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
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movl   $0x0,-0x3c(%rbp)
  movq   $0x0,-0x20(%rbp)
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L21
L6:
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <abs@plt>
  mov    %eax,%edi
  call   <log10@plt>
  mov    %eax,%edi
  call   <floor@plt>
  add    $0x1,%eax
  mov    %eax,-0x3c(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x30(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x30(%rbp)
  jmp    L61
L34:
  mov    -0x30(%rbp),%rax
  cmp    $0x1,%rax
  jne    L35
  mov    -0x3c(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    -0x38(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x30(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x30(%rbp)
  jmp    L60
L35:
  mov    -0x3c(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    -0x38(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,%rbx
  mov    -0x3c(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  imul   -0x20(%rbp),%rax
  lea    (%rbx,%rax,1),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x30(%rbp),%rax
  sub    $0x2,%rax
  mov    %rax,-0x30(%rbp)
L60:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L61:
  mov    -0x30(%rbp),%rax
  test   %rax,%rax
  jne    L34
  mov    -0x38(%rbp),%rax
  mov    -0x28(%rbp),%rcx
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x40(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x40(%rbp)
L21:
  mov    -0x40(%rbp),%eax
  test   %eax,%eax
  jg     L6
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D54
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D45
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D29
type_infered: string

tool: D54: %llu

tool: D45: 
%llu

tool: D29: %llu


assistant: ```c
int main() {
  int test = 0;
  unsigned long long int a = 0, n = 0, m = 0;
  int length = 0;
  unsigned long long int tempa = 0;
  scanf("%llu", &test);
  while (test > 0) {
    scanf("%llu", &a);
    scanf("%llu", &n);
    scanf("%llu", &m);
    length = floor(log10(abs(a))) + 1;
    tempa = a;
    --n;
    while (n > 0) {
      if (n == 1) {
        a = a * pow(10, length);
        a += tempa;
        --n;
      } else {
        a = a * pow(10, length * 2) + tempa * pow(10, length) + tempa;
        n = n - 2;
      }
      printf("\n%llu", a);
    }
    printf("%llu\n", a % m);
    --test;
  }
  return 0;
}
```