user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x7,%esi
  xor    %r8d,%r8d
  mov    $0x4ee728,%edx
  nopw   0x0(%rax,%rax,1)
L36:
  movslq %edx,%rax
  mov    %edx,%ecx
  lea    (%r8,%r8,4),%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdi,2),%r8d
  mov    %eax,%edx
  sub    $0x1,%esi
  jne    L36
  lea    D2(%rip),%rbp
  mov    %r8d,%edx
  mov    $0x4ee728,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x6,%esi
  xor    %r8d,%r8d
  mov    $0x63427,%edx
  nopw   0x0(%rax,%rax,1)
L38:
  movslq %edx,%rax
  mov    %edx,%ecx
  lea    (%r8,%r8,4),%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdi,2),%r8d
  mov    %eax,%edx
  sub    $0x1,%esi
  jne    L38
  mov    %r8d,%edx
  mov    $0x63427,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x6,%esi
  xor    %r8d,%r8d
  mov    $0x5a06e,%edx
  nopl   0x0(%rax,%rax,1)
L16:
  movslq %edx,%rax
  mov    %edx,%ecx
  lea    (%r8,%r8,4),%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdi,2),%r8d
  mov    %eax,%edx
  sub    $0x1,%esi
  jne    L16
  mov    %r8d,%edx
  mov    $0x5a06e,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x7,%esi
  xor    %r8d,%r8d
  mov    $0x2427e6,%edx
  nopl   0x0(%rax,%rax,1)
L1:
  movslq %edx,%rax
  mov    %edx,%ecx
  lea    (%r8,%r8,4),%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdi,2),%r8d
  mov    %eax,%edx
  sub    $0x1,%esi
  jne    L1
  mov    %r8d,%edx
  mov    $0x2427e6,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x7,%esi
  xor    %r8d,%r8d
  mov    $0x21507d,%edx
  nopl   0x0(%rax,%rax,1)
L7:
  movslq %edx,%rax
  mov    %edx,%ecx
  lea    (%r8,%r8,4),%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdi,2),%r8d
  mov    %eax,%edx
  sub    $0x1,%esi
  jne    L7
  mov    %rbp,%rdi
  mov    %r8d,%edx
  mov    $0x21507d,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
data_type: string

tool: D2: %d		%d


assistant: ```c
int main() {
  int num1, num, mun1 = 0;
  num1 = 5170984;
  num = num1;
  while (num1 != 0) {
    mun1 = (mun1 * 10) + (num1 % 10);
    num1 = num1 / 10;
  }
  printf("%d\t\t%d\n", num, mun1);
  num1 = 406567;
  num = num1;
  mun1 = 0;
  while (num1 != 0) {
    mun1 = (mun1 * 10) + (num1 % 10);
    num1 = num1 / 10;
  }
  printf("%d\t\t%d\n", num, mun1);
  num1 = 368750;
  num = num1;
  mun1 = 0;
  while (num1 != 0) {
    mun1 = (mun1 * 10) + (num1 % 10);
    num1 = num1 / 10;
  }
  printf("%d\t\t%d\n", num, mun1);
  num1 = 2369510;
  num = num1;
  mun1 = 0;
  while (num1 != 0) {
    mun1 = (mun1 * 10) + (num1 % 10);
    num1 = num1 / 10;
  }
  printf("%d\t\t%d\n", num, mun1);
  num1 = 2183293;
  num = num1;
  mun1 = 0;
  while (num1 != 0) {
    mun1 = (mun1 * 10) + (num1 % 10);
    num1 = num1 / 10;
  }
  printf("%d\t\t%d\n", num, mun1);
  return 0;
}
```