user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L75:
  movq   $0x1,-0xb0(%rbp)
  movl   $0x0,-0xbc(%rbp)
  jmp    L8
L93:
  mov    -0xbc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0xbc(%rbp),%eax
  cltq
  mov    %edx,-0xa0(%rbp,%rax,4)
  addl   $0x1,-0xbc(%rbp)
L8:
  cmpl   $0x18,-0xbc(%rbp)
  jle    L93
  movq   $0x1,-0xa8(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x30(%rbp),%rax
  lea    D18(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L71
  lea    -0x30(%rbp),%rax
  lea    D37(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L142
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L78
L142:
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xbc(%rbp)
  jmp    L106
L20:
  mov    -0xbc(%rbp),%eax
  cltq
  movzbl -0x30(%rbp,%rax,1),%eax
  cmp    $0x20,%al
  jne    L136
  addl   $0x1,-0xbc(%rbp)
L136:
  movl   $0x0,-0xb8(%rbp)
  mov    -0xbc(%rbp),%eax
  cltq
  movzbl -0x30(%rbp,%rax,1),%eax
  movsbl %al,%eax
  lea    -0xa0(%rbp),%rcx
  mov    -0xbc(%rbp),%edx
  lea    -0x30(%rbp),%rsi
  mov    %eax,%edi
  call   <check@plt>
  mov    %eax,-0xb8(%rbp)
  cmpl   $0x1,-0xb8(%rbp)
  je     L66
  subl   $0x1,-0xbc(%rbp)
L106:
  cmpl   $0x0,-0xbc(%rbp)
  jns    L20
  jmp    L160
L66:
  nop
L160:
  cmpl   $0x1,-0xb8(%rbp)
  je     L149
  movl   $0x0,-0xbc(%rbp)
  jmp    L6
L60:
  mov    -0xbc(%rbp),%eax
  cltq
  movzbl -0x30(%rbp,%rax,1),%eax
  cmp    $0x59,%al
  jne    L134
  mov    -0xbc(%rbp),%eax
  cltq
  mov    -0xa0(%rbp,%rax,4),%eax
  cmp    $0x14,%eax
  je     L134
  mov    -0xbc(%rbp),%eax
  cltq
  mov    -0xa0(%rbp,%rax,4),%edx
  mov    -0xb0(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <lcm@plt>
  mov    %rax,-0xb0(%rbp)
L134:
  addl   $0x1,-0xbc(%rbp)
L6:
  mov    -0xbc(%rbp),%eax
  cmp    -0xb4(%rbp),%eax
  jl     L60
  mov    -0xb0(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L75
L149:
  nop
L78:
  jmp    L75
L71:
  nop
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D31
datatype: string
name: unpack_data
arguments: data_tag: D18
datatype: string
name: unpack_data
arguments: data_tag: D37
datatype: string
name: unpack_data
arguments: data_tag: D36
datatype: string
name: unpack_data
arguments: data_tag: D2
datatype: string

tool: D31: %s

tool: D18: *

tool: D37: N

tool: D36: -1

tool: D2: %llu


assistant: ```c
int main() {
  char input[30];
  int a[25];
  int i, j, c, n;
  unsigned long long int num;
  while (1) {
    unsigned long long int lcms = 1;
    for (i = 0; i < 25; i++) {
      a[i] = (i + 1);
    }
    num = 1;
    scanf("%s", input);
    if (strcmp(input, "*") == 0)
      break;
    if (strcmp(input, "N") == 0) {
      printf("-1\n");
      continue;
    }
    n = strlen(input);
    for (i = n - 1; i >= 0; i--) {
      if (input[i] == ' ')
        i++;
      c = 0;
      c = check(input[i], input, i, a);
      if (c == 1)
        break;
    }
    if (c == 1)
      continue;
    for (i = 0; i < n; i++) {
      if (input[i] == 'Y') {
        if (a[i] != 20)
          lcms = lcm(lcms, a[i]);
      }
    }
    printf("%llu\n", lcms);
  }
  return 0;
}
```