user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D3(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  lea    D25(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L108
  lea    0x10(%rsp),%r12
  xor    %ebx,%ebx
  lea    D10(%rip),%r14
  mov    %r12,%r13
  nopl   (%rax)
L144:
  add    $0x1,%ebx
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  add    $0x4,%r13
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  cmp    %edx,%ebx
  jl     L144
  lea    -0x1(%rdx),%esi
  mov    0x10(%rsp),%r9d
  movslq %esi,%r8
  mov    0x10(%rsp,%r8,4),%ecx
  cmp    $0x1,%edx
  jle    L145
  mov    $0x1,%eax
  jmp    L47
  nopl   0x0(%rax)
L12:
  mov    0x14(%rsp),%eax
  cmp    %ecx,%eax
  jge    L24
  mov    0x10(%rsp,%r8,4),%ecx
  mov    %ecx,0x14(%rsp)
  mov    %eax,%ecx
  mov    %eax,0x10(%rsp,%r8,4)
L24:
  cmp    $0x2,%esi
  je     L179
  mov    0x18(%rsp),%eax
  cmp    %eax,%ecx
  jle    L181
  mov    0x10(%rsp,%r8,4),%ecx
  mov    %ecx,0x18(%rsp)
  mov    %eax,%ecx
  mov    %eax,0x10(%rsp,%r8,4)
L181:
  cmp    $0x3,%esi
  je     L179
  mov    0x1c(%rsp),%eax
  cmp    %eax,%ecx
  jle    L59
  mov    0x10(%rsp,%r8,4),%ecx
  mov    %ecx,0x1c(%rsp)
  mov    %eax,%ecx
  mov    %eax,0x10(%rsp,%r8,4)
L59:
  cmp    $0x4,%esi
  je     L179
  mov    0x20(%rsp),%eax
  cmp    %ecx,%eax
  jge    L179
  mov    0x10(%rsp,%r8,4),%ecx
  mov    %ecx,0x20(%rsp)
  mov    %eax,%ecx
  mov    %eax,0x10(%rsp,%r8,4)
L179:
  movslq %edx,%rax
  cmp    %eax,%edx
  jle    L30
L47:
  mov    0x10(%rsp,%rax,4),%edi
  cmp    %r9d,%edi
  jle    L81
  mov    0x10(%rsp),%r9d
  mov    %edi,0x10(%rsp)
  mov    %r9d,0x10(%rsp,%rax,4)
  mov    %edi,%r9d
L81:
  cmp    $0x1,%esi
  jg     L12
  mov    $0x2,%eax
  cmp    %eax,%edx
  jg     L47
L30:
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L162:
  mov    (%r12,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L162
L108:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L145:
  test   %edx,%edx
  jle    L108
  jmp    L30
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type: string
name: extract_symbolic_data
arguments: data_tag: D25
type: string
name: extract_symbolic_data
arguments: data_tag: D10
type: string

tool: D3: Nhap so phan tu cho chuoi : 

tool: D25: %d

tool: D10: Nhap vao phan tu thu %d cho chuoi : 

assistant: ```c
int main() {
  int arr[5], i, N;
  printf("Nhap so phan tu cho chuoi : ");
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    printf("Nhap vao phan tu thu %d cho chuoi : ", i + 1);
    scanf("%d", &arr[i]);
  }
  int imax = arr[0];
  int imin = arr[N - 1];
  for (i = 1; i <= N - 1; i++) {
    if (imax < arr[i]) {
      imax = arr[i];
      arr[i] = arr[0];
      arr[0] = imax;
    }
    for (i = 1; i < N - 1; i++)
      if (imin > arr[i]) {
        imin = arr[i];
        arr[i] = arr[N - 1];
        arr[N - 1] = imin;
      }
  }
  for (i = 0; i < N; i++) {
    printf("%d", arr[i]);
  }
}
```