user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_get_long_sub_str>:
  endbr64
  push   %r15
  mov    $0xd,%ecx
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %rdi,0x18(%rsp)
  lea    0x20(%rsp),%r13
  mov    %r13,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  rep stos %rax,%es:(%rdi)
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L79
  lea    -0x1(%rax),%eax
  mov    %r15,%rdx
  xor    %ecx,%ecx
  lea    0x1(%r15,%rax,1),%rsi
  nopw   0x0(%rax,%rax,1)
L87:
  movsbl (%rdx),%eax
  sub    $0x61,%eax
  cltq
  cmpl   $0x1,0x20(%rsp,%rax,4)
  adc    $0x0,%ecx
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L87
L30:
  cmp    %ecx,%r12d
  jg     L32
L21:
  mov    0x18(%rsp),%rcx
  movsbl (%rcx),%eax
  sub    $0x61,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,4)
  cmp    $0x1,%ebx
  jle    L10
  lea    -0x2(%rbx),%eax
  mov    $0x1,%r15d
  lea    0x1(%rcx),%rbp
  xor    %ebx,%ebx
  lea    0x2(%rcx,%rax,1),%rax
  movl   $0x0,0x14(%rsp)
  mov    %rax,0x8(%rsp)
  mov    %r15d,%eax
  sub    %ebp,%eax
  mov    %eax,0x10(%rsp)
  nop
L18:
  movsbl 0x0(%rbp),%eax
  sub    $0x61,%eax
  cltq
  addl   $0x1,0x20(%rsp,%rax,4)
  mov    0x10(%rsp),%eax
  lea    (%rax,%rbp,1),%r14d
  jmp    L67
  nopl   0x0(%rax,%rax,1)
L116:
  movsbl 0x0(%rbp),%eax
  add    $0x1,%ebx
  sub    $0x61,%eax
  cltq
  subl   $0x1,0x20(%rsp,%rax,4)
L67:
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <is_valid@plt>
  test   %eax,%eax
  je     L116
  sub    %ebx,%r14d
  cmp    %r15d,%r14d
  jl     L78
  mov    %ebx,0x14(%rsp)
  lea    0x1(%r14),%r15d
L78:
  add    $0x1,%rbp
  cmp    %rbp,0x8(%rsp)
  jne    L18
  mov    0x14(%rsp),%eax
  lea    (%rax,%r15,1),%ecx
  mov    %eax,%edx
L108:
  mov    0x18(%rsp),%rsi
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L80
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L32:
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L21
L79:
  xor    %ecx,%ecx
  jmp    L30
L10:
  movl   $0x0,0x14(%rsp)
  mov    $0x1,%ecx
  mov    0x14(%rsp),%edx
  jmp    L108
L80:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D14
type_infered: string
name: retrieve_data
arguments: tag: D10
type_infered: string

tool: D14: 
Max length sub str of %s starts from %d to %d

tool: D10: 
 Error there are not enough unique characters in the string 

assistant: ```c
void _get_long_sub_str(char *str, int k) {
  int curr_start = 0, curr_end = 0, max_wind_start = 0, max_wind_size = 0;
  int count[26] = {};
  int len = strlen(str);
  int i = 0, unique = 0;
  for (i = 0; i < len; i++) {
    if (count[str[i] - 'a'] == 0) {
      unique++;
    }
  }
  if (unique < k) {
    printf("\n Error there are not enough unique characters in the string ");
    exit(-1);
  }
  max_wind_start = 0;
  max_wind_size = 1;
  curr_start = 0;
  curr_end = 0;
  count[str[0] - 'a']++;
  for (i = 1; i < len; i++) {
    count[str[i] - 'a']++;
    curr_end++;
    while (!is_valid(count, k)) {
      count[str[i] - 'a']--;
      curr_start++;
    }
    if ((curr_end - curr_start + 1) > max_wind_size) {
      max_wind_size = curr_end - curr_start + 1;
      max_wind_start = curr_start;
    }
  }
  printf("\nMax length sub str of %s starts from %d to %d", str, max_wind_start, max_wind_size + max_wind_start);
}
```