user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x18000(%rsp),%r11
L57:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L57
  sub    $0x6e8,%rsp
  lea    D80(%rip),%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x186d8(%rsp)
  xor    %eax,%eax
  lea    0x2c(%rsp),%rax
  movl   $0x1,0x14(%rsp)
  mov    %rax,0x18(%rsp)
  nopw   0x0(%rax,%rax,1)
L66:
  mov    0x18(%rsp),%rsi
  lea    D76(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L118
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x2c(%rsp),%edx
  test   %edx,%edx
  je     L118
  mov    0x14(%rsp),%esi
  lea    D85(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D46(%rip),%rdi
  call   <puts@plt>
  mov    0x2c(%rsp),%eax
  test   %eax,%eax
  jle    L59
  lea    0x31(%rsp),%rax
  xor    %ebp,%ebp
  lea    0x30(%rsp),%r12
  mov    %rax,0x8(%rsp)
  lea    D43(%rip),%r13
  nop
L89:
  mov    %r12,%rdi
  call   <gets@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    D26(%rip),%rdi
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L113
  sub    $0x1,%ebx
  mov    %r12,%r15
  add    0x8(%rsp),%rbx
  jmp    L69
  nopl   0x0(%rax)
L8:
  call   <putchar@plt>
L73:
  add    $0x1,%r15
  cmp    %rbx,%r15
  je     L113
L69:
  movsbl (%r15),%edi
  cmp    $0x22,%dil
  je     L37
  cmp    $0x5c,%dil
  jne    L8
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %rbx,%r15
  jne    L69
L113:
  mov    $0x22,%esi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    %ebp,0x2c(%rsp)
  jg     L89
L59:
  mov    $0x22,%ecx
  mov    $0x5c,%edx
  mov    $0x22,%esi
  xor    %eax,%eax
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  call   <puts@plt>
  addl   $0x1,0x14(%rsp)
  jmp    L66
  nopw   0x0(%rax,%rax,1)
L37:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L73
L118:
  mov    0x186d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x186e8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D80
data_type: string
name: retrieve_data
arguments: label: D76
data_type: string
name: retrieve_data
arguments: label: D85
data_type: string
name: retrieve_data
arguments: label: D46
data_type: string
name: retrieve_data
arguments: label: D43
data_type: string
name: retrieve_data
arguments: label: D26
data_type: string
name: retrieve_data
arguments: label: D22
data_type: string
name: retrieve_data
arguments: label: D42
data_type: string
name: retrieve_data
arguments: label: D20
data_type: string

tool: D80: \"

tool: D76: %d

tool: D85: Case %d:


tool: D46: #include<string.h>
#include<stdio.h>
int main()
{

tool: D43: \\

tool: D26: printf("

tool: D22: \n%c);

tool: D42: printf(%c%cn%c);


tool: D20: return 0;
}

assistant: ```c
int main() {
  int i, len, n, j, l = 1;
  char sp1 = 92, sp2 = 34;
  char a[100000];
  while (scanf("%d", &n) == 1) {
    getchar();
    if (n == 0)
      break;
    printf("Case %d:\n", l);
    printf("#include<string.h>\n#include<stdio.h>\nint main()\n{\n");
    for (i = 0; i < n; i++) {
      gets(a);
      len = strlen(a);
      printf("printf(\"");
      for (j = 0; j < len; j++) {
        if (a[j] == 34)
          printf("\\\"");
        else if (a[j] == 92)
          printf("\\\\");
        else
          printf("%c", a[j]);
      }
      printf("\\n%c);", sp2);
      printf("\n");
    }
    printf("printf(%c%cn%c);\n", sp2, sp1, sp2);
    printf("return 0;\n}\n");
    l++;
  }
  return 0;
}
```