user: 这段汇编代码用C语言表示是什么样的？

```asm
<ubicacionUnicaEnCaja>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %ecx,-0x44(%rbp)
  movl   $0x0,-0x1c(%rbp)
  cmpl   $0x2,-0x3c(%rbp)
  jg     L98
  movl   $0x0,-0x24(%rbp)
L98:
  cmpl   $0x2,-0x3c(%rbp)
  jle    L63
  cmpl   $0x5,-0x3c(%rbp)
  jg     L63
  movl   $0x3,-0x24(%rbp)
L63:
  cmpl   $0x5,-0x3c(%rbp)
  jle    L121
  cmpl   $0x8,-0x3c(%rbp)
  jg     L121
  movl   $0x6,-0x24(%rbp)
L121:
  cmpl   $0x2,-0x40(%rbp)
  jg     L140
  movl   $0x0,-0x20(%rbp)
L140:
  cmpl   $0x2,-0x40(%rbp)
  jle    L78
  cmpl   $0x5,-0x40(%rbp)
  jg     L78
  movl   $0x3,-0x20(%rbp)
L78:
  cmpl   $0x5,-0x40(%rbp)
  jle    L90
  cmpl   $0x8,-0x40(%rbp)
  jg     L90
  movl   $0x6,-0x20(%rbp)
L90:
  mov    -0x24(%rbp),%eax
  add    $0x3,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%eax
  add    $0x3,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L82
L30:
  movl   $0x0,-0xc(%rbp)
  jmp    L160
L137:
  mov    -0x24(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  add    %ecx,%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  test   %eax,%eax
  jne    L44
  mov    -0x24(%rbp),%edx
  mov    -0x10(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x44(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <candidatoCabeEnFila@plt>
  test   %rax,%rax
  je     L44
  mov    -0x20(%rbp),%edx
  mov    -0xc(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x44(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <candidatoCabeEnColumna@plt>
  test   %rax,%rax
  je     L44
  addl   $0x1,-0x1c(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
L44:
  addl   $0x1,-0xc(%rbp)
L160:
  cmpl   $0x2,-0xc(%rbp)
  jle    L137
  addl   $0x1,-0x10(%rbp)
L82:
  cmpl   $0x2,-0x10(%rbp)
  jle    L30
  cmpl   $0x1,-0x1c(%rbp)
  jne    L36
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x14(%rbp),%eax
  cltq
  mov    -0x44(%rbp),%edx
  mov    %edx,(%rcx,%rax,4)
  mov    $0x1,%eax
  jmp    L118
L36:
  mov    $0x0,%eax
L118:
  leave
  ret
```

assistant: ```c
int ubicacionUnicaEnCaja(int s[9][9], int f, int c, int candidato) {
  int f2, c2;
  int cajas1, cajas2;
  int pu = 0;
  int pf;
  int pc;
  if (f < 3)
    f2 = 0;
  if (f >= 3 && f < 6)
    f2 = 3;
  if (f >= 6 && f < 9)
    f2 = 6;
  if (c < 3)
    c2 = 0;
  if (c >= 3 && c < 6)
    c2 = 3;
  if (c >= 6 && c < 9)
    c2 = 6;
  cajas1 = f2 + 3;
  cajas2 = c2 + 3;
  int i, j;
  for (pu = 0, i = 0; i < 3; i++)
    for (j = 0; j < 3; j++) {
      if (s[f2 + i][c2 + j] == 0 && candidatoCabeEnFila(s, f2 + i, candidato) && candidatoCabeEnColumna(s, c2 + j, candidato)) {
        pu++;
        pf = f2 + i;
        pc = c2 + j;
      }
    }
  if (pu == 1) {
    s[pf][pc] = candidato;
    return 1;
  }
  return 0;
}
```