user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x40(%rbp),%rsi
  push   %r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbx
  mov    %r12,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x40(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rdx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L126
L13:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L13
L126:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L22
L81:
  mov    %rsp,%r13
  test   %edx,%edx
  jle    L76
  mov    %r13,%r14
  xor    %ebx,%ebx
  lea    D19(%rip),%r15
  nopl   0x0(%rax,%rax,1)
L130:
  mov    %r14,%rsi
  mov    %r15,%rdi
  add    $0x1,%ebx
  add    $0x4,%r14
  call   <scanf@plt>
  cmp    %ebx,-0x40(%rbp)
  jg     L130
L76:
  lea    -0x3c(%rbp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%ecx
  lea    -0x1(%rcx),%edx
  cmp    %edx,-0x40(%rbp)
  jl     L78
  lea    -0x4(%r13),%r14
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L35:
  lea    (%rcx,%rbx,1),%edi
  xor    %esi,%esi
  cmp    %ebx,%edi
  jle    L5
  cmp    $0x2,%edx
  jbe    L33
  mov    %ecx,%eax
  pxor   %xmm0,%xmm0
  shr    $0x2,%eax
  sub    $0x1,%eax
  shl    $0x4,%rax
  lea    0x10(%r14,%rax,1),%rdx
  mov    %r14,%rax
  xchg   %ax,%ax
L148:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rdx,%rax
  jne    L148
  movdqa %xmm0,%xmm1
  mov    %ecx,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  paddd  %xmm1,%xmm0
  lea    (%rdx,%rbx,1),%eax
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%esi
  cmp    %edx,%ecx
  je     L5
L150:
  movslq %eax,%rdx
  lea    0x1(%rax),%ecx
  lea    0x0(%r13,%rdx,4),%rdx
  add    -0x4(%rdx),%esi
  cmp    %edi,%ecx
  jge    L5
  add    $0x2,%eax
  add    (%rdx),%esi
  cmp    %eax,%edi
  jle    L5
  add    0x4(%rdx),%esi
L5:
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%r14
  call   <printf@plt>
  mov    -0x3c(%rbp),%ecx
  mov    -0x40(%rbp),%eax
  lea    -0x1(%rcx),%edx
  sub    %edx,%eax
  cmp    %ebx,%eax
  jge    L35
L78:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L33:
  mov    %ebx,%eax
  xor    %esi,%esi
  jmp    L150
L22:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L81
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D0
data_type: string
name: unpack_data
arguments: data_tag: D19
data_type: string

tool: D0: %d


tool: D19: %d

assistant: ```c
int main() {
  int n;
  scanf("%d\n", &n);
  int A[n];
  for (int i = 0; i <= n - 1; i++) {
    scanf("%d", &A[i]);
  }
  int k;
  scanf("%d\n", &k);
  for (int a = 0; a <= n - (k - 1); a++) {
    int sum;
    sum = 0;
    for (int b = a; b <= k + a - 1; b++) {
      for (int c = 1; c <= 1; c++)
        sum = sum + A[(b - 1)];
    }
    printf("%d\n", sum);
  }
  return 0;
}
```