user: 下面的汇编指令对应的C源代码是什么：

```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    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3018(%rsp)
  xor    %eax,%eax
  mov    $0x7fff,%edi
  call   <sqrt@plt>
  mov    %eax,%edi
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  call   <find_primes@plt>
  lea    D15(%rip),%rbp
  lea    D26(%rip),%r13
  lea    D17(%rip),%r12
  jmp    L80
L63:
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    %eax,%r14d
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    %eax,%esi
  mov    %r14d,%edi
  call   <pow@plt>
  imul   %eax,%ebx
  mov    $0x0,%edi
L37:
  mov    %rbp,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L63
  lea    0x1010(%rsp),%rcx
  lea    0x10(%rsp),%rsi
  lea    -0x1(%rbx),%edi
  lea    0xc(%rsp),%r8
  mov    0x8(%rsp),%edx
  call   <find_prime_factors@plt>
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  cltq
  mov    0x1010(%rsp,%rax,4),%r14d
  mov    %r14d,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%edx
  cmp    $0x1,%edx
  jle    L12
  movslq %edx,%rax
  shl    $0x2,%rax
  lea    0x1010(%rsp,%rax,1),%rbx
  lea    0x100c(%rsp,%rax,1),%r15
  lea    -0x2(%rdx),%eax
  shl    $0x2,%rax
  sub    %rax,%r15
  mov    $0x1,%esi
  jmp    L0
L107:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbx),%r14d
  mov    %r14d,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
L101:
  sub    $0x4,%rbx
  cmp    %r15,%rbx
  je     L95
L0:
  cmp    %r14d,-0x8(%rbx)
  jne    L107
  add    $0x1,%esi
  jmp    L101
L12:
  mov    $0x1,%esi
L95:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
L80:
  lea    0x2010(%rsp),%rdi
  call   <gets@plt>
  test   %rax,%rax
  je     L108
  cmpb   $0x30,0x2010(%rsp)
  je     L108
  lea    0x2010(%rsp),%rdi
  mov    $0x1,%ebx
  jmp    L37
L108:
  mov    0x3018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  mov    $0x0,%eax
  add    $0x3028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L82:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D15
datatype: string
name: retrieve_data
arguments: data_label: D26
datatype: string
name: retrieve_data
arguments: data_label: D17
datatype: string

tool: D15:  

tool: D26: %d

tool: D17:  %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;
}
```