user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  lea    D6(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%r12d
  test   %r12d,%r12d
  jle    L82
  movdqa D18(%rip),%xmm6
  add    $0x1,%r12d
  mov    $0x2,%ebx
  xor    %r10d,%r10d
  mov    $0x1,%r11d
  mov    $0x1,%ebp
  nopw   0x0(%rax,%rax,1)
L60:
  movd   %r11d,%xmm7
  mov    %ebx,%r8d
  mov    $0x1,%edx
  pshufd $0x0,%xmm7,%xmm5
  jmp    L107
  nopl   0x0(%rax)
L26:
  mov    %r9d,%edx
  add    $0x1,%r8d
  cmp    %ebx,%r9d
  je     L119
L107:
  lea    0x1(%rdx),%r9d
  cmp    $0x1,%r8d
  jle    L26
  cmp    %r8d,%r9d
  mov    %r8d,%edi
  cmovle %r9d,%edi
  lea    -0x1(%rdi),%esi
  cmp    $0x1,%edi
  cmovle %ebp,%esi
  cmp    $0x4,%edi
  jle    L155
  mov    %esi,%ecx
  movd   %edx,%xmm7
  movdqa D26(%rip),%xmm3
  xor    %eax,%eax
  shr    $0x2,%ecx
  pshufd $0x0,%xmm7,%xmm4
  pxor   %xmm2,%xmm2
  nopw   0x0(%rax,%rax,1)
L19:
  movdqa %xmm3,%xmm0
  add    $0x1,%eax
  paddd  %xmm6,%xmm3
  movdqa %xmm0,%xmm1
  paddd  %xmm5,%xmm0
  paddd  %xmm4,%xmm1
  pcmpgtd %xmm4,%xmm0
  pcmpgtd %xmm5,%xmm1
  pand   %xmm1,%xmm0
  psubd  %xmm0,%xmm2
  cmp    %eax,%ecx
  jne    L19
  movdqa %xmm2,%xmm0
  psrldq $0x8,%xmm0
  paddd  %xmm0,%xmm2
  movdqa %xmm2,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm2
  movd   %xmm2,%eax
  add    %eax,%r10d
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  lea    0x1(%rax),%ecx
  cmp    %eax,%esi
  je     L26
L154:
  lea    (%rdx,%rcx,1),%eax
  cmp    %eax,%r11d
  jge    L3
  lea    (%r11,%rcx,1),%eax
  cmp    %edx,%eax
  setg   %al
  movzbl %al,%eax
  add    %eax,%r10d
L3:
  lea    0x1(%rcx),%eax
  cmp    %eax,%edi
  jle    L26
  lea    (%rax,%rdx,1),%esi
  cmp    %r11d,%esi
  jle    L108
  add    %r11d,%eax
  cmp    %edx,%eax
  setg   %al
  movzbl %al,%eax
  add    %eax,%r10d
L108:
  lea    0x2(%rcx),%eax
  cmp    %edi,%eax
  jge    L26
  lea    (%rax,%rdx,1),%ecx
  cmp    %ecx,%r11d
  jge    L26
  add    %r11d,%eax
  cmp    %eax,%edx
  mov    %r9d,%edx
  setl   %al
  add    $0x1,%r8d
  movzbl %al,%eax
  add    %eax,%r10d
  cmp    %ebx,%r9d
  jne    L107
  nopl   0x0(%rax)
L119:
  add    $0x1,%r11d
  add    $0x1,%ebx
  cmp    %r12d,%r11d
  jne    L60
L105:
  mov    %r10d,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L155:
  mov    $0x1,%ecx
  jmp    L154
L82:
  xor    %r10d,%r10d
  jmp    L105
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type: string
name: extract_symbolic_data
arguments: data_label: D18
type: byte[16]
name: extract_symbolic_data
arguments: data_label: D26
type: byte[16]
name: extract_symbolic_data
arguments: data_label: D7
type: string

tool: D6: %d

tool: {"D18": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D26": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D7: Number of possible triangles is %d

assistant: ```c
int main() {
  int n, a = 1, b = 1, c = 1, i;
  scanf("%d", &n);
  i = 0;
  for (a = 1; a <= n; a++) {
    for (b = 1; b <= a; b++) {
      for (c = 1; c <= b; c++) {
        if (a + b > c && b + c > a && c + a > b) {
          i++;
        }
      }
    }
  }
  printf("Number of possible triangles is %d", i);
  return 0;
}
```