user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x3c(%rbp),%rsi
  push   %r13
  lea    D9(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rax
  mov    %rsp,%rdi
  mov    %rax,%rsi
  shl    $0x2,%rax
  lea    0xf(%rax),%rdx
  mov    %rdx,%rcx
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rdi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rdi,%rsp
  je     L108
L185:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdi,%rsp
  jne    L185
L108:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L151
L120:
  lea    0x3(%rsp),%r12
  add    $0xb,%rax
  mov    %rsp,%rcx
  mov    %r12,%rbx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  and    $0xfffffffffffffffc,%r12
  shr    $0x2,%rbx
  and    $0xfffffffffffffff0,%rdx
  sub    %rax,%rcx
  mov    %rbx,-0x48(%rbp)
L99:
  cmp    %rcx,%rsp
  je     L114
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L99
L151:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L120
L114:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L181
L103:
  lea    0x3(%rsp),%rbx
  mov    %rbx,%rax
  and    $0xfffffffffffffffc,%rbx
  shr    $0x2,%rax
  mov    %rax,-0x50(%rbp)
  test   %esi,%esi
  jle    L95
  mov    %r12,%r14
  xor    %r15d,%r15d
  nopl   0x0(%rax)
L100:
  mov    %r14,%rsi
  mov    %r13,%rdi
  add    $0x1,%r15d
  add    $0x4,%r14
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  cmp    %r15d,%eax
  jg     L100
  cmp    $0x1,%eax
  jle    L95
  mov    -0x48(%rbp),%rdx
  lea    -0x2(%rax),%edi
  mov    $0x1,%eax
  add    $0x2,%rdi
  mov    0x0(,%rdx,4),%edx
  nopw   0x0(%rax,%rax,1)
L191:
  mov    %edx,%esi
  mov    (%r12,%rax,4),%edx
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,-0x4(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L191
L95:
  mov    -0x50(%rbp),%rax
  xor    %r14d,%r14d
  mov    0x0(,%rax,4),%r12d
  mov    0x4(,%rax,4),%esi
  mov    %r12d,%edi
  call   <gcd@plt>
  cmpl   $0x1,-0x3c(%rbp)
  mov    %r12d,%esi
  mov    %eax,%edi
  jg     L129
  jmp    L180
  nopw   0x0(%rax,%rax,1)
L13:
  mov    (%rbx,%r14,4),%esi
L129:
  call   <gcd@plt>
  add    $0x1,%r14
  mov    %eax,%edi
  mov    -0x3c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %r14d,%eax
  jg     L13
  test   %eax,%eax
  jle    L180
  lea    0x4(%rbx),%rcx
  lea    (%rbx,%rax,4),%r8
  xor    %esi,%esi
  jmp    L86
  nopw   0x0(%rax,%rax,1)
L112:
  mov    (%rcx),%r12d
  add    $0x4,%rcx
L86:
  mov    %r12d,%eax
  cltd
  idiv   %edi
  lea    -0x1(%rsi,%rax,1),%esi
  cmp    %rcx,%r8
  jne    L112
L164:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L76
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L180:
  xor    %esi,%esi
  jmp    L164
L181:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L103
L76:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D9
type: string

tool: D9: %d

assistant: ```c
int main() {
  int num, i, j, gcd1, sum = 0;
  scanf("%d", &num);
  int input[num];
  int diff[num - 1];
  for (i = 0; i < num; i++) {
    scanf("%d", &input[i]);
  }
  for (i = 0; i < num - 1; i++) {
    diff[i] = input[i + 1] - input[i];
  }
  gcd1 = gcd(diff[0], diff[1]);
  for (i = 0; i < num - 1; i++) {
    gcd1 = gcd(gcd1, diff[i]);
  }
  for (i = 0; i < num - 1; i++) {
    sum += ((diff[i] / gcd1) - 1);
  }
  printf("%d", sum);
  return 0;
}
```