user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D20(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L27
  lea    -0x1(%rdx),%eax
  cmp    $0x6,%eax
  jbe    L87
  mov    %edx,%ecx
  movdqa D40(%rip),%xmm4
  xor    %eax,%eax
  pxor   %xmm3,%xmm3
  movdqa D41(%rip),%xmm6
  movdqa D55(%rip),%xmm5
  shr    $0x2,%ecx
L50:
  movdqa %xmm4,%xmm2
  add    $0x1,%eax
  paddd  %xmm6,%xmm4
  movdqa %xmm2,%xmm0
  paddd  %xmm5,%xmm0
  movdqa %xmm0,%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  psrlq  $0x20,%xmm2
  pmuludq %xmm2,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  paddd  %xmm1,%xmm3
  cmp    %eax,%ecx
  jne    L50
  movdqa %xmm3,%xmm0
  mov    %edx,%ecx
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%ecx
  paddd  %xmm0,%xmm3
  lea    0x1(%rcx,%rcx,1),%eax
  movdqa %xmm3,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm3
  movd   %xmm3,%esi
  cmp    %ecx,%edx
  je     L122
L128:
  lea    0x1(%rax),%edi
  lea    0x2(%rax),%r8d
  imul   %eax,%edi
  add    %edi,%esi
  lea    0x1(%rcx),%edi
  cmp    %edx,%edi
  jge    L122
  lea    0x3(%rax),%edi
  imul   %r8d,%edi
  lea    0x4(%rax),%r8d
  add    %edi,%esi
  lea    0x2(%rcx),%edi
  cmp    %edi,%edx
  jle    L122
  lea    0x5(%rax),%edi
  imul   %r8d,%edi
  lea    0x6(%rax),%r8d
  add    %edi,%esi
  lea    0x3(%rcx),%edi
  cmp    %edi,%edx
  jle    L122
  lea    0x7(%rax),%edi
  imul   %r8d,%edi
  lea    0x8(%rax),%r8d
  add    %edi,%esi
  lea    0x4(%rcx),%edi
  cmp    %edi,%edx
  jle    L122
  lea    0x9(%rax),%edi
  imul   %r8d,%edi
  lea    0xa(%rax),%r8d
  add    %edi,%esi
  lea    0x5(%rcx),%edi
  cmp    %edi,%edx
  jle    L122
  lea    0xb(%rax),%edi
  add    $0x6,%ecx
  imul   %r8d,%edi
  add    %edi,%esi
  lea    0xc(%rax),%edi
  cmp    %ecx,%edx
  jle    L122
  add    $0xd,%eax
  imul   %edi,%eax
  add    %eax,%esi
L122:
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L27:
  xor    %esi,%esi
  jmp    L122
L87:
  xor    %ecx,%ecx
  mov    $0x1,%eax
  xor    %esi,%esi
  jmp    L128
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D20
type: string
name: unpack_data
arguments: data_tag: D24
type: string
name: unpack_data
arguments: data_tag: D40
type: byte[16]
name: unpack_data
arguments: data_tag: D41
type: byte[16]
name: unpack_data
arguments: data_tag: D55
type: byte[16]
name: unpack_data
arguments: data_tag: D7
type: string

tool: D20: Enter the number of terms you want to sum up: 

tool: D24: %d

tool: {"D40": [0x01,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: {"D41": [0x08,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D55": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D7: 
The sum of the series is %d

assistant: ```c
int main(void) {
  int n, sum = 0, num = 1;
  printf("Enter the number of terms you want to sum up: ");
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    sum += (num * (num + 1));
    num += 2;
  }
  printf("\nThe sum of the series is %d", sum);
  return 0;
}
```