user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D21(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <puts@plt>
  movl   $0x1,0xc(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L115:
  mov    0xc(%rsp),%ebx
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  xor    %ebp,%ebp
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %ebx,%edx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L135:
  mov    %edx,%eax
  mov    $0xcccccccd,%ecx
  lea    0x0(%rbp,%rbp,4),%edi
  add    $0x1,%ebx
  imul   %rcx,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdi,2),%ebp
  mov    %eax,%edx
  test   %eax,%eax
  jne    L135
  cmp    $0x2,%ebx
  jne    L47
  nop
L107:
  movl   $0x43,0x8(%rsp)
  mov    $0x4c,%r12d
  mov    $0x58,%r15d
  jmp    L91
  cs nopw 0x0(%rax,%rax,1)
L14:
  cmp    $0x1,%ebx
  mov    $0x58,%eax
  cmovne 0x8(%rsp),%eax
  mov    %eax,0x8(%rsp)
  mov    $0x56,%eax
  cmove  %eax,%r12d
  mov    $0x49,%eax
  cmove  %eax,%r15d
L91:
  mov    %ebp,%edx
  mov    $0xcccccccd,%eax
  imul   %rax,%rdx
  mov    $0x210,%eax
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%ebp
  mov    %ebp,%r13d
  mov    %edx,%ebp
  bt     %r13,%rax
  jae    L74
  mov    0x8(%rsp),%edx
  cmp    $0x9,%r13d
  mov    %r15d,%esi
  lea    D25(%rip),%rdi
  cmovne %r12d,%edx
  xor    %eax,%eax
  call   <printf@plt>
L101:
  sub    $0x1,%ebx
  je     L52
L64:
  cmp    $0x2,%ebx
  je     L107
L47:
  cmp    $0x3,%ebx
  jne    L14
  mov    $0x43,%r15d
  jmp    L91
  nopl   0x0(%rax)
L74:
  cmp    $0x4,%r13d
  jg     L103
L114:
  test   %r13d,%r13d
  je     L101
  xor    %r14d,%r14d
  cs nopw 0x0(%rax,%rax,1)
L90:
  mov    %r15d,%edi
  add    $0x1,%r14d
  call   <putchar@plt>
  cmp    %r14d,%r13d
  jne    L90
  sub    $0x1,%ebx
  jne    L64
L52:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%eax
  cmp    $0x65,%eax
  jne    L115
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L103:
  mov    %r12d,%edi
  sub    $0x5,%r13d
  call   <putchar@plt>
  jmp    L114
```

tool_calls: name: disassemble_data
arguments: data_label: D21
type_infered: string
name: disassemble_data
arguments: data_label: D29
type_infered: string
name: disassemble_data
arguments: data_label: D25
type_infered: string

tool: D21: Decimal:	Romano:

tool: D29: %d		

tool: D25: %c%c

assistant: ```c
int main(void) {
  int num, auxiliar, mirror, lastDigit;
  int sistDecimal;
  int var1, var2, var3;
  int i, j;
  printf("Decimal:\tRomano:\n");
  for (num = 1; num <= 100; num++) {
    printf("%d\t\t", num);
    auxiliar = num;
    sistDecimal = 0;
    mirror = 0;
    while (auxiliar != 0) {
      mirror = mirror * 10 + auxiliar % 10;
      auxiliar /= 10;
      sistDecimal++;
    }
    for (i = sistDecimal; i >= 1; i--) {
      switch (i) {
      case 1:
        var1 = 'I';
        var2 = 'V';
        var3 = 'X';
        break;
      case 2:
        var1 = 'X';
        var2 = 'L';
        var3 = 'C';
        break;
      case 3:
        var1 = 'C';
        break;
      }
      lastDigit = mirror % 10;
      mirror /= 10;
      if (lastDigit == 4 || lastDigit == 9)
        printf("%c%c", var1, lastDigit == 9 ? var3 : var2);
      else {
        if (lastDigit >= 5) {
          printf("%c", var2);
          lastDigit -= 5;
        }
        for (j = 0; j != lastDigit; j++)
          printf("%c", var1);
      }
    }
    putchar('\n');
  }
}
```