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
  xor    %ebp,%ebp
  push   %rbx
  lea    D35(%rip),%rbx
  sub    $0x8,%rsp
  call   <printf@plt>
  nopl   0x0(%rax)
L82:
  mov    %ebp,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    $0x9,%ebp
  jne    L82
  lea    D63(%rip),%rdi
  xor    %eax,%eax
  lea    D71(%rip),%r14
  call   <printf@plt>
  nopl   0x0(%rax)
L61:
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L61
  mov    $0xa,%edi
  mov    $0x1,%ebp
  lea    D46(%rip),%r13
  call   <putchar@plt>
  nopl   (%rax)
L67:
  lea    -0x1(%rbp),%r12d
  mov    %r13,%rdi
  xor    %eax,%eax
  xor    %r15d,%r15d
  mov    %r12d,%esi
  call   <printf@plt>
  nopl   0x0(%rax)
L48:
  mov    %r15d,%esi
  mov    %r12d,%edi
  add    $0x1,%r15d
  call   <comb@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %ebp,%r15d
  jne    L48
  mov    $0xa,%edi
  lea    0x1(%r15),%ebp
  call   <putchar@plt>
  cmp    $0x9,%r15d
  jne    L67
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <printf@plt>
  nopl   0x0(%rax)
L64:
  mov    %ebp,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    $0x9,%ebp
  jne    L64
  lea    D63(%rip),%rdi
  xor    %eax,%eax
  mov    $0x9,%ebx
  call   <printf@plt>
  nop
L77:
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L77
  mov    $0xa,%edi
  mov    $0x1,%ebp
  lea    D56(%rip),%r14
  call   <putchar@plt>
  nopl   (%rax)
L73:
  lea    -0x1(%rbp),%r12d
  mov    %r13,%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  mov    %r12d,%esi
  call   <printf@plt>
  nopl   0x0(%rax,%rax,1)
L45:
  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    L45
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    $0xa,%ebp
  jne    L73
  xor    %ebx,%ebx
  lea    D15(%rip),%rbp
  nopl   0x0(%rax,%rax,1)
L76:
  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    L76
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: tag: D38
data_type: string
name: parse_data
arguments: tag: D35
data_type: string
name: parse_data
arguments: tag: D63
data_type: string
name: parse_data
arguments: tag: D71
data_type: string
name: parse_data
arguments: tag: D46
data_type: string
name: parse_data
arguments: tag: D56
data_type: string
name: parse_data
arguments: tag: D15
data_type: string

tool: D38: 
  k

tool: D35: %10d

tool: D63: 
n  

tool: D71: ------

tool: D46: %10d |

tool: D56: %10lu

tool: D15: 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;
}
```