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    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xea8,%rsp
  lea    D11(%rip),%r12
  lea    D9(%rip),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x3e98(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  nop
L68:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L77
  xor    %ebx,%ebx
  lea    0x10(%rsp),%r14
  xchg   %ax,%ax
L74:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %eax,(%r14,%rbx,4)
  mov    %ebx,%ebp
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  add    $0x1,%rbx
  test   %eax,%eax
  jne    L74
  test   %ebp,%ebp
  je     L49
  mov    $0x1,%edi
  mov    0x10(%rsp),%ecx
  cmovg  %ebp,%edi
  cmp    $0x3,%ebp
  jle    L40
  movd   %ecx,%xmm7
  lea    0x1f50(%rsp),%r8
  mov    %edi,%esi
  shr    $0x2,%esi
  pshufd $0x0,%xmm7,%xmm4
  lea    0x14(%rsp),%rdx
  mov    %r8,%rax
  sub    $0x1,%esi
  pxor   %xmm5,%xmm5
  shl    $0x4,%rsi
  lea    0x24(%rsp,%rsi,1),%rsi
  nopl   0x0(%rax,%rax,1)
L37:
  movdqu (%rdx),%xmm3
  movdqa %xmm5,%xmm0
  movdqu (%rdx),%xmm6
  movdqa %xmm4,%xmm2
  add    $0x10,%rdx
  add    $0x10,%rax
  psubd  %xmm4,%xmm3
  psubd  %xmm6,%xmm2
  pcmpgtd %xmm3,%xmm0
  movdqa %xmm0,%xmm1
  pandn  -0x10(%rax),%xmm1
  pand   %xmm0,%xmm2
  por    %xmm2,%xmm1
  pand   %xmm0,%xmm1
  pandn  %xmm3,%xmm0
  por    %xmm1,%xmm0
  movaps %xmm0,-0x10(%rax)
  cmp    %rsi,%rdx
  jne    L37
  mov    %edi,%edx
  and    $0xfffffffc,%edx
  lea    0x1(%rdx),%eax
  cmp    %edx,%edi
  je     L2
L111:
  movslq %eax,%rdi
  lea    -0x1(%rax),%edx
  mov    0x10(%rsp,%rdi,4),%esi
  movslq %edx,%rdx
  mov    %esi,%r9d
  sub    %ecx,%r9d
  js     L139
  mov    %r9d,0x1f50(%rsp,%rdx,4)
L121:
  lea    0x1(%rax),%r10d
  cmp    %eax,%ebp
  jle    L2
  movslq %r10d,%rsi
  mov    %ecx,%r9d
  mov    0x10(%rsp,%rsi,4),%edx
  sub    %edx,%r9d
  sub    %ecx,%edx
  cmovs  %r9d,%edx
  add    $0x2,%eax
  mov    %edx,0x1f50(%rsp,%rdi,4)
  cmp    %r10d,%ebp
  jle    L2
  cltq
  mov    %ecx,%edx
  mov    0x10(%rsp,%rax,4),%eax
  sub    %eax,%edx
  sub    %ecx,%eax
  cmovs  %edx,%eax
  mov    %eax,0x1f50(%rsp,%rsi,4)
L2:
  mov    %ebp,%esi
  mov    %r8,%rdi
  call   <gcdOfArr@plt>
  mov    %r15,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  jmp    L68
  nopl   0x0(%rax)
L139:
  mov    %ecx,%r9d
  sub    %esi,%r9d
  mov    %r9d,0x1f50(%rsp,%rdx,4)
  jmp    L121
L77:
  mov    0x3e98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L118
  add    $0x3ea8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L49:
  lea    0x1f50(%rsp),%r8
  jmp    L2
L40:
  mov    $0x1,%eax
  lea    0x1f50(%rsp),%r8
  jmp    L111
L118:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D11
type_infered: string
name: parse_data
arguments: data_tag: D9
type_infered: string

tool: D11: %d

tool: D9: %d


assistant: ```c
int main(void) {
  int arr[2000], diffArr[2000], num, mSize, i;
  while (1) {
    mSize = 0;
    scanf("%d", &num);
    if (num == 0)
      break;
    do {
      arr[mSize] = num;
      mSize++;
      scanf("%d", &num);
    } while (num != 0);
    for (i = 1; i < mSize; i++) {
      diffArr[i - 1] = arr[i] - arr[0];
      if (diffArr[i - 1] < 0)
        diffArr[i - 1] = -diffArr[i - 1];
    }
    printf("%d\n", gcdOfArr(diffArr, mSize - 1));
  }
  return 0;
}
```