) {
    printf("Number %d is prime\n\n", number);
  }
  is_prime = 1;
  denominator = 2;
  while (denominator <= (int)sqrt(number)) {
    if ((number % denominator) == 0) {
      printf("Number %d is not prime, denominator is %d\n\n", number, denominator);
      is_prime = 0;
      break;
    }
    denominator++;
  }
  if (is_prime) {
    printf("Number %d is prime\n\n", number);
  }
  is_prime = 1;
  denominator = 2;
  do {
    if ((number % denominator) == 0) {
      printf("Number %d is not prime, denominator is %d\n\n", number, denominator);
      is_prime = 0;
      break;
    }
    denominator++;
  } while (denominator <= (int)sqrt(number));
  if (is_prime) {
    printf("Number %d is prime\n\n", number);
  }
  return 0;
}
```