user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D39(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L122
  movss  D59(%rip),%xmm5
  add    $0x1,%edi
  pxor   %xmm4,%xmm4
  mov    $0x1,%edx
  movdqa D21(%rip),%xmm3
  nopl   (%rax)
L2:
  lea    -0x1(%rdx),%eax
  cmp    $0x9,%eax
  jbe    L156
  mov    %edx,%ecx
  movdqa D42(%rip),%xmm1
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  nop
L92:
  movdqa %xmm1,%xmm2
  add    $0x1,%eax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %eax,%ecx
  jne    L92
  movdqa %xmm0,%xmm1
  mov    %edx,%esi
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%esi
  paddd  %xmm1,%xmm0
  lea    0x1(%rsi),%ecx
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %edx,%esi
  je     L115
L71:
  lea    0x1(%rcx),%esi
  add    %ecx,%eax
  cmp    %esi,%edx
  jl     L115
  add    %esi,%eax
  lea    0x2(%rcx),%esi
  cmp    %esi,%edx
  jl     L115
  add    %esi,%eax
  lea    0x3(%rcx),%esi
  cmp    %esi,%edx
  jl     L115
  add    %esi,%eax
  lea    0x4(%rcx),%esi
  cmp    %esi,%edx
  jl     L115
  add    %esi,%eax
  lea    0x5(%rcx),%esi
  cmp    %esi,%edx
  jl     L115
  add    %esi,%eax
  lea    0x6(%rcx),%esi
  cmp    %esi,%edx
  jl     L115
  add    %esi,%eax
  lea    0x7(%rcx),%esi
  cmp    %esi,%edx
  jl     L115
  add    %esi,%eax
  lea    0x8(%rcx),%esi
  cmp    %esi,%edx
  jl     L115
  add    %esi,%eax
  add    $0x9,%ecx
  lea    (%rax,%rcx,1),%esi
  cmp    %ecx,%edx
  cmovge %esi,%eax
L115:
  pxor   %xmm1,%xmm1
  movaps %xmm5,%xmm0
  add    $0x1,%edx
  cvtsi2ss %eax,%xmm1
  divss  %xmm1,%xmm0
  addss  %xmm0,%xmm4
  cmp    %edi,%edx
  jne    L2
L144:
  pxor   %xmm0,%xmm0
  lea    D27(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm4,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L156:
  xor    %eax,%eax
  mov    $0x1,%ecx
  jmp    L71
L122:
  pxor   %xmm4,%xmm4
  jmp    L144
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D14
type_infered: string
name: extract_symbolic_data
arguments: label: D39
type_infered: string
name: extract_symbolic_data
arguments: label: D59
type_infered: byte[4]
name: extract_symbolic_data
arguments: label: D21
type_infered: byte[16]
name: extract_symbolic_data
arguments: label: D42
type_infered: byte[16]
name: extract_symbolic_data
arguments: label: D27
type_infered: string

tool: D14: Please input the value of n:

tool: D39: %d

tool: {"D59": [0x00,0x00,0x80,0x3f]}

tool: {"D21": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D42": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D27: sum=%.6f


assistant: ```c
int main() {
  int n, i, j, s;
  float sum = 0;
  printf("Please input the value of n:\n");
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    s = 0;
    for (j = 1; j <= i; j++)
      s += j;
    sum += 1 / (float)s;
  }
  printf("sum=%.6f\n", sum);
  return 0;
}
```