user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x6000(%rsp),%r11
L42:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L42
  sub    $0x958,%rsp
  mov    $0x1,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x6948(%rsp)
  xor    %eax,%eax
  lea    0x2a(%rsp),%rax
  lea    0x2b(%rsp),%rbx
  mov    %rax,0x8(%rsp)
  lea    0x2c(%rsp),%rax
  mov    %rax,0x10(%rsp)
  lea    0x30(%rsp),%rax
  mov    %rax,0x18(%rsp)
  nopl   0x0(%rax)
L44:
  mov    0x8(%rsp),%rdx
  mov    0x10(%rsp),%rsi
  mov    %rbx,%rcx
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x3,%eax
  jne    L111
  cmp    $0x1,%ebp
  je     L17
  mov    $0xa,%edi
  call   <putchar@plt>
L17:
  movslq 0x2c(%rsp),%rbp
  test   %ebp,%ebp
  jle    L74
  mov    %ebp,%edi
  mov    0x18(%rsp),%rsi
  lea    -0x1(%rbp),%r13d
  xor    %r8d,%r8d
  sar    %edi
  shr    %r13d
  movzbl 0x2b(%rsp),%r15d
  movsbl 0x2a(%rsp),%r14d
  add    $0x1,%edi
  add    $0x2,%r13d
  mov    $0x1,%r11d
  movslq %edi,%rdi
  imul   $0x14c,%rdi,%rax
  imul   $0xfffffffffffffeb8,%rdi,%rdi
  lea    (%rsi,%rax,1),%r12
  lea    0x34(%rsp,%rax,1),%r9
  mov    %rdi,%r10
  neg    %r10
  nopw   0x0(%rax,%rax,1)
L30:
  test   $0x1,%r11b
  movsbl %r15b,%edx
  lea    -0x4(%r9),%rcx
  mov    %r12,%rax
  cmovne %r14d,%edx
  nopl   0x0(%rax,%rax,1)
L117:
  mov    %edx,(%rcx,%r8,4)
  lea    (%rax,%rdi,1),%rsi
  add    $0x4,%rax
  add    $0x148,%rcx
  mov    %edx,-0x4(%rax)
  mov    %edx,(%rsi,%r10,1)
  mov    %edx,-0x148(%rcx)
  cmp    %r9,%rax
  jne    L117
  add    $0x1,%r11d
  sub    $0x14c,%r12
  sub    $0x2,%r8
  add    $0x148,%rdi
  add    $0x148,%r10
  lea    -0x144(%rax),%r9
  cmp    %r13d,%r11d
  jne    L30
  cmp    $0x1,%ebp
  je     L82
  movl   $0x20,0x17c(%rsp)
  movslq %ebp,%rax
  lea    (%rax,%rax,4),%rdx
  lea    (%rax,%rdx,8),%rdx
  movl   $0x20,0x34(%rsp,%rdx,8)
  movl   $0x20,0x178(%rsp,%rax,4)
  imul   $0x14c,%rax,%rax
  movl   $0x20,0x30(%rsp,%rax,1)
L82:
  lea    0x178(%rsp),%r12
  mov    $0x1,%r13d
  nopl   0x0(%rax)
L87:
  mov    $0x1,%r14d
  test   %ebp,%ebp
  jle    L10
  nopw   0x0(%rax,%rax,1)
L43:
  mov    (%r12,%r14,4),%edi
  add    $0x1,%r14
  call   <putchar@plt>
  cmp    %r14d,0x2c(%rsp)
  jge    L43
L10:
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    $0x148,%r12
  call   <putchar@plt>
  mov    0x2c(%rsp),%ebp
  cmp    %r13d,%ebp
  jge    L87
  xor    %ebp,%ebp
  jmp    L44
L111:
  mov    0x6948(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x6958,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L74:
  movl   $0x20,0x17c(%rsp)
  lea    0x0(%rbp,%rbp,4),%rax
  lea    0x0(%rbp,%rax,8),%rax
  movl   $0x20,0x34(%rsp,%rax,8)
  movl   $0x20,0x178(%rsp,%rbp,4)
  imul   $0x14c,%rbp,%rbp
  movl   $0x20,0x30(%rsp,%rbp,1)
  xor    %ebp,%ebp
  jmp    L44
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D6
type_guessed: string

tool: D6: %d %c %c

assistant: ```c
int main() {
  int outPutBuf[82][82], n, i, j, k, x, y, firstCase;
  char a, b, c;
  firstCase = 1;
  while (scanf("%d %c %c", &n, &a, &b) == 3) {
    if (firstCase == 1)
      firstCase = 0;
    else
      printf("\n");
    for (i = 1, j = 1; i <= n; i += 2, j++) {
      x = n / 2 + 1;
      y = x;
      x -= j - 1;
      y -= j - 1;
      c = j % 2 == 1 ? a : b;
      for (k = 1; k <= i; k++) {
        outPutBuf[x + k - 1][y] = c;
        outPutBuf[x][y + k - 1] = c;
        outPutBuf[x + i - 1][y + k - 1] = c;
        outPutBuf[x + k - 1][y + i - 1] = c;
      }
    }
    if (n != 1) {
      outPutBuf[1][1] = ' ';
      outPutBuf[n][1] = ' ';
      outPutBuf[1][n] = ' ';
      outPutBuf[n][n] = ' ';
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        printf("%c", outPutBuf[i][j]);
      }
      printf("\n");
    }
  }
  return 0;
}
```