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
  sub    $0xa8,%rsp
  mov    %rsi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L62
  mov    0x8(%rsp),%rcx
  lea    0x8(%rcx),%rbx
  lea    -0x2(%rdi),%eax
  lea    0x10(%rcx,%rax,8),%r13
  mov    $0x0,%r12d
  lea    D4(%rip),%r14
  lea    D56(%rip),%r15
  jmp    L29
L62:
  mov    (%rsi),%rsi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L28:
  mov    (%rbx),%rsi
  lea    D63(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L45
L24:
  lea    D72(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L33:
  lea    0x10(%rsp),%rsi
  lea    -0x1(%rbp),%edi
  mov    $0x0,%eax
  call   <scrn_passwd@plt>
  test   %eax,%eax
  je     L101
  lea    0x10(%rsp),%rdx
  mov    %ebp,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L45:
  add    $0x8,%rbx
  cmp    %r13,%rbx
  je     L9
L29:
  mov    (%rbx),%rdi
  mov    $0x0,%eax
  call   <atoi@plt>
  test   %eax,%eax
  jle    L28
  mov    (%rbx),%rdi
  mov    $0x0,%eax
  call   <atoi@plt>
  mov    %eax,%ebp
  cmp    $0x63,%eax
  jg     L33
  test   %r12d,%r12d
  jne    L33
  mov    %r15,%rdi
  call   <puts@plt>
  lea    D50(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x30(%rsp),%r12
  mov    %r12,%rsi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0xa,%edx
  lea    D69(%rip),%rsi
  mov    %r12,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L24
  mov    $0x1,%r12d
  jmp    L33
L101:
  lea    D0(%rip),%rdx
  mov    0x8(%rsp),%rax
  mov    (%rax),%rsi
  lea    D82(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L9:
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L106
  mov    $0x0,%eax
  add    $0xa8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L106:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D4
type: string
name: retrieve_data
arguments: tag: D56
type: string
name: retrieve_data
arguments: tag: D14
type: string
name: retrieve_data
arguments: tag: D63
type: string
name: retrieve_data
arguments: tag: D72
type: string
name: retrieve_data
arguments: tag: D50
type: string
name: retrieve_data
arguments: tag: D17
type: string
name: retrieve_data
arguments: tag: D69
type: string
name: retrieve_data
arguments: tag: D0
type: string
name: retrieve_data
arguments: tag: D82
type: string

tool: D4: Screen %d password is '%s'.


tool: D56: You need clearance to see passwords below 100.

tool: D14: Usage: %s screen1 screen2 ...


tool: D63: Option not known: %s.


tool: D72: Foo, charlatan!

tool: D50: Enter master password:

tool: D17: %s

tool: D69: poppycrash

tool: D0: /usr/share/dict/words

tool: D82: %s: Cannot access %s.


assistant: ```c
main(argc, argv) int argc;
char *argv[];
{
  int scr, position, mpw = 0;
  char pass[20];
  char pw[100];
  if (argc < 2) {
    printf("Usage: %s screen1 screen2 ...\n", argv[0]);
    exit(-1);
  }
  position = 0;
  while (++position < argc) {
    if (atoi(argv[position]) < 1) {
      printf("Option not known: %s.\n", argv[position]);
      continue;
    }
    scr = atoi(argv[position]);
    if ((scr < 100) && (mpw == 0)) {
      printf("You need clearance to see passwords below 100.\n");
      printf("Enter master password:");
      scanf("%s", pw);
      if (strncmp(pw, "poppycrash", strlen("poppycrash"))) {
        printf("Foo, charlatan!\n");
        exit(-1);
      }
      mpw = 1;
    }
    if (!scrn_passwd((scr - 1), pass)) {
      printf("%s: Cannot access %s.\n", argv[0], "/usr/share/dict/words");
      exit(-1);
    }
    printf("Screen %d password is '%s'.\n", scr, pass);
  }
}
```