user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movq   $0x0,-0xe0(%rbp)
  movq   $0x0,-0xd8(%rbp)
  movq   $0x1,-0xd0(%rbp)
  movq   $0x2,-0xc8(%rbp)
  movq   $0x2,-0xc0(%rbp)
  movq   $0x3,-0xb8(%rbp)
  movq   $0x4,-0xb0(%rbp)
  movq   $0x4,-0xa8(%rbp)
  movq   $0x5,-0xa0(%rbp)
  movq   $0x6,-0x98(%rbp)
  movq   $0x6,-0x90(%rbp)
  lea    -0xe0(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xb,%edi
  call   <dotestJohn@plt>
  movq   $0x0,-0x80(%rbp)
  movq   $0x0,-0x78(%rbp)
  movq   $0x1,-0x70(%rbp)
  movq   $0x2,-0x68(%rbp)
  movq   $0x2,-0x60(%rbp)
  movq   $0x3,-0x58(%rbp)
  movq   $0x4,-0x50(%rbp)
  movq   $0x4,-0x48(%rbp)
  movq   $0x5,-0x40(%rbp)
  movq   $0x6,-0x38(%rbp)
  movq   $0x6,-0x30(%rbp)
  movq   $0x7,-0x28(%rbp)
  movq   $0x7,-0x20(%rbp)
  movq   $0x8,-0x18(%rbp)
  lea    -0x80(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xe,%edi
  call   <dotestJohn@plt>
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movq   $0x1,-0xe0(%rbp)
  movq   $0x1,-0xd8(%rbp)
  movq   $0x2,-0xd0(%rbp)
  movq   $0x2,-0xc8(%rbp)
  movq   $0x3,-0xc0(%rbp)
  movq   $0x3,-0xb8(%rbp)
  lea    -0xe0(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x6,%edi
  call   <dotestAnn@plt>
  movq   $0x1,-0x80(%rbp)
  movq   $0x1,-0x78(%rbp)
  movq   $0x2,-0x70(%rbp)
  movq   $0x2,-0x68(%rbp)
  movq   $0x3,-0x60(%rbp)
  movq   $0x3,-0x58(%rbp)
  movq   $0x4,-0x50(%rbp)
  movq   $0x5,-0x48(%rbp)
  movq   $0x5,-0x40(%rbp)
  movq   $0x6,-0x38(%rbp)
  movq   $0x6,-0x30(%rbp)
  movq   $0x7,-0x28(%rbp)
  movq   $0x8,-0x20(%rbp)
  movq   $0x8,-0x18(%rbp)
  movq   $0x9,-0x10(%rbp)
  lea    -0x80(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xf,%edi
  call   <dotestAnn@plt>
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x6b8,%esi
  mov    $0x4b,%edi
  call   <dotestsumJohn@plt>
  mov    $0x745,%esi
  mov    $0x4e,%edi
  call   <dotestsumJohn@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xfe6,%esi
  mov    $0x73,%edi
  call   <dotestsumAnn@plt>
  mov    $0x1b12,%esi
  mov    $0x96,%edi
  call   <dotestsumAnn@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D17
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D24
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D7
type_infered: string

tool: D6: john


tool: D17: ann


tool: D24: sumJohn


tool: D7: sumAnn


assistant: ```c
int main() {
  printf("john\n");
  {
    long long sol1[11] = {0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6};
    dotestJohn(11, sol1);
    long long sol2[14] = {0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8};
    dotestJohn(14, sol2);
  }
  printf("ann\n");
  {
    long long sol1[6] = {1, 1, 2, 2, 3, 3};
    dotestAnn(6, sol1);
    long long sol2[15] = {1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9};
    dotestAnn(15, sol2);
  }
  printf("sumJohn\n");
  {
    dotestsumJohn(75, 1720);
    dotestsumJohn(78, 1861);
  }
  printf("sumAnn\n");
  {
    dotestsumAnn(115, 4070);
    dotestsumAnn(150, 6930);
  }
}
```