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,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x2c(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L116
L82:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x2aaaaaab,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  mov    %eax,%ecx
  sub    %esi,%ecx
  mov    %ecx,%eax
  add    %eax,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,%ecx
  lea    0x1(%rcx),%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x1,-0xc(%rbp)
  jne    L39
  addl   $0x1,-0x24(%rbp)
  jmp    L83
L39:
  cmpl   $0x2,-0xc(%rbp)
  jne    L119
  addl   $0x1,-0x20(%rbp)
  jmp    L83
L119:
  cmpl   $0x3,-0xc(%rbp)
  jne    L63
  addl   $0x1,-0x1c(%rbp)
  jmp    L83
L63:
  cmpl   $0x4,-0xc(%rbp)
  jne    L27
  addl   $0x1,-0x18(%rbp)
  jmp    L83
L27:
  cmpl   $0x5,-0xc(%rbp)
  jne    L26
  addl   $0x1,-0x14(%rbp)
  jmp    L83
L26:
  cmpl   $0x6,-0xc(%rbp)
  jne    L83
  addl   $0x1,-0x10(%rbp)
L83:
  addl   $0x1,-0x28(%rbp)
L116:
  mov    -0x2c(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jl     L82
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D78(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D2
data_type: string
name: read_data
arguments: label: D39
data_type: string
name: read_data
arguments: label: D48
data_type: string
name: read_data
arguments: label: D79
data_type: string
name: read_data
arguments: label: D20
data_type: string
name: read_data
arguments: label: D38
data_type: string
name: read_data
arguments: label: D78
data_type: string
name: read_data
arguments: label: D21
data_type: string

tool: D2:  How many times you want to roll a die?

tool: D39: %d

tool: D48: Ones = %d


tool: D79: Twos = %d


tool: D20: Threes = %d


tool: D38: Fours = %d


tool: D78: Fives = %d


tool: D21: Sixes = %d


assistant: ```c
int main() {
  int x;
  int counter = 0;
  int roll;
  int ones = 0;
  int twos = 0;
  int threes = 0;
  int fours = 0;
  int fives = 0;
  int sixes = 0;
  srand(time(NULL));
  printf(" How many times you want to roll a die?\n");
  scanf("%d", &x);
  while (counter < x) {
    roll = rand() % 6 + 1;
    if (roll == 1) {
      ones++;
    } else if (roll == 2) {
      twos++;
    } else if (roll == 3) {
      threes++;
    } else if (roll == 4) {
      fours++;
    } else if (roll == 5) {
      fives++;
    } else if (roll == 6) {
      sixes++;
    }
    counter++;
  }
  printf("Ones = %d\n", ones);
  printf("Twos = %d\n", twos);
  printf("Threes = %d\n", threes);
  printf("Fours = %d\n", fours);
  printf("Fives = %d\n", fives);
  printf("Sixes = %d\n", sixes);
  return 0;
}
```