user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    D17(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D27(%rip),%rbx
  sub    $0x228,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x218(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L138:
  xor    %eax,%eax
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L108
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L108
  lea    0x10(%rsp),%r15
  mov    %r13,%rdi
  xor    %eax,%eax
  mov    %r15,%rsi
  call   <scanf@plt>
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    0xc(%rsp),%esi
  mov    %eax,%r14d
  test   %eax,%eax
  jle    L104
  xor    %ecx,%ecx
  jmp    L148
  nopl   0x0(%rax,%rax,1)
L57:
  mov    %ecx,%eax
  add    $0x1,%ecx
L20:
  cltq
  movzbl 0x10(%rsp,%rax,1),%edx
  mov    %dl,0x110(%rsp,%rax,1)
  cmp    %ecx,%r14d
  jle    L104
L148:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L57
  test   %ecx,%ecx
  je     L74
  lea    (%rsi,%rcx,1),%eax
  lea    0x1(%rax),%ecx
  jmp    L20
  nopl   (%rax)
L74:
  xor    %eax,%eax
  mov    $0x1,%ecx
  jmp    L20
  nopl   0x0(%rax)
L104:
  lea    (%rsi,%rsi,1),%edx
  mov    %esi,%eax
  lea    (%rdx,%rsi,1),%ecx
  mov    %ecx,%r8d
  cmp    %r14d,%esi
  jl     L88
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L139:
  lea    (%rdx,%rsi,1),%ecx
L88:
  cmp    %eax,%ecx
  jne    L137
  add    %esi,%eax
  add    %r8d,%edx
L137:
  movslq %eax,%rdi
  sub    $0x1,%edx
  add    $0x1,%eax
  movzbl 0x10(%rsp,%rdi,1),%edi
  movslq %edx,%rcx
  mov    %dil,0x110(%rsp,%rcx,1)
  cmp    %eax,%r14d
  jg     L139
L9:
  lea    0x110(%rsp),%r12
  mov    %r15,%rdi
  mov    %r12,%rsi
  call   <strcpy@plt>
  movslq 0xc(%rsp),%rdi
  test   %edi,%edi
  jle    L158
  xor    %r10d,%r10d
  xor    %esi,%esi
  lea    0x10f(%rsp),%r9
  nop
L196:
  lea    0x1(%rsi),%eax
  mov    %r10,%rdx
  cltq
  cmp    %r10d,%r14d
  jle    L184
  nopl   (%rax)
L23:
  movzbl (%r15,%rdx,1),%ecx
  add    %rdi,%rdx
  movslq %eax,%rsi
  mov    %cl,(%r9,%rax,1)
  add    $0x1,%rax
  cmp    %edx,%r14d
  jg     L23
L184:
  add    $0x1,%r10
  cmp    %rdi,%r10
  jne    L196
L52:
  movb   $0x0,0x110(%rsp,%rsi,1)
  lea    D19(%rip),%rdi
  mov    %r12,%rsi
  call   <printf@plt>
  jmp    L138
L108:
  mov    0x218(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L134
  add    $0x228,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L158:
  xor    %esi,%esi
  jmp    L52
L134:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D17
data_type: string
name: unpack_data
arguments: label: D27
data_type: string
name: unpack_data
arguments: label: D19
data_type: string

tool: D17: %s

tool: D27: %d

tool: D19: %s


assistant: ```c
int main(int argc, char **argv) {
  int cur, i, len, offset, range;
  char code[255], solution[255];
  while (scanf("%d", &range) && range) {
    scanf("%s", code);
    len = strlen(code);
    for (cur = 0, i = 0; i < len; i++) {
      if (i % range == 0 && i) {
        i += range;
        cur += range;
      }
      solution[cur++] = code[i];
    }
    for (cur = range * 2, i = range; i < len; ++i) {
      if (cur + range == i) {
        i += range;
        cur += range * 3;
      }
      solution[--cur] = code[i];
    }
    strcpy(code, solution);
    for (cur = 0, offset = 0; offset < range; offset++) {
      for (i = offset; i < len; i += range) {
        solution[cur++] = code[i];
      }
    }
    solution[cur] = '\0';
    printf("%s\n", solution);
  }
  return 0;
}
```