user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1d0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1b4(%rbp)
  lea    -0x1c4(%rbp),%rdx
  lea    -0x1c8(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L123
L34:
  addl   $0x1,-0x1b4(%rbp)
  mov    -0x1b4(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x1ac(%rbp)
  jmp    L92
L65:
  lea    -0x1a0(%rbp),%rdx
  mov    -0x1ac(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x1ac(%rbp)
L92:
  mov    -0x1c8(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L65
  movl   $0x0,-0x1ac(%rbp)
  jmp    L99
L127:
  lea    -0x1bc(%rbp),%rdx
  lea    -0x1c0(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x1b0(%rbp)
  movl   $0x0,-0x1a8(%rbp)
  jmp    L96
L21:
  lea    -0x1b8(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x1b8(%rbp),%edx
  mov    -0x1bc(%rbp),%eax
  cmp    %eax,%edx
  jne    L129
  addl   $0x1,-0x1b0(%rbp)
  jmp    L58
L129:
  mov    -0x1b0(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x1a4(%rbp)
  jmp    L52
L28:
  mov    -0x1a4(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  movslq %ecx,%rax
  mov    %edx,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x1a4(%rbp)
L52:
  mov    -0x1c0(%rbp),%eax
  cmp    %eax,-0x1a4(%rbp)
  jl     L28
L58:
  addl   $0x1,-0x1a8(%rbp)
L96:
  mov    -0x1c0(%rbp),%eax
  cmp    %eax,-0x1a8(%rbp)
  jl     L21
  addl   $0x1,-0x1ac(%rbp)
L99:
  mov    -0x1c4(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L127
  mov    -0x1a0(%rbp),%eax
  mov    %eax,%esi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x1c4(%rbp),%rdx
  lea    -0x1c8(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
L123:
  mov    -0x1c8(%rbp),%eax
  test   %eax,%eax
  jne    L34
  mov    -0x1c4(%rbp),%eax
  test   %eax,%eax
  jne    L34
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D5
type_guessed: string
name: read_data
arguments: data_label: D12
type_guessed: string
name: read_data
arguments: data_label: D7
type_guessed: string
name: read_data
arguments: data_label: D44
type_guessed: string

tool: D5: %d %d

tool: D12: Teste %d


tool: D7: %d

tool: D44: %d



assistant: ```c
int main() {
  int teste = 0;
  int participantes, rodadas;
  int part[100];
  int npart;
  int ordem;
  int acao;
  int posacao;
  int i, j, k;
  scanf("%d %d", &participantes, &rodadas);
  while (participantes || rodadas) {
    printf("Teste %d\n", ++teste);
    for (i = 0; i < participantes; i++)
      scanf("%d", &(part[i]));
    for (i = 0; i < rodadas; i++) {
      scanf("%d %d", &npart, &ordem);
      posacao = 0;
      for (j = 0; j < npart; j++) {
        scanf("%d", &acao);
        if (acao == ordem)
          posacao++;
        else
          for (k = posacao + 1; k < npart; k++)
            part[k - 1] = part[k];
      }
    }
    printf("%d\n\n", part[0]);
    scanf("%d %d", &participantes, &rodadas);
  }
  return 0;
}
```