user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D10(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    -0x3c(%rbp),%rsi
  lea    D19(%rip),%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    -0x3c(%rbp),%edi
  call   <fact@plt>
  mov    %rsp,%rcx
  mov    %eax,%r12d
  movslq -0x3c(%rbp),%rax
  imul   %eax,%r12d
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L33
L46:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L46
L33:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L115
L2:
  lea    0x0(,%r12,4),%r14d
  mov    %rsp,%r15
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    -0x3c(%rbp),%edx
  mov    %rax,%r13
  test   %edx,%edx
  jle    L53
  movslq %edx,%rcx
  xor    %eax,%eax
  nopl   (%rax)
L127:
  mov    %eax,(%r15,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L127
L53:
  sub    $0x1,%edx
  xor    %esi,%esi
  mov    %r13,%rcx
  mov    %r15,%rdi
  call   <permute@plt>
  test   %r12d,%r12d
  jle    L5
  xor    %ebx,%ebx
  lea    D37(%rip),%r15
  jmp    L119
  nopl   (%rax)
L78:
  mov    %ecx,-0x44(%rbp)
  test   %ebx,%ebx
  jne    L94
  mov    $0x1,%ebx
L27:
  cmp    %ebx,%r12d
  jle    L5
L119:
  movslq %ebx,%rax
  mov    %r15,%rdi
  mov    0x0(%r13,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x1(%rbx),%ecx
  mov    %ecx,%eax
  cltd
  idivl  -0x3c(%rbp)
  test   %edx,%edx
  je     L78
  mov    %ecx,%ebx
  cmp    %ebx,%r12d
  jg     L119
L5:
  mov    %r13,%rdi
  call   <free@plt>
  movslq %r14d,%rdx
  mov    %r14d,%eax
  lea    0x3ff(%r14),%esi
  imul   $0x7ff2c15f,%rdx,%rdx
  sar    $0x1f,%eax
  lea    D55(%rip),%rdi
  sar    $0x33,%rdx
  sub    %eax,%edx
  test   %r14d,%r14d
  cmovns %r14d,%esi
  xor    %eax,%eax
  sar    $0xa,%esi
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L68
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax)
L94:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x44(%rbp),%ebx
  jmp    L27
L115:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L2
L68:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D10
data_type: string
name: unpack_data
arguments: data_tag: D19
data_type: string
name: unpack_data
arguments: data_tag: D11
data_type: string
name: unpack_data
arguments: data_tag: D37
data_type: string
name: unpack_data
arguments: data_tag: D55
data_type: string
name: unpack_data
arguments: data_tag: D5
data_type: string

tool: D10: Enter a number for permutation.

tool: D19: %d

tool: D11: 
PERMUTING %d...


tool: D37: %d 

tool: D55: 
MEMORY USED: %d Kb (about %d Mb)


tool: D5: 
Number of Permutations: %d



assistant: ```c
int main(void) {
  int numm, i, j, *nummptr;
  printf("Enter a number for permutation.\n");
  scanf("%d", &numm);
  printf("\nPERMUTING %d...\n", numm);
  nummptr = &numm;
  int permutations = fact(numm) * numm;
  int *permptr;
  permptr = &permutations;
  int *perm_array;
  int allele_array[numm];
  perm_array = (int *)malloc(permutations * sizeof(int));
  int mem = permutations * sizeof(int);
  for (i = 0; i < numm; i++) {
    allele_array[i] = i;
  }
  permute(allele_array, 0, numm - 1, perm_array);
  for (i = 0; i < permutations; i++) {
    printf("%d ", perm_array[i]);
    int n = i;
    n++;
    if (n % numm == 0 && i != 0) {
      printf("\n");
    }
  }
  free(perm_array);
  printf("\nMEMORY USED: %d Kb (about %d Mb)\n", mem / 1024, mem / 1049000);
  printf("\nNumber of Permutations: %d\n\n", permutations);
  return 0;
}
```