user: 这段汇编代码用C语言表示是什么样的:

```asm
<contagem_maos>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rdx,0x18(%rsp)
  mov    %r9,0x10(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  mov    $0x1c,%eax
  cltd
  idiv   %r8d
  test   %r8d,%r8d
  jle    L57
  mov    %rsi,%r15
  mov    %rcx,%r14
  mov    %r8d,%ebp
  mov    %eax,%r13d
  lea    -0x1(%rax),%eax
  mov    $0x1,%esi
  movl   $0x1,0x4(%rsp)
  mov    $0x2710,%r11d
  lea    0x4(,%rax,4),%rbx
  jmp    L120
L169:
  cmp    $0x4,%esi
  cmove  %r14,%r10
  jmp    L174
L80:
  mov    0x18(%rsp),%r10
  jmp    L174
L39:
  mov    0x8(%rsp),%r10
  jmp    L174
L173:
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%r9d
  sar    $0x1f,%r9d
  sub    %r9d,%eax
  imul   $0xfffffff7,%eax,%eax
  lea    (%rax,%rcx,1),%eax
  add    %eax,0x1c(%rsp,%rsi,4)
L143:
  add    $0x4,%rdx
  cmp    %rdi,%rdx
  je     L75
L156:
  mov    (%rdx),%ecx
  cmp    $0x63,%ecx
  jne    L173
  jmp    L143
L75:
  mov    0x1c(%rsp,%rsi,4),%eax
  cmp    %r11d,%eax
  jge    L226
  mov    %r8d,0x4(%rsp)
  mov    %eax,%r11d
L226:
  add    $0x1,%rsi
  cmp    %esi,%ebp
  jl     L133
L120:
  mov    %esi,%r8d
  cmp    $0x3,%esi
  je     L80
  jg     L169
  cmp    $0x1,%esi
  je     L39
  cmp    $0x2,%esi
  cmove  %r15,%r10
L174:
  test   %r13d,%r13d
  jle    L75
  mov    %r10,%rdx
  lea    (%rbx,%r10,1),%rdi
  jmp    L156
L133:
  mov    0x10(%rsp),%r12
  mov    $0x1,%ebx
  jmp    L177
L196:
  mov    %r13d,%esi
  mov    0x8(%rsp),%rdi
  call   <mostra_mao@plt>
  jmp    L214
L129:
  mov    %r13d,%esi
  mov    %r15,%rdi
  call   <mostra_mao@plt>
  jmp    L214
L189:
  mov    %r13d,%esi
  mov    0x18(%rsp),%rdi
  call   <mostra_mao@plt>
L214:
  mov    0x1c(%rsp,%rbx,4),%esi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  add    $0x14,%r12
  cmp    %ebx,%ebp
  jl     L50
L177:
  mov    %r12,%rdi
  call   <puts@plt>
  cmp    $0x1,%ebx
  je     L196
  cmp    $0x2,%ebx
  je     L129
  cmp    $0x3,%ebx
  je     L189
  mov    %r13d,%esi
  mov    %r14,%rdi
  call   <mostra_mao@plt>
  jmp    L214
L57:
  movl   $0x1,0x4(%rsp)
L50:
  movslq 0x4(%rsp),%rax
  lea    (%rax,%rax,4),%rax
  mov    0x10(%rsp),%rbx
  lea    -0x14(%rbx,%rax,4),%rsi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D11
type_infered: string
name: disassemble_data
arguments: data_tag: D17
type_infered: string

tool: D11: Soma =%d



tool: D17: 
%s VENCEU!


assistant: ```c
int contagem_maos(int mao_1[], int mao_2[], int mao_3[], int mao_4[], int n_jog, char nomes[][20]) {
  int menor = 10000;
  int jog_ven = 1;
  int soma_m[4] = {0};
  int pp_1, pp_2;
  int j;
  int i;
  int t_mao = 28 / n_jog;
  int *mao_c;
  for (j = 1; j <= n_jog; j++) {
    switch (j) {
    case 1:
      mao_c = mao_1;
      break;
    case 2:
      mao_c = mao_2;
      break;
    case 3:
      mao_c = mao_3;
      break;
    case 4:
      mao_c = mao_4;
      break;
    default:
      break;
    }
    for (i = 0; i < t_mao; i++)
      if (mao_c[i] != 99) {
        pp_1 = mao_c[i] / 10;
        pp_2 = mao_c[i] - pp_1 * 10;
        soma_m[j - 1] += pp_1 + pp_2;
      }
    if (menor > soma_m[j - 1]) {
      menor = soma_m[j - 1];
      jog_ven = j;
    }
  }
  for (j = 1; j <= n_jog; j++) {
    printf("%s\n", nomes[j - 1]);
    if (j == 1)
      mostra_mao(mao_1, t_mao);
    else if (j == 2)
      mostra_mao(mao_2, t_mao);
    else if (j == 3)
      mostra_mao(mao_3, t_mao);
    else
      mostra_mao(mao_4, t_mao);
    printf("Soma =%d\n\n", soma_m[j - 1]);
  }
  printf("\n%s VENCEU!\n", nomes[jog_ven - 1]);
  return 0;
}
```