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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x2c(%rbp)
  jmp    L16
L54:
  lea    D80(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x28(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D90(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D101(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x2c(%rbp)
L16:
  cmpl   $0x1,-0x2c(%rbp)
  jle    L54
  movl   $0x0,-0x2c(%rbp)
  jmp    L48
L10:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  cmp    $0x63,%eax
  jle    L22
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x28(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D61(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x18(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D98(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L22:
  addl   $0x1,-0x2c(%rbp)
L48:
  cmpl   $0x1,-0x2c(%rbp)
  jle    L10
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D80
data_type: string
name: parse_data
arguments: data_tag: D49
data_type: string
name: parse_data
arguments: data_tag: D56
data_type: string
name: parse_data
arguments: data_tag: D90
data_type: string
name: parse_data
arguments: data_tag: D101
data_type: string
name: parse_data
arguments: data_tag: D61
data_type: string
name: parse_data
arguments: data_tag: D32
data_type: string
name: parse_data
arguments: data_tag: D74
data_type: string
name: parse_data
arguments: data_tag: D98
data_type: string

tool: D80: Digite a matricula do aluno

tool: D49: %d

tool: D56: Digite a quantidade de acertos 

tool: D90: Digite a quantidade de erros

tool: D101: Digite a quantidade de questoes em brancos

tool: D61: matricula : %d acertou mais de 50 por cento das questoes


tool: D32: com %d acertos


tool: D74: errou %d questoes


tool: D98: com %d questoes nao respondidas

assistant: ```c
int main() {
  int i, c, matricula[2], qtdacerto[2], qtderros[2], qtdbranco[2];
  for (c = 0; c < 2; c++) {
    printf("Digite a matricula do aluno");
    scanf("%d", &matricula[c]);
    printf("Digite a quantidade de acertos ");
    scanf("%d", &qtdacerto[c]);
    printf("Digite a quantidade de erros");
    scanf("%d", &qtderros[c]);
    printf("Digite a quantidade de questoes em brancos");
    scanf("%d", &qtdbranco[c]);
  }
  for (c = 0; c < 2; c++) {
    if (qtdacerto[c] >= 100) {
      printf("matricula : %d acertou mais de 50 por cento das questoes\n", matricula[c]);
      printf("com %d acertos\n", qtdacerto[c]);
      printf("errou %d questoes\n", qtderros[c]);
      printf("com %d questoes nao respondidas", qtdbranco[c]);
    }
  }
}
```