user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %edi,-0x64(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x30(%rbp)
  movl   $0x3,-0x2c(%rbp)
  movl   $0x4,-0x28(%rbp)
  movl   $0x2,-0x24(%rbp)
  movl   $0x4,-0x20(%rbp)
  movl   $0x4,-0x1c(%rbp)
  movl   $0x2,-0x18(%rbp)
  movl   $0x4,-0x14(%rbp)
  movl   $0x4,-0x10(%rbp)
  movl   $0x9,-0x34(%rbp)
  movl   $0x0,-0x48(%rbp)
  jmp    L167
L114:
  cmpl   $0x0,-0x58(%rbp)
  jle    L122
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,-0x50(%rbp)
  jne    L122
  addl   $0x1,-0x58(%rbp)
  jmp    L48
L122:
  cmpl   $0x0,-0x54(%rbp)
  jle    L51
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,-0x4c(%rbp)
  jne    L51
  addl   $0x1,-0x54(%rbp)
  jmp    L48
L51:
  cmpl   $0x0,-0x58(%rbp)
  jne    L43
  movl   $0x0,-0x58(%rbp)
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x50(%rbp)
  jmp    L48
L43:
  cmpl   $0x0,-0x54(%rbp)
  jne    L160
  movl   $0x0,-0x54(%rbp)
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x4c(%rbp)
  jmp    L48
L160:
  subl   $0x1,-0x58(%rbp)
  subl   $0x1,-0x54(%rbp)
L48:
  addl   $0x1,-0x48(%rbp)
L167:
  mov    -0x48(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L114
  movl   $0x0,-0x44(%rbp)
  movl   $0x0,-0x40(%rbp)
  movl   $0x0,-0x3c(%rbp)
  jmp    L172
L83:
  mov    -0x3c(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,-0x50(%rbp)
  jne    L196
  addl   $0x1,-0x44(%rbp)
L196:
  addl   $0x1,-0x3c(%rbp)
L172:
  mov    -0x3c(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L83
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cmp    %eax,-0x44(%rbp)
  jle    L69
  mov    -0x50(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L69:
  movl   $0x0,-0x38(%rbp)
  jmp    L139
L95:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,-0x4c(%rbp)
  jne    L2
  addl   $0x1,-0x40(%rbp)
L2:
  addl   $0x1,-0x38(%rbp)
L139:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L95
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cmp    %eax,-0x40(%rbp)
  jle    L8
  mov    -0x4c(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L8:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L100
  call   <__stack_chk_fail@plt>
L100:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D14
datatype: string

tool: D14: Majority element can be  %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int a[] = {3, 3, 4, 2, 4, 4, 2, 4, 4};
  int size = sizeof(a) / sizeof(a[0]);
  int c1, c2;
  int A, B;
  for (int i = 0; i < size; i++) {
    if (c1 > 0 && A == a[i])
      c1++;
    else if (c2 > 0 && B == a[i])
      c2++;
    else if (c1 == 0) {
      c1 = 0;
      A = a[i];
    } else if (c2 == 0) {
      c2 = 0;
      B = a[i];
    } else {
      c1--;
      c2--;
    }
  }
  int count1 = 0, count2 = 0;
  for (int i = 0; i < size; i++) {
    if (a[i] == A) {
      count1++;
    }
  }
  if (count1 >= ((size / 3) + 1))
    printf("Majority element can be  %d\n", A);
  for (int i = 0; i < size; i++) {
    if (a[i] == B) {
      count2++;
    }
  }
  if (count2 >= ((size / 3) + 1))
    printf("Majority element can be  %d\n", B);
  return 0;
}
```