user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x38,%rsp
  mov    $0x7fff,%edi
  lea    D4(%rip),%rbp
  lea    D6(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x3028(%rsp)
  xor    %eax,%eax
  lea    0x2020(%rsp),%r13
  call   <sqrt@plt>
  lea    0x18(%rsp),%rdx
  mov    %eax,%edi
  lea    0x20(%rsp),%rax
  mov    %rax,%rsi
  mov    %rax,(%rsp)
  call   <find_primes@plt>
  lea    0x1c(%rsp),%rax
  mov    %r13,%rdi
  mov    %rax,0x8(%rsp)
  call   <gets@plt>
  test   %rax,%rax
  je     L27
  nopl   0x0(%rax)
L64:
  cmpb   $0x30,0x2020(%rsp)
  je     L27
  mov    %r13,%rdi
  mov    $0x1,%ebx
  jmp    L86
  nopl   0x0(%rax,%rax,1)
L28:
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  mov    %rbp,%rsi
  xor    %edi,%edi
  mov    %eax,%r14d
  call   <strtok@plt>
  mov    $0xa,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %r14d,%edi
  mov    %eax,%esi
  call   <pow@plt>
  xor    %edi,%edi
  imul   %eax,%ebx
L86:
  mov    %rbp,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L28
  mov    0x8(%rsp),%r8
  mov    0x18(%rsp),%edx
  lea    -0x1(%rbx),%edi
  lea    0x1020(%rsp),%r15
  mov    (%rsp),%rsi
  mov    %r15,%rcx
  call   <find_prime_factors@plt>
  mov    0x1c(%rsp),%eax
  lea    D14(%rip),%rdi
  sub    $0x1,%eax
  cltq
  mov    0x1020(%rsp,%rax,4),%r14d
  xor    %eax,%eax
  mov    %r14d,%esi
  call   <printf@plt>
  mov    0x1c(%rsp),%esi
  cmp    $0x1,%esi
  jle    L103
  movslq %esi,%rax
  shl    $0x2,%rax
  lea    0x101c(%rsp,%rax,1),%rbx
  add    %rax,%r15
  lea    -0x2(%rsi),%eax
  shl    $0x2,%rax
  sub    %rax,%rbx
L75:
  mov    $0x1,%esi
L12:
  cmp    %r14d,-0x8(%r15)
  jne    L89
  sub    $0x4,%r15
  add    $0x1,%esi
  cmp    %r15,%rbx
  jne    L12
L94:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r13,%rdi
  call   <gets@plt>
  test   %rax,%rax
  jne    L64
L27:
  mov    0x3028(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L92
  add    $0x3038,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L89:
  mov    %r12,%rdi
  xor    %eax,%eax
  sub    $0x4,%r15
  call   <printf@plt>
  mov    -0x4(%r15),%r14d
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    %r14d,%esi
  call   <printf@plt>
  cmp    %rbx,%r15
  jne    L75
L103:
  mov    $0x1,%esi
  jmp    L94
L92:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D4
type: string
name: parse_data
arguments: data_label: D6
type: string
name: parse_data
arguments: data_label: D14
type: string

tool: D4:  

tool: D6:  %d

tool: D14: %d

assistant: ```c
int main() {
  int num, base, exp;
  int primes[1 << 10], num_primes;
  int factors[1 << 10], num_factors;
  char line[4096];
  int i, count, prev;
  find_primes(sqrt(32767), primes, &num_primes);
  while (gets(line) != NULL && line[0] != '0') {
    char *tok;
    char *ptr = &line[0];
    num = 1;
    while ((tok = strtok(ptr, " ")) != NULL) {
      base = strtol(tok, NULL, 10);
      tok = strtok(NULL, " ");
      exp = strtol(tok, NULL, 10);
      num *= pow(base, exp);
      ptr = NULL;
    }
    --num;
    find_prime_factors(num, primes, num_primes, factors, &num_factors);
    prev = factors[num_factors - 1];
    count = 1;
    printf("%d", prev);
    for (i = num_factors - 2; i >= 0; i--) {
      if (factors[i] == prev) {
        count++;
      } else {
        printf(" %d", count);
        prev = factors[i];
        printf(" %d", prev);
        count = 1;
      }
    }
    printf(" %d", count);
    printf("\n");
  }
  return 0;
}
```