user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x3c(%rbp),%rsi
  push   %r13
  push   %r12
  lea    D17(%rip),%r12
  push   %rbx
  mov    %r12,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rbx
  mov    %rsp,%rsi
  mov    %rbx,%rax
  mov    %rbx,%rcx
  imul   %rbx,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L151
L108:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L108
L151:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L50
L0:
  mov    %rsp,-0x48(%rbp)
  test   %ecx,%ecx
  jle    L165
  mov    %rsp,%r13
  xor    %r15d,%r15d
  nopw   0x0(%rax,%rax,1)
L184:
  mov    %r13,%rsi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <scanf@plt>
  mov    -0x3c(%rbp),%r14d
  add    %rbx,%r13
  cmp    %r15d,%r14d
  jg     L184
  lea    -0x1(%r14),%r8d
  cmp    $0x1,%r8d
  jle    L193
  mov    -0x48(%rbp),%rax
  lea    (%rbx,%rbx,1),%rdi
  mov    %rbx,%rsi
  xor    %r10d,%r10d
  mov    $0x1,%r13d
  mov    $0x1,%r15d
  lea    (%rax,%rbx,1),%r11
  nopl   0x0(%rax)
L145:
  mov    %r15d,%r9d
  add    $0x1,%r13d
  mov    %r11,%rax
  sub    %r11d,%r9d
  nopl   (%rax)
L123:
  mov    %rax,%rcx
  movzbl 0x1(%rax),%edx
  sub    %rsi,%rcx
  cmp    0x1(%rcx,%rdi,1),%dl
  jle    L27
  cmp    0x1(%rcx,%r10,1),%dl
  jle    L27
  cmp    (%rax),%dl
  jle    L27
  cmp    0x2(%rax),%dl
  jle    L27
  movb   $0x64,0x1(%rax)
  nopl   0x0(%rax)
L27:
  add    $0x1,%rax
  lea    (%r9,%rax,1),%edx
  cmp    %edx,%r8d
  jg     L123
  add    %rbx,%rsi
  add    %rbx,%r10
  add    %rbx,%r11
  add    %rbx,%rdi
  cmp    %r8d,%r13d
  jl     L145
L193:
  test   %r14d,%r14d
  jle    L165
  mov    -0x48(%rbp),%r13
  xor    %r14d,%r14d
  nopw   0x0(%rax,%rax,1)
L40:
  xor    %r15d,%r15d
  nopl   0x0(%rax,%rax,1)
L120:
  movsbl 0x0(%r13,%r15,1),%esi
  cmp    $0x64,%sil
  je     L126
  sub    $0x30,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15d,-0x3c(%rbp)
  jg     L120
L159:
  mov    $0xa,%edi
  add    $0x1,%r14d
  add    %rbx,%r13
  call   <putchar@plt>
  cmp    %r14d,-0x3c(%rbp)
  jg     L40
L165:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L126:
  mov    $0x58,%edi
  add    $0x1,%r15
  call   <putchar@plt>
  cmp    %r15d,-0x3c(%rbp)
  jg     L120
  jmp    L159
L50:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L0
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D17
datatype: string
name: retrieve_data
arguments: data_tag: D9
datatype: string

tool: D17: %d

tool: D9: %s

assistant: ```c
int main() {
  int n;
  char ch;
  scanf("%d", &n);
  char grid[n][n];
  int i = 0, j = 0;
  for (int k = 0; k < n; k++) {
    scanf("%s", grid[k]);
  }
  for (int i = 1; i < n - 1; i++) {
    for (int j = 1; j < n - 1; j++) {
      if (grid[i][j] > grid[i + 1][j] && grid[i][j] > grid[i - 1][j] && grid[i][j] > grid[i][j - 1] && grid[i][j] > grid[i][j + 1])
        grid[i][j] = 100;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (grid[i][j] == 100)
        printf("X");
      else
        printf("%d", grid[i][j] - '0');
    }
    printf("\n");
  }
  return 0;
}
```