user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D36(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x438,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x428(%rsp)
  xor    %eax,%eax
  lea    0x38(%rsp),%rsi
  call   <scanf@plt>
  mov    0x38(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x38(%rsp)
  test   %eax,%eax
  je     L66
  lea    0x40(%rsp),%rax
  movl   $0x1,0x14(%rsp)
  lea    0x3c(%rsp),%r13
  xor    %r15d,%r15d
  mov    %rax,0x28(%rsp)
  lea    0x190(%rsp),%rax
  mov    %rax,0x18(%rsp)
  lea    0x2e0(%rsp),%rax
  mov    %rax,0x20(%rsp)
  nopl   0x0(%rax,%rax,1)
L46:
  mov    0x28(%rsp),%rdi
  xor    %eax,%eax
  mov    $0x28,%ecx
  xor    %ebx,%ebx
  mov    $0x1,%r14d
  rep stos %rax,%es:(%rdi)
  mov    $0x28,%ecx
  movl   $0x0,(%rdi)
  mov    0x18(%rsp),%rdi
  rep stos %rax,%es:(%rdi)
  mov    $0x28,%ecx
  movl   $0x0,(%rdi)
  mov    0x20(%rsp),%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  nopl   0x0(%rax,%rax,1)
L10:
  movslq %ebx,%rax
  xor    %ebp,%ebp
  lea    (%rax,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  xchg   %ax,%ax
L7:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <getQuad@plt>
  mov    0x3c(%rsp),%ecx
  movslq %ebp,%rsi
  lea    (%rsi,%rsi,8),%rsi
  cltq
  lea    -0x1(%rcx),%edx
  mov    0x8(%rsp),%rcx
  lea    (%rax,%rax,8),%rax
  movslq %edx,%rdx
  lea    (%rcx,%rdx,1),%rdi
  mov    0x190(%rsp,%rdi,4),%ecx
  movl   $0x1,0x190(%rsp,%rdi,4)
  test   %ecx,%ecx
  cmovne %r15d,%r14d
  add    %rdx,%rsi
  mov    0x40(%rsp,%rsi,4),%r8d
  movl   $0x1,0x40(%rsp,%rsi,4)
  test   %r8d,%r8d
  cmovne %r15d,%r14d
  add    %rdx,%rax
  mov    0x2e0(%rsp,%rax,4),%r9d
  movl   $0x1,0x2e0(%rsp,%rax,4)
  test   %r9d,%r9d
  cmovne %r15d,%r14d
  add    $0x1,%ebp
  cmp    $0x9,%ebp
  jne    L7
  add    $0x1,%ebx
  cmp    $0x9,%ebx
  jne    L10
  mov    0x14(%rsp),%esi
  xor    %eax,%eax
  lea    D33(%rip),%rdi
  lea    0x1(%rsi),%ebx
  call   <printf@plt>
  test   %r14d,%r14d
  je     L62
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    0x38(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x38(%rsp)
  test   %eax,%eax
  je     L66
L35:
  mov    %ebx,0x14(%rsp)
  jmp    L46
L62:
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    0x38(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x38(%rsp)
  test   %eax,%eax
  jne    L35
L66:
  mov    0x428(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  add    $0x438,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D36
type_infered: string
name: extract_symbolic_data
arguments: label: D33
type_infered: string
name: extract_symbolic_data
arguments: label: D11
type_infered: string
name: extract_symbolic_data
arguments: label: D10
type_infered: string

tool: D36: %d

tool: D33: Instancia %d


tool: D11: SIM


tool: D10: NAO


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;
}
```