user: 请从给定的汇编指令推导出C源代码:

```asm
<quick>:
  endbr64
  push   %r15
  mov    %esi,%r8d
  mov    %edx,%r15d
  push   %r14
  mov    $0xfffffffe,%r14d
  push   %r13
  lea    0x4(%rdi),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x18,%rsp
L84:
  lea    (%r8,%r15,1),%edx
  mov    %r15d,%esi
  mov    %r8d,%ebp
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  cltq
  mov    (%r12,%rax,4),%edi
  cmp    %r15d,%r8d
  jg     L51
  nopw   0x0(%rax,%rax,1)
L99:
  mov    (%rbx),%r9d
  mov    %ebp,%ecx
  lea    0x1(%r9),%eax
  mov    %eax,(%rbx)
  test   %eax,%eax
  je     L67
  mov    %r14d,%eax
  movslq %ebp,%rcx
  sub    %r9d,%eax
  add    $0x2,%r9d
  lea    (%r12,%rcx,4),%rdx
  add    %rcx,%rax
  sub    %ebp,%r9d
  mov    %ebp,%ecx
  lea    0x0(%r13,%rax,4),%r10
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L27:
  add    $0x4,%rdx
  mov    %eax,(%rbx)
  add    $0x1,%ecx
  cmp    %rdx,%r10
  je     L67
L36:
  lea    (%r9,%rcx,1),%eax
  cmp    %edi,(%rdx)
  jl     L27
  mov    %eax,(%rbx)
  test   %eax,%eax
  je     L229
L225:
  movslq %esi,%rdx
  lea    (%r12,%rdx,4),%rdx
  jmp    L143
  nopl   (%rax)
L110:
  mov    (%rbx),%eax
  sub    $0x1,%esi
  sub    $0x4,%rdx
  add    $0x1,%eax
  mov    %eax,(%rbx)
  test   %eax,%eax
  je     L229
L143:
  cmp    %edi,(%rdx)
  jg     L110
  cmp    %esi,%ecx
  jge    L133
  add    $0x1,%eax
  jne    L201
  movslq %ecx,%rax
  movl   $0x1,(%rbx)
  mov    %ecx,%ebp
  mov    (%r12,%rax,4),%eax
  cmp    %eax,(%rdx)
  jne    L99
L93:
  mov    %esi,%eax
  mov    %ecx,%esi
L273:
  lea    0x1(%rsi),%ebp
  lea    -0x1(%rax),%esi
L198:
  cmp    %ebp,%esi
  jge    L99
L51:
  lea    -0x1(%rbp),%edx
  cmp    %r8d,%edx
  jg     L127
L155:
  cmp    %ebp,%r15d
  jle    L275
  cmp    %r15d,%r8d
  jg     L128
  mov    %ebp,%r8d
  jmp    L84
L128:
  jmp    L128
  nopl   0x0(%rax)
L229:
  cmp    %ecx,%esi
  jle    L133
  movl   $0x1,(%rbx)
  mov    $0x1,%eax
L137:
  movslq %ecx,%rdx
  lea    (%r12,%rdx,4),%r10
  movslq %esi,%rdx
  lea    (%r12,%rdx,4),%rdx
  mov    (%r10),%r9d
  mov    (%rdx),%r11d
  cmp    %r11d,%r9d
  je     L39
  mov    %r11d,(%r10)
  add    $0x1,%ecx
  sub    $0x1,%esi
  mov    %r9d,(%rdx)
  addl   $0x1,0x4(%rbx)
L133:
  mov    %esi,%eax
  cmp    %esi,%ecx
  je     L273
  mov    (%rbx),%eax
  mov    %ecx,%ebp
L215:
  add    $0x1,%eax
  mov    %eax,(%rbx)
  je     L198
  movslq %esi,%rdx
  movslq %ecx,%rax
  mov    (%r12,%rdx,4),%edx
  cmp    %edx,(%r12,%rax,4)
  jne    L198
  jmp    L93
  nopw   0x0(%rax,%rax,1)
L67:
  add    $0x1,%eax
  mov    %eax,(%rbx)
  jmp    L225
L39:
  mov    %ecx,%ebp
  jmp    L215
L275:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L127:
  mov    %r8d,%esi
  mov    %rbx,%rcx
  mov    %r12,%rdi
  mov    %r8d,0xc(%rsp)
  call   <quick>
  mov    0xc(%rsp),%r8d
  jmp    L155
L201:
  mov    %eax,(%rbx)
  jmp    L137
```

assistant: ```c
void quick(int n[], int start, int end, int time[]) {
  int temp;
  int pivo;
  int i;
  int j;
  i = start;
  j = end;
  pivo = n[(start + end) / 2];
  while (i <= j) {
    while (++time[0] && n[i] < pivo)
      i++;
    while (++time[0] && pivo < n[j])
      j--;
    if (i < j && ++time[0] && n[i] != n[j]) {
      temp = n[i];
      n[i] = n[j];
      n[j] = temp;
      time[1]++;
      i++;
      j--;
    }
    if (i == j || (++time[0] && n[i] == n[j])) {
      i++;
      j--;
    }
  }
  if (start < i - 1)
    quick(n, start, i - 1, time);
  if (i < end)
    quick(n, i, end, time);
}
```