user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  pxor   %xmm10,%xmm10
  movdqa D1(%rip),%xmm7
  xor    %eax,%eax
  movdqa %xmm10,%xmm0
  movdqa %xmm10,%xmm11
  movdqa D50(%rip),%xmm8
  movdqa D25(%rip),%xmm9
  pcmpgtd %xmm7,%xmm0
  movdqa %xmm10,%xmm12
  movdqa D8(%rip),%xmm13
  pcmpgtd %xmm8,%xmm11
  movaps %xmm0,(%rsp)
  xchg   %ax,%ax
L6:
  movdqa %xmm9,%xmm0
  movdqa %xmm12,%xmm4
  movdqa %xmm11,%xmm3
  add    $0x1,%eax
  movdqa %xmm0,%xmm1
  movdqa %xmm11,%xmm15
  paddd  D21(%rip),%xmm9
  punpckldq %xmm0,%xmm1
  pcmpgtd %xmm1,%xmm4
  pmuludq %xmm1,%xmm3
  movdqa %xmm1,%xmm5
  pmuludq %xmm8,%xmm5
  movdqa %xmm4,%xmm2
  pmuludq %xmm7,%xmm4
  pmuludq %xmm8,%xmm2
  paddq  %xmm3,%xmm2
  movdqa %xmm12,%xmm3
  psllq  $0x20,%xmm2
  paddq  %xmm2,%xmm5
  movdqa %xmm0,%xmm2
  punpckhdq %xmm0,%xmm2
  pcmpgtd %xmm2,%xmm3
  pmuludq %xmm2,%xmm15
  movdqa %xmm2,%xmm14
  pmuludq %xmm8,%xmm14
  movdqa %xmm3,%xmm6
  pmuludq %xmm7,%xmm3
  pmuludq %xmm8,%xmm6
  paddq  %xmm15,%xmm6
  movdqa (%rsp),%xmm15
  psllq  $0x20,%xmm6
  paddq  %xmm6,%xmm14
  shufps $0xdd,%xmm14,%xmm5
  psrad  $0x1,%xmm5
  movdqa %xmm5,%xmm6
  pslld  $0x2,%xmm6
  paddd  %xmm6,%xmm5
  movdqa %xmm15,%xmm6
  pmuludq %xmm1,%xmm6
  pmuludq %xmm7,%xmm1
  pcmpeqd %xmm0,%xmm5
  pand   %xmm13,%xmm5
  paddq  %xmm6,%xmm4
  psllq  $0x20,%xmm4
  paddq  %xmm4,%xmm1
  movdqa %xmm15,%xmm4
  pmuludq %xmm2,%xmm4
  pmuludq %xmm7,%xmm2
  paddq  %xmm4,%xmm3
  psllq  $0x20,%xmm3
  paddq  %xmm3,%xmm2
  movdqa %xmm1,%xmm3
  shufps $0xdd,%xmm2,%xmm3
  movdqa %xmm3,%xmm1
  movdqa %xmm13,%xmm2
  pslld  $0x1,%xmm1
  paddd  %xmm3,%xmm1
  pcmpeqd %xmm0,%xmm1
  pand   %xmm1,%xmm2
  pandn  %xmm5,%xmm1
  por    %xmm2,%xmm1
  movdqa %xmm0,%xmm2
  pmuludq %xmm1,%xmm2
  psrlq  $0x20,%xmm0
  psrlq  $0x20,%xmm1
  pmuludq %xmm1,%xmm0
  pshufd $0x8,%xmm2,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  paddd  %xmm1,%xmm10
  cmp    $0xf9,%eax
  jne    L6
  movdqa %xmm10,%xmm0
  lea    D3(%rip),%rdi
  psrldq $0x8,%xmm0
  paddd  %xmm0,%xmm10
  movdqa %xmm10,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm10
  movd   %xmm10,%esi
  add    $0x3e7,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_infered: byte[16]
name: extract_symbolic_data
arguments: data_tag: D50
type_infered: byte[16]
name: extract_symbolic_data
arguments: data_tag: D25
type_infered: byte[16]
name: extract_symbolic_data
arguments: data_tag: D8
type_infered: byte[16]
name: extract_symbolic_data
arguments: data_tag: D21
type_infered: byte[16]
name: extract_symbolic_data
arguments: data_tag: D3
type_infered: string

tool: {"D1": [0x56,0x55,0x55,0x55,0x56,0x55,0x55,0x55,0x56,0x55,0x55,0x55,0x56,0x55,0x55,0x55]}

tool: {"D50": [0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66]}

tool: {"D25": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D8": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D21": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D3: %d


assistant: ```c
int main() {
  int i, sum;
  for (i = 0, sum = 0; i != 1000; i++) {
    sum += (i % 3 == 0 || i % 5 == 0) * i;
  }
  printf("%d\n", sum);
  return 0;
}
```