user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D37(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x40(%rbp),%rsi
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x40(%rbp)
  call   <scanf@plt>
  movslq -0x40(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L60
L77:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L77
L60:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L8
L88:
  mov    %rsp,%r13
  test   %ecx,%ecx
  jle    L85
  mov    %r13,%r14
  xor    %r12d,%r12d
  lea    D1(%rip),%rbx
  nopl   (%rax)
L110:
  mov    %r14,%rsi
  mov    %rbx,%rdi
  add    $0x1,%r12d
  add    $0x4,%r14
  call   <scanf@plt>
  cmp    %r12d,-0x40(%rbp)
  jg     L110
L85:
  lea    -0x3c(%rbp),%rsi
  lea    D20(%rip),%rdi
  movl   $0x0,-0x3c(%rbp)
  call   <scanf@plt>
  mov    -0x40(%rbp),%esi
  mov    -0x3c(%rbp),%edi
  call   <sumc@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jle    L40
  mov    -0x3c(%rbp),%edx
  mov    %r13,%r15
  mov    $0x1,%r14d
  lea    D1(%rip),%rbx
  nopl   0x0(%rax)
L168:
  lea    -0x1(%r14),%ecx
  cmp    %ecx,%edx
  jle    L9
  mov    %edx,%eax
  sub    %r14d,%eax
  lea    0x1(%rax),%edi
  cmp    $0x2,%eax
  jbe    L94
  mov    %edi,%eax
  pxor   %xmm0,%xmm0
  shr    $0x2,%eax
  sub    $0x1,%eax
  shl    $0x4,%rax
  lea    0x10(%r15,%rax,1),%rsi
  mov    %r15,%rax
  nopl   (%rax)
L111:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rsi,%rax
  jne    L111
  movdqa %xmm0,%xmm1
  mov    %edi,%eax
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%eax
  paddd  %xmm1,%xmm0
  add    %eax,%ecx
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%esi
  cmp    %eax,%edi
  je     L80
L141:
  movslq %ecx,%rax
  lea    0x1(%rcx),%edi
  lea    0x0(%r13,%rax,4),%rax
  add    (%rax),%esi
  cmp    %edx,%edi
  jge    L80
  add    $0x2,%ecx
  add    0x4(%rax),%esi
  cmp    %ecx,%edx
  jle    L80
  add    0x8(%rax),%esi
L80:
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r14d
  add    $0x4,%r15
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x3c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x3c(%rbp)
  cmp    %r14d,%r12d
  jge    L168
L40:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L150
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  xchg   %ax,%ax
L9:
  xor    %esi,%esi
  jmp    L80
L94:
  xor    %esi,%esi
  jmp    L141
L8:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L88
L150:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D37
type_guessed: string
name: retrieve_data
arguments: data_tag: D1
type_guessed: string
name: retrieve_data
arguments: data_tag: D20
type_guessed: string

tool: D37: %d 


tool: D1: %d

tool: D20: 
 %d

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