user: 请从给定的汇编指令推导出C源代码:

```asm
<teste_permutacao>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L176
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rcx
  movdqa D3(%rip),%xmm4
  xor    %ebp,%ebp
  mov    %rax,%r10
  lea    (%rax,%rax,4),%rax
  pxor   %xmm3,%xmm3
  mov    %esi,%r8d
  lea    (%rax,%rax,4),%rax
  shl    $0x4,%rax
  lea    0x190(%rdi,%rax,1),%r11
  mov    %esi,%eax
  mov    %esi,%edi
  xor    %esi,%esi
  shr    $0x2,%eax
  and    $0xfffffffc,%edi
  lea    -0x1(%rax),%r9d
  add    $0x1,%r9
  shl    $0x4,%r9
  cs nopw 0x0(%rax,%rax,1)
L6:
  cmp    $0x3,%r10d
  jbe    L14
  pxor   %xmm1,%xmm1
  lea    (%r9,%rcx,1),%rdx
  mov    %rcx,%rax
  movdqa %xmm1,%xmm2
  nopl   0x0(%rax)
L36:
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  pcmpeqd %xmm4,%xmm0
  psubd  %xmm0,%xmm1
  movdqu -0x10(%rax),%xmm0
  pcmpeqd %xmm3,%xmm0
  psubd  %xmm0,%xmm2
  cmp    %rdx,%rax
  jne    L36
  movdqa %xmm2,%xmm0
  psrldq $0x8,%xmm0
  paddd  %xmm0,%xmm2
  movdqa %xmm2,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm2
  movdqa %xmm1,%xmm0
  psrldq $0x8,%xmm0
  movd   %xmm2,%eax
  paddd  %xmm0,%xmm1
  add    %eax,%esi
  movdqa %xmm1,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm1
  movd   %xmm1,%eax
  add    %eax,%ebp
  mov    %edi,%eax
  cmp    %edi,%r8d
  je     L39
L83:
  movslq %eax,%rdx
  lea    (%rcx,%rdx,4),%rdx
  mov    (%rdx),%ebx
  test   %ebx,%ebx
  je     L102
  cmp    $0x1,%ebx
  sete   %bl
  movzbl %bl,%ebx
  add    %ebx,%ebp
L96:
  lea    0x1(%rax),%ebx
  cmp    %ebx,%r8d
  jle    L39
  mov    0x4(%rdx),%ebx
  test   %ebx,%ebx
  je     L8
  cmp    $0x1,%ebx
  sete   %bl
  movzbl %bl,%ebx
  add    %ebx,%ebp
L26:
  lea    0x2(%rax),%ebx
  cmp    %ebx,%r8d
  jle    L39
  mov    0x8(%rdx),%ebx
  test   %ebx,%ebx
  je     L146
  cmp    $0x1,%ebx
  sete   %bl
  movzbl %bl,%ebx
  add    %ebx,%ebp
L154:
  add    $0x3,%eax
  cmp    %eax,%r8d
  jle    L39
  mov    0xc(%rdx),%eax
  test   %eax,%eax
  jne    L50
  add    $0x1,%esi
L39:
  add    $0x190,%rcx
  cmp    %r11,%rcx
  jne    L6
L121:
  lea    D0(%rip),%r12
  mov    %r12,%rdi
  call   <printf@plt>
  pop    %rbx
  mov    %ebp,%esi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L102:
  add    $0x1,%esi
  jmp    L96
  nopl   (%rax)
L8:
  add    $0x1,%esi
  jmp    L26
  nopl   (%rax)
L146:
  add    $0x1,%esi
  jmp    L154
  nopl   (%rax)
L50:
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%ebp
  jmp    L39
  nopl   (%rax)
L14:
  xor    %eax,%eax
  jmp    L83
L176:
  xor    %ebp,%ebp
  xor    %esi,%esi
  jmp    L121
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type_infered: byte[16]
name: extract_symbolic_data
arguments: label: D0
type_infered: string

tool: {"D3": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D0: %d


assistant: ```c
void teste_permutacao(int M[][100], int ordem) {
  int i, j, cont1 = 0, cont2 = 0, aux = 1;
  for (i = 0; i < ordem; i++) {
    for (j = 0; j < ordem; j++) {
      if (M[i][j] == 0) {
        cont1++;
      }
      if (M[i][j] == 1) {
        cont2++;
      }
    }
  }
  printf("%d\n", cont1);
  printf("%d\n", cont2);
}
```