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
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%rsi
  lea    D49(%rip),%rdi
  call   <scanf@plt>
  mov    0x18(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x18(%rsp)
  movl   $0x1,0xc(%rsp)
  mov    $0x0,%r12d
  test   %eax,%eax
  jne    L15
L56:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  mov    $0x0,%eax
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L76:
  lea    D42(%rip),%rdi
  call   <puts@plt>
L39:
  mov    0x18(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x18(%rsp)
  test   %eax,%eax
  je     L56
  mov    %ebx,0xc(%rsp)
L15:
  lea    0x20(%rsp),%rdx
  lea    0x170(%rsp),%rsi
  lea    0x2c0(%rsp),%rcx
  lea    0x164(%rsp),%rdi
L36:
  mov    $0x0,%eax
L86:
  movl   $0x0,(%rdx,%rax,1)
  movl   $0x0,(%rsi,%rax,1)
  movl   $0x0,(%rcx,%rax,1)
  add    $0x4,%rax
  cmp    $0x24,%rax
  jne    L86
  add    $0x24,%rdx
  add    $0x24,%rsi
  add    $0x24,%rcx
  cmp    %rdi,%rdx
  jne    L36
  mov    $0x1,%ebp
  mov    $0x0,%r14d
  lea    0x1c(%rsp),%r15
L26:
  mov    $0x0,%ebx
  movslq %r14d,%rax
  lea    (%rax,%rax,8),%r13
L33:
  mov    %r15,%rsi
  lea    D49(%rip),%rdi
  call   <scanf@plt>
  mov    %ebx,%esi
  mov    %r14d,%edi
  call   <getQuad@plt>
  mov    %eax,%edx
  mov    0x1c(%rsp),%eax
  sub    $0x1,%eax
  movslq %eax,%rcx
  add    %r13,%rcx
  cmpl   $0x0,0x170(%rsp,%rcx,4)
  cmovne %r12d,%ebp
  movslq %eax,%rsi
  movslq %ebx,%rcx
  lea    (%rcx,%rcx,8),%rcx
  add    %rsi,%rcx
  cmpl   $0x0,0x20(%rsp,%rcx,4)
  cmovne %r12d,%ebp
  movslq %edx,%rcx
  lea    (%rcx,%rcx,8),%rcx
  add    %rsi,%rcx
  cmpl   $0x0,0x2c0(%rsp,%rcx,4)
  cmovne %r12d,%ebp
  mov    %rsi,%rax
  lea    0x0(%r13,%rsi,1),%rcx
  movl   $0x1,0x170(%rsp,%rcx,4)
  movslq %ebx,%rcx
  lea    (%rcx,%rcx,8),%rcx
  add    %rsi,%rcx
  movl   $0x1,0x20(%rsp,%rcx,4)
  movslq %edx,%rdx
  lea    (%rdx,%rdx,8),%rdx
  add    %rdx,%rax
  movl   $0x1,0x2c0(%rsp,%rax,4)
  add    $0x1,%ebx
  cmp    $0x9,%ebx
  jne    L33
  add    $0x1,%r14d
  cmp    $0x9,%r14d
  jne    L26
  mov    0xc(%rsp),%esi
  lea    0x1(%rsi),%ebx
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  je     L76
  lea    D3(%rip),%rdi
  call   <puts@plt>
  jmp    L39
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D49
datatype: string
name: retrieve_data
arguments: data_label: D42
datatype: string
name: retrieve_data
arguments: data_label: D1
datatype: string
name: retrieve_data
arguments: data_label: D3
datatype: string

tool: D49: %d

tool: D42: NAO


tool: D1: Instancia %d


tool: D3: SIM


assistant: ```c
int main() {
  int mat[9][9], columns[9][9], rows[9][9], quads[9][9];
  int i, j, n, k = 1, c, flag, nQuad;
  scanf("%d", &n);
  while (n--) {
    flag = 1;
    for (i = 0; i < 9; i++) {
      for (j = 0; j < 9; j++) {
        columns[i][j] = 0;
        rows[i][j] = 0;
        quads[i][j] = 0;
      }
    }
    for (i = 0; i < 9; i++) {
      for (j = 0; j < 9; j++) {
        scanf("%d", &c);
        nQuad = getQuad(i, j);
        if (rows[i][c - 1]) {
          flag = 0;
        }
        if (columns[j][c - 1]) {
          flag = 0;
        }
        if (quads[nQuad][c - 1]) {
          flag = 0;
        }
        rows[i][c - 1] = 1;
        columns[j][c - 1] = 1;
        quads[nQuad][c - 1] = 1;
      }
    }
    printf("Instancia %d\n", k++);
    if (flag) {
      printf("SIM\n\n");
    } else {
      printf("NAO\n\n");
    }
  }
  return 0;
}
```