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
  lea    -0x4c(%rbp),%rax
  mov    %rax,%rsi
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L65
L43:
  lea    -0x48(%rbp),%rax
  mov    %rax,%rsi
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x38(%rbp),%rax
  mov    -0x48(%rbp),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rdx,%rcx
  cqto
  idiv   %rcx
  mov    %rax,-0x28(%rbp)
  shlq   -0x28(%rbp)
  mov    -0x38(%rbp),%rax
  mov    -0x48(%rbp),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rdx,%rcx
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  je     L70
  addq   $0x1,-0x28(%rbp)
L70:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x40(%rbp),%rax
  mov    -0x48(%rbp),%rdx
  sub    %rdx,%rax
  mov    -0x28(%rbp),%rdx
  lea    -0x5(%rdx),%rcx
  cqto
  idiv   %rcx
  mov    %rax,-0x20(%rbp)
  mov    -0x48(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  add    %rdx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L59
  movq   $0x1,-0x30(%rbp)
  jmp    L52
L13:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addq   $0x1,-0x30(%rbp)
L52:
  mov    -0x30(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jle    L13
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x4c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4c(%rbp)
  jmp    L65
L59:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  jmp    L36
L35:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  add    %rax,-0x30(%rbp)
L36:
  mov    -0x30(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jne    L35
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x4c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4c(%rbp)
L65:
  mov    -0x4c(%rbp),%eax
  test   %eax,%eax
  jne    L43
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L58
  call   <__stack_chk_fail@plt>
L58:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D70
data_type: string
name: disassemble_data
arguments: label: D79
data_type: string
name: disassemble_data
arguments: label: D0
data_type: string
name: disassemble_data
arguments: label: D4
data_type: string

tool: D70: %d

tool: D79: %lld

tool: D0: %lld


tool: D4: %lld 

assistant: ```c
int main() {
  long long int a, b, S, i;
  int t;
  scanf("%d", &t);
  while (t != 0) {
    scanf("%lld", &a);
    scanf("%lld", &b);
    scanf("%lld", &S);
    long long int n = S / (a + b);
    n *= 2;
    if (S % (a + b) != 0) {
      n++;
    }
    printf("%lld\n", n);
    long long int d = (b - a) / (n - 5);
    long long int p = a - 2 * d, q = b + 2 * d;
    if (d == 0) {
      for (i = 1; i <= n; i++) {
        printf("%lld ", p);
      }
      printf("\n");
      t--;
      continue;
    }
    for (i = p; i != q; i += d) {
      printf("%lld ", i);
    }
    printf("%lld ", i);
    printf("\n");
    t--;
  }
  return 0;
}
```