user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x2710,%edi
  lea    D6(%rip),%r12
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  lea    D14(%rip),%rdi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <scanf@plt>
  mov    %rbp,%rbx
  xchg   %ax,%ax
L21:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  sub    %ebp,%eax
  cmp    %eax,%r8d
  jle    L53
  movzbl (%rbx),%eax
  sub    $0x61,%eax
  cmp    $0x19,%al
  ja     L223
  movzbl %al,%eax
  movslq (%r12,%rax,4),%rax
  add    %r12,%rax
  notrack jmp *%rax
  movb   $0x5a,(%rbx)
  xchg   %ax,%ax
L223:
  add    $0x1,%rbx
  jmp    L21
  cs nopw 0x0(%rax,%rax,1)
L53:
  xor    %ebx,%ebx
  lea    D0(%rip),%r12
  jmp    L47
  nopl   0x0(%rax,%rax,1)
L201:
  movsbl 0x0(%rbp,%rbx,1),%esi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
L47:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L201
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  movb   $0x59,(%rbx)
  jmp    L223
  movb   $0x58,(%rbx)
  jmp    L223
  movb   $0x57,(%rbx)
  jmp    L223
  movb   $0x56,(%rbx)
  jmp    L223
  movb   $0x55,(%rbx)
  jmp    L223
  movb   $0x54,(%rbx)
  jmp    L223
  movb   $0x53,(%rbx)
  jmp    L223
  movb   $0x52,(%rbx)
  jmp    L223
  movb   $0x51,(%rbx)
  jmp    L223
  movb   $0x50,(%rbx)
  jmp    L223
  movb   $0x4f,(%rbx)
  jmp    L223
  movb   $0x4e,(%rbx)
  jmp    L223
  movb   $0x4d,(%rbx)
  jmp    L223
  movb   $0x4c,(%rbx)
  jmp    L223
  movb   $0x4b,(%rbx)
  jmp    L223
  movb   $0x4a,(%rbx)
  jmp    L223
  movb   $0x49,(%rbx)
  jmp    L223
  movb   $0x48,(%rbx)
  jmp    L223
  movb   $0x47,(%rbx)
  jmp    L223
  movb   $0x46,(%rbx)
  jmp    L223
  movb   $0x45,(%rbx)
  jmp    L223
  movb   $0x44,(%rbx)
  jmp    L223
  movb   $0x43,(%rbx)
  jmp    L223
  movb   $0x42,(%rbx)
  jmp    L223
  movb   $0x41,(%rbx)
  jmp    L223
```

tool_calls: name: parse_data
arguments: tag: D14
type: string
name: parse_data
arguments: tag: D0
type: string

tool: D14: %s

tool: D0: %c

assistant: ```c
int main() {
  char *seq = malloc(10000);
  int i, j;
  scanf("%s", seq);
  for (i = 0; i < strlen(seq); i++) {
    switch (seq[i]) {
    case 'a':
      seq[i] = 'A';
      break;
    case 'b':
      seq[i] = 'B';
      break;
    case 'c':
      seq[i] = 'C';
      break;
    case 'd':
      seq[i] = 'D';
      break;
    case 'e':
      seq[i] = 'E';
      break;
    case 'f':
      seq[i] = 'F';
      break;
    case 'g':
      seq[i] = 'G';
      break;
    case 'h':
      seq[i] = 'H';
      break;
    case 'i':
      seq[i] = 'I';
      break;
    case 'j':
      seq[i] = 'J';
      break;
    case 'k':
      seq[i] = 'K';
      break;
    case 'l':
      seq[i] = 'L';
      break;
    case 'm':
      seq[i] = 'M';
      break;
    case 'n':
      seq[i] = 'N';
      break;
    case 'o':
      seq[i] = 'O';
      break;
    case 'p':
      seq[i] = 'P';
      break;
    case 'q':
      seq[i] = 'Q';
      break;
    case 'r':
      seq[i] = 'R';
      break;
    case 's':
      seq[i] = 'S';
      break;
    case 't':
      seq[i] = 'T';
      break;
    case 'u':
      seq[i] = 'U';
      break;
    case 'v':
      seq[i] = 'V';
      break;
    case 'x':
      seq[i] = 'X';
      break;
    case 'y':
      seq[i] = 'Y';
      break;
    case 'w':
      seq[i] = 'W';
      break;
    case 'z':
      seq[i] = 'Z';
      break;
    }
  }
  for (i = 0; i < strlen(seq); i++) {
    printf("%c", seq[i]);
  }
  return 0;
}
```