user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<findMaxPldrmSubstr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x4(,%rax,8),%edi
  call   <malloc@plt>
  mov    %rax,%r10
  movl   $0x1,(%rax)
  mov    $0x1,%ebp
  cmpb   $0x0,(%rbx)
  je     L100
  lea    0x4(%rax),%rdi
  mov    $0x1,%ebp
  mov    $0x0,%r8d
  mov    $0x0,%r9d
  mov    $0x1,%ecx
  mov    $0x1,%r12d
  mov    $0x3,%edx
  mov    $0x0,%eax
  mov    $0x0,%r11d
  jmp    L53
L111:
  lea    0x4(%rax),%edx
  add    $0x1,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  lea    0x1(%rax,%rax,1),%edx
  jmp    L8
L50:
  lea    -0x1(%rdx,%rdx,1),%edx
  mov    %edx,(%rdi)
  cmp    %edx,%ebp
  cmovl  %edx,%ebp
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  add    %ecx,%eax
  cmp    %r8d,%eax
  jle    L117
  mov    %eax,%r8d
  mov    %ecx,%r9d
L117:
  add    $0x1,%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  cltq
  cmpb   $0x0,(%rbx,%rax,1)
  je     L100
  mov    %ecx,%r12d
  and    $0x1,%r12d
  cmp    $0x1,%r12d
  sbb    %edx,%edx
  and    $0xfffffffe,%edx
  add    $0x3,%edx
  add    $0x4,%rdi
  mov    %r11d,%eax
  cmp    %ecx,%r8d
  jle    L53
  movslq %r9d,%rax
  mov    %ecx,%esi
  sub    %r9d,%esi
  add    %esi,%esi
  mov    (%r10,%rax,4),%eax
  sub    %esi,%eax
  lea    (%r9,%r9,1),%esi
  sub    %ecx,%esi
  movslq %esi,%rsi
  mov    (%r10,%rsi,4),%esi
  cmp    %esi,%eax
  cmovg  %esi,%eax
L53:
  cmp    %edx,%eax
  cmovl  %edx,%eax
  test   %r12d,%r12d
  je     L111
  lea    0x6(%rax),%edx
  add    $0x3,%eax
  cmovns %eax,%edx
  sar    $0x2,%edx
  add    %edx,%edx
L8:
  mov    %ecx,%esi
  cmp    %edx,%ecx
  jl     L50
L90:
  mov    %ecx,%r12d
  sub    %edx,%r12d
  lea    (%rdx,%rcx,1),%r13d
  mov    %r13d,%eax
  shr    $0x1f,%eax
  add    %r13d,%eax
  sar    %eax
  cltq
  movzbl (%rbx,%rax,1),%r13d
  test   %r13b,%r13b
  je     L50
  mov    %r12d,%eax
  shr    $0x1f,%eax
  add    %r12d,%eax
  sar    %eax
  cltq
  cmp    (%rbx,%rax,1),%r13b
  jne    L50
  add    $0x2,%edx
  cmp    %edx,%esi
  jns    L90
  jmp    L50
L100:
  mov    %r10,%rdi
  call   <free@plt>
  sub    $0x1,%ebp
  mov    %ebp,%eax
  shr    $0x1f,%eax
  add    %ebp,%eax
  sar    %eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int findMaxPldrmSubstr(char *str) {
  int length = strlen(str) * 2 + 1;
  int *f = (int *)malloc(sizeof(int) * length);
  int i, k, t, position = 1, j = 0, maxIcalcedByJ = 0, maxPldrmStrLength = 1;
  f[0] = 1;
  for (i = 1; str[i / 2]; i++) {
    k = i % 2 ? 3 : 1;
    t = 0;
    if (i < maxIcalcedByJ) {
      t = ((f[2 * j - i]) < (f[j] - 2 * (i - j)) ? (f[2 * j - i]) : (f[j] - 2 * (i - j)));
    }
    k = ((k) > (t) ? (k) : (t));
    k = i % 2 ? (k + 3) / 4 * 2 : (k + 1) / 4 * 2 + 1;
    while (i - k >= 0 && str[(i + k) / 2] && str[(i - k) / 2] == str[(i + k) / 2])
      k += 2;
    f[i] = k * 2 - 1;
    if (maxPldrmStrLength < f[i]) {
      position = i;
      maxPldrmStrLength = f[i];
    }
    if (i + f[i] / 2 > maxIcalcedByJ) {
      j = i;
      maxIcalcedByJ = i + f[i] / 2;
    }
  }
  free(f);
  return (maxPldrmStrLength - 1) / 2;
}
```