user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    D22(%rip),%r13
  push   %r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x10(%rsp),%r15
  nopl   0x0(%rax,%rax,1)
L35:
  mov    %r12,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    (%rsp),%esi
  test   %esi,%esi
  je     L139
  pxor   %xmm0,%xmm0
  lea    0x8(%rsp),%rdx
  lea    0xc(%rsp),%rcx
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  mov    %r15,%r8
  lea    D18(%rip),%rdi
  movq   $0x0,0x30(%rsp)
  movaps %xmm0,0x20(%rsp)
  call   <scanf@plt>
  mov    (%rsp),%eax
  sub    $0x1,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,8)
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,8)
  mov    0x8(%rsp),%eax
  sub    $0x1,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,8)
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,8)
  mov    0x10(%rsp),%eax
  sub    $0x1,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,8)
  mov    0x20(%rsp),%eax
  mov    0x28(%rsp),%edx
  test   %eax,%eax
  jle    L76
  test   %edx,%edx
  jle    L26
  mov    0x30(%rsp),%ecx
  test   %ecx,%ecx
  jle    L26
L75:
  movl   $0x3,0x24(%rsp)
  movl   $0x3,0x2c(%rsp)
  movl   $0x3,0x34(%rsp)
L2:
  mov    %r12,%r14
  lea    0x14(%rsp),%rbx
L48:
  mov    (%r14),%eax
  mov    %rbp,%rdi
  add    $0x4,%r14
  sub    $0x1,%eax
  cltq
  mov    0x24(%rsp,%rax,8),%esi
  call   <printf@plt>
  cmp    %r14,%rbx
  jne    L48
  jmp    L35
  nopl   0x0(%rax)
L26:
  cmp    $0x5,%eax
  je     L75
L76:
  cmp    $0x5,%edx
  je     L75
  mov    0x30(%rsp),%ecx
  cmp    $0x5,%ecx
  je     L75
  test   %edx,%edx
  setg   %dl
  test   %eax,%eax
  setg   %al
  jle    L148
  test   %dl,%dl
  je     L148
  movl   $0x1,0x24(%rsp)
  movl   $0x2,0x2c(%rsp)
  jmp    L2
  nopl   (%rax)
L148:
  test   %ecx,%ecx
  jle    L2
  test   %dl,%dl
  je     L114
  movl   $0x1,0x2c(%rsp)
  movl   $0x2,0x34(%rsp)
  jmp    L2
  nopl   (%rax)
L139:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L66
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L114:
  test   %al,%al
  je     L2
  movl   $0x1,0x34(%rsp)
  movl   $0x2,0x24(%rsp)
  jmp    L2
L66:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D22
type_infered: string
name: parse_data
arguments: tag: D3
type_infered: string
name: parse_data
arguments: tag: D18
type_infered: string

tool: D22: %d

tool: D3: %d


tool: D18: %d %d %d %d

assistant: ```c
int main(void) {
  int hand[5], kind[3][2], i, num;
  while (1) {
    scanf("%d", &hand[0]);
    if (hand[0] == 0)
      break;
    for (i = 0; i < 3; i++)
      kind[i][0] = 0, kind[i][1] = 0;
    scanf("%d %d %d %d", &hand[1], &hand[2], &hand[3], &hand[4]);
    for (i = 0; i < 5; i++)
      kind[hand[i] - 1][0]++;
    if ((kind[0][0] > 0 && kind[1][0] > 0 && kind[2][0] > 0) || kind[0][0] == 5 || kind[1][0] == 5 || kind[2][0] == 5) {
      kind[0][1] = 3, kind[1][1] = 3, kind[2][1] = 3;
    } else if (kind[0][0] > 0 && kind[1][0] > 0) {
      kind[0][1] = 1, kind[1][1] = 2;
    } else if (kind[1][0] > 0 && kind[2][0] > 0) {
      kind[1][1] = 1, kind[2][1] = 2;
    } else if (kind[2][0] > 0 && kind[0][0] > 0) {
      kind[2][1] = 1, kind[0][1] = 2;
    }
    for (i = 0; i < 5; i++) {
      printf("%d\n", kind[hand[i] - 1][1]);
    }
  }
  return 0;
}
```