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
  sub    $0x1c0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x1ac(%rbp)
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1b4(%rbp),%rax
  mov    %rax,%rsi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x1b0(%rbp)
  jmp    L43
L20:
  mov    -0x1b0(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1a0(%rbp),%rdx
  mov    -0x1b0(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x1b0(%rbp),%eax
  cltq
  movss  -0x1a0(%rbp,%rax,4),%xmm0
  movss  -0x1ac(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x1ac(%rbp)
  addl   $0x1,-0x1b0(%rbp)
L43:
  mov    -0x1b4(%rbp),%eax
  cmp    %eax,-0x1b0(%rbp)
  jl     L20
  mov    -0x1b4(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  movss  -0x1ac(%rbp),%xmm0
  divss  %xmm1,%xmm0
  movss  %xmm0,-0x1a8(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0x1a8(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movl   $0x0,-0x1b0(%rbp)
  jmp    L33
L49:
  mov    -0x1b0(%rbp),%eax
  cltq
  movss  -0x1a0(%rbp,%rax,4),%xmm0
  subss  -0x1a8(%rbp),%xmm0
  movss  %xmm0,-0x1a4(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x1a4(%rbp),%xmm0
  mov    -0x1b0(%rbp),%eax
  cltq
  movss  -0x1a0(%rbp,%rax,4),%xmm1
  pxor   %xmm3,%xmm3
  cvtss2sd %xmm1,%xmm3
  movq   %xmm3,%rax
  mov    -0x1b0(%rbp),%edx
  add    $0x1,%edx
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  mov    %edx,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  addl   $0x1,-0x1b0(%rbp)
L33:
  mov    -0x1b4(%rbp),%eax
  cmp    %eax,-0x1b0(%rbp)
  jl     L49
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D43
type_guessed: string
name: parse_data
arguments: tag: D46
type_guessed: string
name: parse_data
arguments: tag: D55
type_guessed: string
name: parse_data
arguments: tag: D34
type_guessed: string
name: parse_data
arguments: tag: D12
type_guessed: string
name: parse_data
arguments: tag: D48
type_guessed: string

tool: D43: How many numbers? 

tool: D46: %d

tool: D55: i = %d x = 

tool: D34: %f

tool: D12: 
Average is: %5.2f



tool: D48: i = %d x = %5.2f d = %5.2f


assistant: ```c
int main() {
  int n, count;
  float avg, d, sum = 0;
  float num[100];
  printf("How many numbers? ");
  scanf("%d", &n);
  printf("\n");
  for (count = 0; count < n; count++) {
    printf("i = %d x = ", count + 1);
    scanf("%f", &num[count]);
    sum += num[count];
  }
  avg = sum / n;
  printf("\nAverage is: %5.2f\n\n", avg);
  for (count = 0; count < n; count++) {
    d = num[count] - avg;
    printf("i = %d x = %5.2f d = %5.2f\n", count + 1, num[count], d);
  }
  return 0;
}
```