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
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x738,%rsp
  lea    D13(%rip),%rbx
  mov    %rbx,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x2728(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0xc(%rsp),%r12
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  je     L1
L102:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L153
  lea    0x10(%rsp),%r13
  xor    %r15d,%r15d
  lea    D3(%rip),%r14
  mov    %r13,%rbp
  nopw   0x0(%rax,%rax,1)
L115:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <scanf@plt>
  mov    0xc(%rsp),%r9d
  add    $0x64,%rbp
  cmp    %r15d,%r9d
  jg     L115
  test   %r9d,%r9d
  jle    L153
  xor    %r14d,%r14d
  xor    %r11d,%r11d
  lea    0x1(%r13),%r15
  nopw   0x0(%rax,%rax,1)
L77:
  lea    (%r15,%r14,1),%r8
  mov    $0x1,%edi
  movslq %r11d,%rax
  cmp    $0x1,%r9d
  je     L147
  lea    (%rax,%rax,4),%rax
  lea    (%rax,%rax,4),%rax
  lea    0x2730(%rsp,%rax,4),%r10
  nopw   0x0(%rax,%rax,1)
L83:
  movzbl (%r8),%esi
  mov    %edi,%eax
  mov    %r8,%rdx
  nopl   0x0(%rax)
L59:
  movzbl -0x1(%rdx),%ecx
  cmp    %sil,%cl
  jle    L75
  mov    %cl,(%rdx)
  sub    $0x1,%rdx
  sub    $0x1,%eax
  jne    L59
  xor    %eax,%eax
L75:
  cltq
  add    $0x1,%edi
  add    $0x1,%r8
  mov    %sil,-0x2720(%rax,%r10,1)
  cmp    %edi,%r9d
  jne    L83
L147:
  add    $0x1,%r11d
  add    $0x64,%r14
  cmp    %r11d,%r9d
  jne    L77
  lea    -0x1(%r9),%r8d
  mov    %r8,%r9
  add    %r15,%r8
  nopl   0x0(%rax,%rax,1)
L156:
  test   %r9d,%r9d
  je     L117
  movzbl 0x0(%r13),%esi
  mov    %r13,%rcx
  xor    %edi,%edi
  xor    %eax,%eax
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L130:
  cmp    %eax,%r9d
  jle    L19
L158:
  add    $0x64,%rcx
  mov    %edx,%esi
L34:
  movzbl 0x64(%rcx),%edx
  add    $0x1,%eax
  cmp    %sil,%dl
  jge    L130
  cmp    %eax,%r9d
  jle    L39
  mov    $0x1,%edi
  jmp    L158
  nopl   0x0(%rax,%rax,1)
L19:
  cmp    $0x1,%edi
  je     L39
L117:
  add    $0x1,%r13
  cmp    %r8,%r13
  jne    L156
L153:
  lea    D14(%rip),%rdi
  call   <printf@plt>
L178:
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  jne    L102
L1:
  mov    0x2728(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x2738,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L39:
  lea    D28(%rip),%rdi
  call   <printf@plt>
  jmp    L178
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D13
datatype: string
name: unpack_data
arguments: label: D3
datatype: string
name: unpack_data
arguments: label: D14
datatype: string
name: unpack_data
arguments: label: D28
datatype: string

tool: D13: %d

tool: D3: %s

tool: D14: YES


tool: D28: NO


assistant: ```c
int main() {
  int i, j, k, t, n, flag = 0;
  char str[100][100];
  scanf("%d", &t);
  while (t--) {
    flag = 0;
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
      scanf("%s", str[i]);
    }
    for (k = 0; k < n; k++) {
      char current;
      for (i = 1; i < n; i++) {
        current = str[k][i];
        for (j = i - 1; j >= 0; j--) {
          if (current < str[k][j])
            str[k][j + 1] = str[k][j];
          else
            break;
        }
        str[k][j + 1] = current;
      }
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < n - 1; j++) {
        if (str[j][i] <= str[j + 1][i]) {
        } else {
          flag = 1;
        }
      }
      if (flag == 1) {
        break;
      }
    }
    if (flag == 0) {
      printf("YES\n");
    } else
      printf("NO\n");
  }
  return 0;
}
```