user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  lea    D22(%rip),%r14
  push   %r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  lea    D36(%rip),%rbp
  push   %rbx
  mov    $0x5,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
L146:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  movdqa D37(%rip),%xmm3
  test   %esi,%esi
  jle    L42
  lea    -0x1(%rsi),%eax
  cmp    $0x9,%eax
  jbe    L3
  mov    %esi,%edx
  movdqa D6(%rip),%xmm1
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  shr    $0x2,%edx
  nopw   0x0(%rax,%rax,1)
L80:
  movdqa %xmm1,%xmm2
  add    $0x1,%eax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %eax,%edx
  jne    L80
  movdqa %xmm0,%xmm1
  mov    %esi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  lea    0x1(%rcx),%eax
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%edx
  cmp    %ecx,%esi
  je     L98
L159:
  lea    0x1(%rax),%ecx
  add    %eax,%edx
  cmp    %ecx,%esi
  jl     L98
  add    %ecx,%edx
  lea    0x2(%rax),%ecx
  cmp    %esi,%ecx
  jg     L98
  add    %ecx,%edx
  lea    0x3(%rax),%ecx
  cmp    %ecx,%esi
  jl     L98
  add    %ecx,%edx
  lea    0x4(%rax),%ecx
  cmp    %ecx,%esi
  jl     L98
  add    %ecx,%edx
  lea    0x5(%rax),%ecx
  cmp    %ecx,%esi
  jl     L98
  add    %ecx,%edx
  lea    0x6(%rax),%ecx
  cmp    %ecx,%esi
  jl     L98
  add    %ecx,%edx
  lea    0x7(%rax),%ecx
  cmp    %ecx,%esi
  jl     L98
  add    %ecx,%edx
  lea    0x8(%rax),%ecx
  cmp    %ecx,%esi
  jl     L98
  add    %ecx,%edx
  add    $0x9,%eax
  lea    (%rdx,%rax,1),%ecx
  cmp    %eax,%esi
  cmovge %ecx,%edx
L98:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L146
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  xor    %edx,%edx
  jmp    L98
L3:
  xor    %edx,%edx
  mov    $0x1,%eax
  jmp    L159
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D22
type_infered: string
name: unpack_data
arguments: data_label: D10
type_infered: string
name: unpack_data
arguments: data_label: D36
type_infered: string
name: unpack_data
arguments: data_label: D37
type_infered: byte[16]
name: unpack_data
arguments: data_label: D6
type_infered: byte[16]

tool: D22: Que numero triangular quieres: 

tool: D10: %i

tool: D36: Numero triangular %i es %i



tool: {"D37": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D6": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int main(void) {
  int n, numero, numeroTriangular, contador;
  for (contador = 1; contador <= 5; contador++) {
    printf("Que numero triangular quieres: ");
    scanf("%i", &numero);
    numeroTriangular = 0;
    for (n = 1; n <= numero; n++) {
      numeroTriangular += n;
    }
    printf("Numero triangular %i es %i\n\n", numero, numeroTriangular);
  }
  return 0;
}
```