user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x1f4,%ecx
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfc8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfb8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  lea    0x4(%rsp),%rdx
  rep stos %rax,%es:(%rdi)
  lea    D15(%rip),%rdi
  lea    0x8(%rsp),%rcx
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  mov    0x8(%rsp),%esi
  mov    0xc(%rsp),%r8d
  cmp    %esi,%edi
  mov    %esi,%r9d
  cmovle %edi,%r9d
  cmp    %r8d,%r9d
  jg     L78
  cmp    $0x1,%esi
  jne    L46
  cmp    $0x1,%edi
  lea    0x14(%rsp),%rdx
  mov    $0x2,%eax
  setne  %cl
  test   %cl,%cl
  jne    L46
  nopl   0x0(%rax,%rax,1)
L104:
  mov    %eax,%ecx
  movl   $0x1,(%rdx)
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %ecx,%r8d
  jge    L104
L78:
  mov    $0x1,%ebp
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L9:
  mov    0x4(%rsp),%edi
  mov    0x8(%rsp),%esi
  mov    %edi,%ebx
  imul   %esi,%ebx
  call   <gy@plt>
  mov    0x4(%rsp),%edi
  mov    0x8(%rsp),%esi
  mov    %eax,%r8d
  imul   %ebp,%ebx
  add    $0x1,%ebp
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  cltq
  movl   $0x0,0x10(%rsp,%rax,4)
L1:
  mov    %esi,%ebx
  imul   %edi,%ebx
  call   <gy@plt>
  mov    0xc(%rsp),%ecx
  mov    %eax,%r8d
  imul   %ebp,%ebx
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  cmp    %ecx,%eax
  jle    L9
  movslq %ecx,%rax
  test   %ecx,%ecx
  jg     L158
  jmp    L113
  nopw   0x0(%rax,%rax,1)
L101:
  sub    $0x1,%rax
  test   %eax,%eax
  jle    L45
L158:
  cmpl   $0x1,(%r12,%rax,4)
  mov    %eax,%ebp
  jne    L101
L45:
  mov    $0x1,%ebx
  lea    D12(%rip),%r13
  jmp    L70
  nop
L119:
  add    $0x1,%rbx
  cmp    %ebx,%ecx
  jl     L113
L70:
  cmpl   $0x1,(%r12,%rbx,4)
  mov    %ebx,%esi
  jne    L119
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    %ebx,%ebp
  je     L64
  mov    $0x20,%edi
  call   <putchar@plt>
L64:
  mov    0xc(%rsp),%ecx
  add    $0x1,%rbx
  cmp    %ebx,%ecx
  jge    L70
  nopl   0x0(%rax,%rax,1)
L113:
  mov    0xfb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L154
  add    $0xfc8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  movslq %edi,%rbx
  movslq %esi,%r11
  lea    (%r9,%r9,1),%eax
  shl    $0x2,%rbx
  shl    $0x2,%r11
  lea    (%r12,%rbx,1),%rcx
  lea    (%r12,%r11,1),%rdx
  nopl   (%rax)
L67:
  mov    %eax,%r10d
  movl   $0x1,(%rcx)
  add    %r9d,%eax
  add    %rbx,%rcx
  movl   $0x1,(%rdx)
  add    %r11,%rdx
  cmp    %r10d,%r8d
  jge    L67
  jmp    L78
L154:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D15
type_guessed: string
name: read_data
arguments: data_tag: D12
type_guessed: string

tool: D15: %d%d%d

tool: D12: %d

assistant: ```c
main() {
  int m, n, k, i, j, leng;
  int a[1000] = {0};
  scanf("%d%d%d", &k, &m, &n);
  i = 1;
  j = m > n ? n : m;
  while (i * j <= k) {
    a[i * m] = 1;
    a[i * n] = 1;
    i++;
  }
  for (i = 1; m * n * i / gy(m, n) <= k; i++) {
    a[m * n * i / gy(m, n)] = 0;
  }
  for (i = k; i >= 1; i--)
    if (a[i] == 1) {
      leng = i;
      break;
    }
  for (i = 1; i <= k; i++)
    if (a[i] == 1) {
      printf("%d", i);
      if (i != leng)
        putchar(' ');
    }
  return 0;
}
```