user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  lea    -0x64(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x64(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x48(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L22:
  cmp    %rdx,%rsp
  je     L58
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L22
L58:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L70
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L70:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x40(%rbp)
  movl   $0x0,-0x5c(%rbp)
  jmp    L23
L56:
  mov    -0x5c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x5c(%rbp)
L23:
  mov    -0x64(%rbp),%eax
  cmp    %eax,-0x5c(%rbp)
  jl     L56
  lea    -0x60(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x58(%rbp)
  jmp    L13
L11:
  movl   $0x0,-0x54(%rbp)
  mov    -0x58(%rbp),%eax
  mov    %eax,-0x50(%rbp)
  jmp    L72
L68:
  movl   $0x1,-0x4c(%rbp)
  jmp    L57
L39:
  mov    -0x50(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x40(%rbp),%rax
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  add    %eax,-0x54(%rbp)
  addl   $0x1,-0x4c(%rbp)
L57:
  cmpl   $0x1,-0x4c(%rbp)
  jle    L39
  addl   $0x1,-0x50(%rbp)
L72:
  mov    -0x60(%rbp),%edx
  mov    -0x58(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x50(%rbp)
  jl     L68
  mov    -0x54(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x58(%rbp)
L13:
  mov    -0x64(%rbp),%eax
  mov    -0x60(%rbp),%edx
  sub    $0x1,%edx
  sub    %edx,%eax
  cmp    %eax,-0x58(%rbp)
  jle    L11
  mov    $0x0,%eax
  mov    %rbx,%rsp
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D3
type_infered: string
name: parse_data
arguments: label: D16
type_infered: string

tool: D3: %d


tool: D16: %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;
}
```