user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D74(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  nopl   0x0(%rax)
L27:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x9,%ebx
  jne    L27
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  lea    D61(%rip),%r14
  call   <printf@plt>
  nopl   0x0(%rax)
L71:
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L71
  mov    $0xa,%edi
  mov    $0x1,%ebx
  lea    D88(%rip),%r13
  call   <putchar@plt>
  nopl   (%rax)
L15:
  lea    -0x1(%rbx),%r12d
  mov    %r13,%rdi
  xor    %eax,%eax
  xor    %r15d,%r15d
  mov    %r12d,%esi
  call   <printf@plt>
  nopl   0x0(%rax)
L37:
  mov    %r15d,%esi
  mov    %r12d,%edi
  add    $0x1,%r15d
  call   <comb@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r15d,%ebx
  jne    L37
  mov    $0xa,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    $0xa,%ebx
  jne    L15
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L63:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x9,%ebx
  jne    L63
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L76:
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L76
  mov    $0xa,%edi
  mov    $0x1,%ebp
  lea    D58(%rip),%r14
  call   <putchar@plt>
  nopl   (%rax)
L46:
  lea    -0x1(%rbp),%r12d
  mov    %r13,%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  mov    %r12d,%esi
  call   <printf@plt>
  nopl   0x0(%rax,%rax,1)
L67:
  mov    %ebx,%esi
  mov    %r12d,%edi
  add    $0x1,%ebx
  call   <combination@plt>
  mov    %r14,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L67
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    $0xa,%ebp
  jne    L46
  xor    %ebx,%ebx
  lea    D49(%rip),%rbp
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %ebx,%esi
  mov    $0x22,%edi
  call   <combination@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    $0x12,%ebx
  jne    L4
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D38
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D74
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D1
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D61
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D88
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D58
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D49
type_guessed: string

tool: D38: 
  k

tool: D74: %10d

tool: D1: 
n  

tool: D61: ------

tool: D88: %10d |

tool: D58: %10lu

tool: D49: 34C%10d = %10lu


assistant: ```c
int main() {
  int n, k;
  printf("\n  k");
  for (k = 0; k <= 8; k++)
    printf("%10d", k);
  printf("\nn  ");
  for (k = 0; k <= 8; k++)
    printf("------");
  printf("\n");
  for (n = 0; n <= 8; n++) {
    printf("%10d |", n);
    for (k = 0; k <= n; k++)
      printf("%10d", comb(n, k));
    printf("\n");
  }
  printf("\n  k");
  for (k = 0; k <= 8; k++)
    printf("%10d", k);
  printf("\nn  ");
  for (k = 0; k <= 8; k++)
    printf("------");
  printf("\n");
  for (n = 0; n <= 8; n++) {
    printf("%10d |", n);
    for (k = 0; k <= n; k++)
      printf("%10lu", combination(n, k));
    printf("\n");
  }
  for (k = 0; k <= 17; k++)
    printf("34C%10d = %10lu\n", k, combination(34, k));
  return 0;
}
```