user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    -0x3c(%rbp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  movslq %eax,%rdx
  shl    $0x2,%rdx
  lea    0xf(%rdx),%rcx
  mov    %rcx,%rdi
  and    $0xfffffffffffffff0,%rdi
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rsi
  sub    %rcx,%rsi
L149:
  cmp    %rsi,%rsp
  je     L65
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L149
L65:
  mov    %rdi,%rcx
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  je     L18
  orq    $0x0,-0x8(%rsp,%rcx,1)
L18:
  mov    %rsp,%r14
  add    $0xb,%rdx
  mov    %rdx,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rdx
  mov    %rsp,%rcx
  sub    %rdx,%rcx
L110:
  cmp    %rcx,%rsp
  je     L163
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L110
L163:
  mov    %rsi,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L38
  orq    $0x0,-0x8(%rsp,%rdx,1)
L38:
  lea    0x3(%rsp),%r13
  mov    %r13,%rbx
  shr    $0x2,%rbx
  mov    %rbx,-0x48(%rbp)
  and    $0xfffffffffffffffc,%r13
  test   %eax,%eax
  jle    L159
  mov    %r14,%r12
  mov    $0x0,%ebx
  lea    D18(%rip),%r15
L69:
  mov    %r12,%rsi
  mov    %r15,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    -0x3c(%rbp),%eax
  add    $0x4,%r12
  cmp    %ebx,%eax
  jg     L69
L159:
  cmp    $0x1,%eax
  jle    L105
  lea    -0x2(%rax),%ecx
  mov    $0x0,%eax
L102:
  mov    0x4(%r14,%rax,4),%edx
  sub    (%r14,%rax,4),%edx
  mov    %edx,0x0(%r13,%rax,4)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rcx,%rdx
  jne    L102
L105:
  mov    -0x48(%rbp),%rax
  mov    0x4(,%rax,4),%esi
  mov    0x0(,%rax,4),%edi
  call   <gcd@plt>
  mov    %eax,%edi
  cmpl   $0x1,-0x3c(%rbp)
  jle    L76
  mov    $0x0,%ebx
L77:
  mov    0x0(%r13,%rbx,4),%esi
  call   <gcd@plt>
  mov    %eax,%edi
  mov    -0x3c(%rbp),%eax
  lea    -0x1(%rax),%edx
  add    $0x1,%rbx
  cmp    %ebx,%edx
  jg     L77
  test   %edx,%edx
  jle    L14
  mov    %r13,%rcx
  lea    -0x1(%rax),%eax
  lea    0x0(%r13,%rax,4),%r8
  mov    $0x0,%esi
L139:
  mov    (%rcx),%eax
  cltd
  idiv   %edi
  lea    -0x1(%rsi,%rax,1),%esi
  add    $0x4,%rcx
  cmp    %rcx,%r8
  jne    L139
L117:
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L88
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L76:
  mov    $0x0,%esi
  jmp    L117
L14:
  mov    $0x0,%esi
  jmp    L117
L88:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
datatype: string

tool: D18: %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;
}
```