user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  lea    D38(%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    L60
  movdqa D22(%rip),%xmm6
  add    $0x2,%r12d
  mov    $0x2,%ebx
  xor    %r10d,%r10d
  mov    $0x1,%ebp
  nopl   0x0(%rax)
L45:
  lea    -0x1(%rbx),%r11d
  mov    %ebx,%r8d
  mov    $0x1,%edx
  movd   %r11d,%xmm7
  pshufd $0x0,%xmm7,%xmm5
  jmp    L149
L84:
  mov    %r9d,%edx
  add    $0x1,%r8d
  cmp    %ebx,%r9d
  je     L123
L149:
  lea    0x1(%rdx),%r9d
  cmp    $0x1,%r8d
  jle    L84
  cmp    %r8d,%r9d
  mov    %r8d,%edi
  cmovle %r9d,%edi
  lea    -0x1(%rdi),%esi
  cmp    $0x1,%edi
  cmovle %ebp,%esi
  cmp    $0x4,%edi
  jle    L70
  mov    %esi,%ecx
  movd   %edx,%xmm7
  movdqa D28(%rip),%xmm3
  xor    %eax,%eax
  shr    $0x2,%ecx
  pshufd $0x0,%xmm7,%xmm4
  pxor   %xmm2,%xmm2
  nopw   0x0(%rax,%rax,1)
L166:
  movdqa %xmm3,%xmm0
  add    $0x1,%eax
  paddd  %xmm6,%xmm3
  movdqa %xmm0,%xmm1
  paddd  %xmm4,%xmm0
  paddd  %xmm5,%xmm1
  pcmpgtd %xmm5,%xmm0
  pcmpgtd %xmm4,%xmm1
  pand   %xmm1,%xmm0
  psubd  %xmm0,%xmm2
  cmp    %eax,%ecx
  jne    L166
  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     L84
L80:
  lea    (%r11,%rcx,1),%eax
  cmp    %eax,%edx
  jge    L109
  lea    (%rdx,%rcx,1),%eax
  cmp    %r11d,%eax
  setg   %al
  movzbl %al,%eax
  add    %eax,%r10d
L109:
  lea    0x1(%rcx),%eax
  cmp    %eax,%edi
  jle    L84
  lea    (%rax,%r11,1),%esi
  cmp    %edx,%esi
  jle    L17
  add    %edx,%eax
  cmp    %r11d,%eax
  setg   %al
  movzbl %al,%eax
  add    %eax,%r10d
L17:
  lea    0x2(%rcx),%eax
  cmp    %edi,%eax
  jge    L84
  lea    (%rax,%r11,1),%ecx
  cmp    %ecx,%edx
  jge    L84
  add    %edx,%eax
  mov    %r9d,%edx
  cmp    %eax,%r11d
  setl   %al
  add    $0x1,%r8d
  movzbl %al,%eax
  add    %eax,%r10d
  cmp    %ebx,%r9d
  jne    L149
  nopl   0x0(%rax)
L123:
  add    $0x1,%ebx
  cmp    %r12d,%ebx
  jne    L45
L132:
  mov    %r10d,%esi
  lea    D35(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L90
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L70:
  mov    $0x1,%ecx
  jmp    L80
L60:
  xor    %r10d,%r10d
  jmp    L132
L90:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D38
type_guessed: string
name: extract_symbolic_data
arguments: label: D22
type_guessed: byte[16]
name: extract_symbolic_data
arguments: label: D28
type_guessed: byte[16]
name: extract_symbolic_data
arguments: label: D35
type_guessed: string

tool: D38: %d

tool: {"D22": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D28": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D35: Number of possible triangles is %d

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