user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  lea    D10(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  movabs $0x346dc5d63886594b,%rdx
  movabs $0x20c49ba5e353f7cf,%rcx
  mov    %rdi,%rax
  mov    %rdi,%r11
  mov    %rdi,%rsi
  imul   %rdx
  sar    $0x3f,%r11
  sar    $0xb,%rdx
  sub    %r11,%rdx
  imul   $0x2710,%rdx,%rax
  mov    %rdx,%r9
  sub    %rax,%rsi
  mov    %rsi,%rax
  sar    $0x3f,%rsi
  imul   %rcx
  mov    %rdi,%rax
  sar    $0x7,%rdx
  sub    %rsi,%rdx
  mov    %rdi,%rsi
  mov    %rdx,%r8
  imul   %rcx
  movabs $0xa3d70a3d70a3d70b,%rcx
  sar    $0x7,%rdx
  sub    %r11,%rdx
  imul   $0x3e8,%rdx,%rdx
  sub    %rdx,%rsi
  mov    %rsi,%rax
  imul   %rcx
  mov    %rdi,%rax
  lea    (%rdx,%rsi,1),%r10
  sar    $0x3f,%rsi
  imul   %rcx
  sar    $0x6,%r10
  sub    %rsi,%r10
  movabs $0x6666666666666667,%rsi
  lea    (%rdx,%rdi,1),%rcx
  sar    $0x6,%rcx
  sub    %r11,%rcx
  lea    (%rcx,%rcx,4),%rax
  mov    %rdi,%rcx
  lea    (%rax,%rax,4),%rax
  shl    $0x2,%rax
  sub    %rax,%rcx
  mov    %rcx,%rax
  sar    $0x3f,%rcx
  imul   %rsi
  mov    %rdi,%rax
  sar    $0x2,%rdx
  sub    %rcx,%rdx
  mov    %rdx,%r12
  imul   %rsi
  mov    %rdx,%rsi
  sar    $0x2,%rsi
  sub    %r11,%rsi
  lea    (%rsi,%rsi,4),%rax
  add    %rax,%rax
  sub    %rax,%rdi
  mov    %rdi,%rsi
  test   %r9,%r9
  jne    L25
  test   %r8,%r8
  jne    L31
  test   %r10,%r10
  jne    L32
  test   %r12,%r12
  je     L74
  mov    %r12,%rdx
  lea    D53(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L50
L31:
  mov    %r10,%rcx
  mov    %r12,%rdx
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L50:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L86
  add    $0x10,%rsp
  pop    %r12
  ret
L25:
  mov    %r10,%rcx
  mov    %r12,%rdx
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L50
L74:
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L50
L32:
  mov    %r10,%rcx
  mov    %r12,%rdx
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L50
L86:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D10
type_infered: string
name: read_data
arguments: data_tag: D53
type_infered: string
name: read_data
arguments: data_tag: D14
type_infered: string
name: read_data
arguments: data_tag: D25
type_infered: string
name: read_data
arguments: data_tag: D17
type_infered: string
name: read_data
arguments: data_tag: D40
type_infered: string

tool: D10: %ld

tool: D53: There are two:%d,%d


tool: D14: There are four:%d,%d,%d,%d


tool: D25: There are five:%d,%d,%d,%d,%d


tool: D17: There are one:%d


tool: D40: There are three:%d,%d,%d


assistant: ```c
void main() {
  long int a, b, c, d, e;
  long int number;
  scanf("%ld", &number);
  a = number / 10000;
  b = number % 10000 / 1000;
  c = number % 1000 / 100;
  d = number % 100 / 10;
  e = number % 10;
  if (0 != a)
    printf("There are five:%d,%d,%d,%d,%d\n", e, d, c, b, a);
  else if (0 != b)
    printf("There are four:%d,%d,%d,%d\n", e, d, c, b);
  else if (0 != c)
    printf("There are three:%d,%d,%d\n", e, d, c);
  else if (0 != d)
    printf("There are two:%d,%d\n", e, d);
  else
    printf("There are one:%d\n", e);
}
```