user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x44(%rbp)
  lea    -0x48(%rbp),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L125
L23:
  movl   $0x0,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  mov    -0x38(%rbp),%eax
  mov    %eax,-0x3c(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0x40(%rbp)
  lea    -0x4c(%rbp),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L42
L49:
  lea    -0x30(%rbp),%rdx
  mov    -0x40(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x40(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  add    %eax,-0x3c(%rbp)
  mov    -0x40(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jle    L80
  mov    -0x40(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L73
L80:
  mov    -0x40(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jne    L73
  mov    -0x40(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  je     L73
  movl   $0x1,-0x34(%rbp)
L73:
  addl   $0x1,-0x40(%rbp)
L42:
  mov    -0x4c(%rbp),%eax
  cmp    %eax,-0x40(%rbp)
  jl     L49
  cmpl   $0x0,-0x34(%rbp)
  je     L29
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L56
L29:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x3c(%rbp),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  cmp    %eax,%edx
  jle    L69
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L56
L69:
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L56:
  addl   $0x1,-0x44(%rbp)
L125:
  mov    -0x48(%rbp),%eax
  cmp    %eax,-0x44(%rbp)
  jl     L23
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L81
  call   <__stack_chk_fail@plt>
L81:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D40
datatype: string
name: extract_symbolic_data
arguments: tag: D48
datatype: string
name: extract_symbolic_data
arguments: tag: D22
datatype: string
name: extract_symbolic_data
arguments: tag: D27
datatype: string

tool: D40: %d

tool: D48: no winner

tool: D22: majority winner %d


tool: D27: minority winner %d


assistant: ```c
int main() {
  int i, j, c, t, n, tot, win, tie;
  int votes[10];
  for (i = 0, scanf("%d", &t); i < t; i++) {
    for (j = tot = win = tie = 0, scanf("%d", &c); j < c; j++) {
      scanf("%d", &votes[j]);
      tot += votes[j];
      if (votes[j] > votes[win]) {
        win = j;
        tie = 0;
      } else if (votes[j] == votes[win] && j != win) {
        tie = 1;
      }
    }
    if (tie) {
      printf("no winner\n");
    } else if (votes[win] > tot / 2)
      printf("majority winner %d\n", win + 1);
    else
      printf("minority winner %d\n", win + 1);
  }
  return 0;
}
```