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
  sub    $0xe8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  lea    0x2c(%rsp),%rax
  mov    %rax,0x18(%rsp)
  lea    0x28(%rsp),%rax
  mov    %rax,0x10(%rsp)
  nopl   0x0(%rax)
L43:
  mov    0x18(%rsp),%rdx
  mov    0x10(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  cmp    $0xffffffff,%eax
  je     L44
  mov    0x28(%rsp),%eax
  mov    0x2c(%rsp),%edx
  mov    %eax,%ebx
  or     %edx,%ebx
  je     L44
  xor    %ecx,%ecx
  lea    0x30(%rsp),%rbx
  lea    0x80(%rsp),%r13
L99:
  movl   $0x1,(%rcx,%rbx,1)
  add    $0x4,%rcx
  movl   $0x1,-0x4(%rcx,%r13,1)
  cmp    $0x50,%rcx
  jne    L99
  mov    %eax,%ecx
  mov    %eax,%edi
  shr    $0x1f,%ecx
  sub    %edx,%edi
  add    %eax,%ecx
  sar    %ecx
  cmp    %ecx,%edx
  jge    L81
  mov    %edx,%ecx
  mov    %edi,0x2c(%rsp)
  mov    %edi,%edx
  mov    %ecx,%edi
L81:
  lea    0x1(%rdx),%ecx
  cmp    %ecx,%eax
  jl     L152
  movslq %edx,%rsi
  movslq %ecx,%rcx
  neg    %rsi
  lea    (%rbx,%rsi,4),%rsi
  nopl   0x0(%rax)
L28:
  mov    %ecx,-0x4(%rsi,%rcx,4)
  add    $0x1,%rcx
  cmp    %ecx,%eax
  jge    L28
L152:
  lea    0x1(%rdi),%esi
  mov    $0x2,%ecx
  cmp    $0x1,%edi
  jle    L57
  nopl   (%rax)
L36:
  mov    %ecx,0x0(%r13,%rcx,4)
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  jne    L36
L72:
  movl   $0x0,0xc(%rsp)
  mov    %rbx,%rbp
  nopl   0x0(%rax)
L58:
  cmp    %edx,%eax
  jl     L84
  mov    0x0(%rbp),%r15d
  xor    %r14d,%r14d
  nopl   0x0(%rax,%rax,1)
L93:
  mov    0x0(%r13,%r14,4),%r12d
  mov    %r15d,%edi
  mov    %r12d,%esi
  call   <bcd@plt>
  mov    %eax,%ecx
  mov    %r15d,%eax
  cltd
  idiv   %ecx
  mov    %eax,0x0(%rbp)
  mov    %eax,%r15d
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  mov    0x2c(%rsp),%edx
  mov    %eax,0x0(%r13,%r14,4)
  mov    0x28(%rsp),%eax
  add    $0x1,%r14
  mov    %eax,%ecx
  sub    %edx,%ecx
  cmp    %r14d,%ecx
  jge    L93
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%edi
  add    $0x4,%rbp
  cmp    %edi,%ecx
  jge    L58
  nop
L84:
  mov    $0x1,%esi
  nopl   (%rax)
L145:
  imul   (%rbx),%esi
  add    $0x4,%rbx
  cmp    %r13,%rbx
  jne    L145
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L43
L57:
  test   %edi,%edi
  js     L84
  jmp    L72
  xchg   %ax,%ax
L44:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0xe8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D0
type_infered: string
name: parse_data
arguments: data_label: D3
type_infered: string

tool: D0: %d%d

tool: D3: %d


assistant: ```c
int main() {
  int n, k, i, j;
  int res[20], ma[20], r, bc;
  while (scanf("%d%d", &n, &k) != -1) {
    if (n == 0 && k == 0)
      break;
    for (i = 0; i < 20; i++) {
      res[i] = 1;
      ma[i] = 1;
    }
    if (k < n / 2)
      k = n - k;
    for (i = k + 1; i <= n; i++)
      res[i - k - 1] = i;
    for (i = 2; i <= n - k; i++)
      ma[i] = i;
    for (i = 0; i <= n - k; i++)
      for (j = 0; j <= n - k; j++) {
        bc = bcd(res[i], ma[j]);
        res[i] /= bc;
        ma[j] /= bc;
      }
    r = 1;
    for (i = 0; i < 20; i++)
      r *= res[i];
    printf("%d\n", r);
  }
  return 0;
}
```