user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D16(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D31(%rip),%r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x48(%rbp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x40(%rbp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D86(%rip),%r12
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x50(%rbp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x4c(%rbp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x40(%rbp),%rax
  cmp    %rax,-0x48(%rbp)
  jg     L196
L110:
  movslq -0x50(%rbp),%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,8),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L1
L203:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L203
L1:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L80
L49:
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <clock@plt>
  mov    %eax,-0x60(%rbp)
  mov    -0x50(%rbp),%eax
  lea    -0x1(%rax),%ecx
  test   %ecx,%ecx
  jle    L106
  cmp    $0x2,%eax
  je     L41
  mov    %ecx,%edx
  movdqa D87(%rip),%xmm0
  mov    %rbx,%rax
  shr    %edx
  shl    $0x4,%rdx
  add    %rbx,%rdx
  nopl   0x0(%rax)
L241:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L241
  mov    %ecx,%eax
  and    $0xfffffffe,%eax
  test   $0x1,%cl
  je     L106
L136:
  cltq
  movq   $0xffffffffffffff9c,(%rbx,%rax,8)
L106:
  mov    -0x48(%rbp),%r8
  movslq %ecx,%rcx
  movq   $0x2,(%rbx,%rcx,8)
  test   $0x1,%r8b
  jne    L128
  add    $0x1,%r8
  mov    %r8,-0x48(%rbp)
L128:
  cmp    -0x40(%rbp),%r8
  movl   $0x0,-0x5c(%rbp)
  mov    %r8,%r15
  jg     L138
  movabs $0x2aaaaaaaaaaaaaaa,%r13
  nopl   0x0(%rax,%rax,1)
L156:
  mov    %r15,%rdi
  call   <sqrt@plt>
  mov    %rax,%rsi
  cmp    $0x2,%rax
  jle    L12
  movabs $0xaaaaaaaaaaaaaaab,%rax
  movabs $0x5555555555555554,%rdi
  imul   %r15,%rax
  add    %r13,%rax
  cmp    %rdi,%rax
  jbe    L192
  mov    $0x3,%ecx
  jmp    L237
  nopl   0x0(%rax,%rax,1)
L44:
  mov    %r15,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  je     L192
L237:
  add    $0x2,%rcx
  cmp    %rcx,%rsi
  jge    L44
L12:
  mov    -0x50(%rbp),%r12d
  lea    -0x1(%r12),%r14d
  test   %r14d,%r14d
  jle    L147
  lea    -0x2(%r12),%eax
  lea    0x8(%rbx),%rsi
  mov    %rbx,%rdi
  lea    0x8(,%rax,8),%rdx
  call   <memmove@plt>
L147:
  movslq %r14d,%rcx
  movslq -0x4c(%rbp),%rax
  mov    %r15,%rdx
  mov    %r15,(%rbx,%rcx,8)
  sub    (%rbx),%rdx
  cmp    %rax,%rdx
  je     L100
L192:
  add    $0x2,%r15
  cmp    %r15,-0x40(%rbp)
  jge    L156
L138:
  mov    -0x5c(%rbp),%esi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <clock@plt>
  sub    -0x60(%rbp),%eax
  lea    D27(%rip),%rdi
  movslq %eax,%rsi
  cltd
  imul   $0x10624dd3,%rsi,%rsi
  sar    $0x26,%rsi
  sub    %edx,%esi
  imul   $0x3e8,%esi,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L67
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L196:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L110
L100:
  test   %r12d,%r12d
  jle    L93
  xor    %edx,%edx
  nop
L20:
  mov    (%rbx,%rdx,8),%rsi
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  mov    %rdx,-0x58(%rbp)
  call   <printf@plt>
  mov    -0x58(%rbp),%rdx
  add    $0x1,%rdx
  cmp    %edx,-0x50(%rbp)
  jg     L20
L93:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x5c(%rbp)
  jmp    L192
L80:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L49
L41:
  xor    %eax,%eax
  jmp    L136
L67:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D16
data_type: string
name: retrieve_data
arguments: label: D31
data_type: string
name: retrieve_data
arguments: label: D42
data_type: string
name: retrieve_data
arguments: label: D86
data_type: string
name: retrieve_data
arguments: label: D14
data_type: string
name: retrieve_data
arguments: label: D38
data_type: string
name: retrieve_data
arguments: label: D87
data_type: byte[16]
name: retrieve_data
arguments: label: D7
data_type: string
name: retrieve_data
arguments: label: D27
data_type: string
name: retrieve_data
arguments: label: D44
data_type: string

tool: D16: Enter your starting number: 

tool: D31: %lld

tool: D42: Enter your finishing number: 

tool: D86: %d

tool: D14: Enter your number of primes: 

tool: D38: Enter the distance: 

tool: {"D87": [0x9c,0xff,0xff,0xff,0xff,0xff,0xff,0xff,0x9c,0xff,0xff,0xff,0xff,0xff,0xff,0xff]}

tool: D7: 
Number of tuples: %d


tool: D27: Time: %d.%.3d sec 



tool: D44: %7d

assistant: ```c
int main() {
  clock_t start, finish;
  long long int a, b;
  printf("Enter your starting number: ");
  scanf("%lld", &a);
  printf("Enter your finishing number: ");
  scanf("%lld", &b);
  int k, s;
  printf("Enter your number of primes: ");
  scanf("%d", &k);
  printf("Enter the distance: ");
  scanf("%d", &s);
  printf("\n");
  if (a > b)
    exit(1);
  long long int tup[k];
  start = clock();
  int i;
  for (i = 0; i < k - 1; i++)
    tup[i] = -100;
  tup[k - 1] = 2;
  if (a % 2 == 0)
    a = (a + 1);
  long long int x;
  long long int y, x2;
  char fl;
  int ans = 0;
  for (x = a; x <= b; x = x + 2) {
    x2 = sqrt(x);
    fl = 1;
    for (y = 3; y <= x2; y = y + 2)
      if (x % y == 0) {
        fl = 0;
        break;
      }
    if (fl == 1) {
      for (i = 0; i < (k - 1); i++)
        tup[i] = tup[i + 1];
      tup[k - 1] = x;
      if ((tup[k - 1] - tup[0]) == s) {
        int j;
        for (j = 0; j < k; j++)
          printf("%7d", tup[j]);
        printf("\n");
        ans++;
      }
    }
  }
  printf("\nNumber of tuples: %d\n", ans);
  finish = clock();
  printf("Time: %d.%.3d sec \n\n", (finish - start) / 1000, (finish - start) % 1000);
  return 0;
}
```