user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x12(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x12(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L126
L57:
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L195
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x56,%al
  jne    L195
  addl   $0x4,-0x1c(%rbp)
  addl   $0x1,-0x20(%rbp)
  jmp    L136
L195:
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x56,%al
  jne    L136
  addl   $0x5,-0x1c(%rbp)
L136:
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L52
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x58,%al
  jne    L52
  addl   $0x9,-0x1c(%rbp)
  addl   $0x1,-0x20(%rbp)
  jmp    L244
L52:
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x58,%al
  jne    L244
  addl   $0xa,-0x1c(%rbp)
L244:
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L80
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x4c,%al
  jne    L80
  addl   $0x30,-0x1c(%rbp)
  addl   $0x1,-0x20(%rbp)
  jmp    L123
L80:
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x4c,%al
  jne    L123
  addl   $0x32,-0x1c(%rbp)
L123:
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L1
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x43,%al
  jne    L1
  addl   $0x63,-0x1c(%rbp)
  addl   $0x1,-0x20(%rbp)
  jmp    L59
L1:
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x43,%al
  jne    L59
  addl   $0x64,-0x1c(%rbp)
L59:
  addl   $0x1,-0x20(%rbp)
L126:
  mov    -0x20(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L57
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L87
  mov    -0x18(%rbp),%eax
  sub    $0x2,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L87
  mov    -0x18(%rbp),%eax
  sub    $0x3,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L87
  addl   $0x3,-0x1c(%rbp)
  jmp    L169
L87:
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L169
  mov    -0x18(%rbp),%eax
  sub    $0x2,%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  cmp    $0x49,%al
  jne    L169
  addl   $0x2,-0x1c(%rbp)
L169:
  cmpl   $0xff,-0x1c(%rbp)
  jg     L187
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L133
L187:
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L133:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L260
  call   <__stack_chk_fail@plt>
L260:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D37
type_infered: string
name: parse_data
arguments: data_label: D17
type_infered: string
name: parse_data
arguments: data_label: D25
type_infered: string
name: parse_data
arguments: data_label: D24
type_infered: string

tool: D37: Strange letters number:

tool: D17: %s

tool: D25: Normal number with digits: %d

tool: D24: Sorry bro, it's toooo big

assistant: ```c
int main() {
  char a[10];
  int n, l, sum = 0;
  printf("Strange letters number:");
  scanf("%s", a);
  l = strlen(a);
  for (n = 0; n < l; n++) {
    if (a[n] == 'I' && a[n + 1] == 'V') {
      sum = sum + 4;
      n++;
    } else if (a[n] == 'V')
      sum = sum + 5;
    if (a[n] == 'I' && a[n + 1] == 'X') {
      sum = sum + 9;
      n++;
    } else if (a[n] == 'X')
      sum = sum + 10;
    if (a[n] == 'I' && a[n + 1] == 'L') {
      sum = sum + 48;
      n++;
    } else if (a[n] == 'L')
      sum = sum + 50;
    if (a[n] == 'I' && a[n + 1] == 'C') {
      sum = sum + 99;
      n++;
    } else if (a[n] == 'C')
      sum = sum + 100;
  }
  if (a[l - 1] == 'I' && a[l - 2] == 'I' && a[l - 3] == 'I') {
    sum = sum + 3;
  } else if (a[l - 1] == 'I' && a[l - 2] == 'I')
    sum = sum + 2;
  if (sum < 256) {
    printf("Normal number with digits: %d", sum);
  } else
    printf("Sorry bro, it's toooo big");
  return 0;
}
```