user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x9c40,%edi
  push   %r12
  push   %rbp
  sub    $0x10,%rsp
  call   <malloc@plt>
  mov    $0x9c40,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    $0x9c40,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  lea    0x4(%r12),%rsi
  lea    0x9c40(%r12),%rcx
  mov    %rax,%r13
  mov    %rbp,%rax
  sub    %rsi,%rax
  cmp    $0x8,%rax
  mov    %r12,%rax
  jbe    L77
  mov    %rbp,%rdx
  movdqa D2(%rip),%xmm0
  sub    %r12,%rdx
  nopw   0x0(%rax,%rax,1)
L78:
  movups %xmm0,(%rax)
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L78
L19:
  lea    0x4(%rbp),%rdx
  mov    %r13,%rax
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L67
  mov    %r13,%rax
  sub    %rsi,%rax
  cmp    $0x8,%rax
  jbe    L67
  mov    %r12,%rcx
  mov    %r13,%rdx
  lea    0x9c40(%rbp),%rsi
  mov    %rbp,%rax
  sub    %rbp,%rcx
  sub    %rbp,%rdx
  nopw   0x0(%rax,%rax,1)
L98:
  movdqu (%rcx,%rax,1),%xmm0
  movdqu (%rax),%xmm3
  paddd  %xmm3,%xmm0
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rsi
  jne    L98
L69:
  mov    %r13,%rax
  lea    0x9c40(%r13),%rdx
  pxor   %xmm0,%xmm0
L70:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rax,%rdx
  jne    L70
  movdqa %xmm0,%xmm1
  mov    %rbp,%rdi
  psrldq $0x8,%xmm1
  paddd  %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  movdqa (%rsp),%xmm0
  mov    $0x2710,%esi
  lea    D15(%rip),%rdi
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%edx
  call   <printf@plt>
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L77:
  mov    %rbp,%rdx
  sub    %r12,%rdx
  nopl   0x0(%rax,%rax,1)
L94:
  movl   $0x1,(%rax)
  movl   $0x1,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L94
  jmp    L19
L67:
  mov    %r13,%rsi
  mov    %r12,%rcx
  lea    0x9c40(%rbp),%rdi
  mov    %rbp,%rax
  sub    %rbp,%rsi
  sub    %rbp,%rcx
  nopl   0x0(%rax)
L60:
  mov    (%rax),%edx
  add    (%rcx,%rax,1),%edx
  mov    %edx,(%rsi,%rax,1)
  add    $0x4,%rax
  cmp    %rdi,%rax
  jne    L60
  jmp    L69
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type_guessed: byte[16]
name: extract_symbolic_data
arguments: label: D15
type_guessed: string

tool: {"D2": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D15: Sample size: %d | Sum: %d


assistant: ```c
int main(int argc, char **argv) {
  const int sample = 10000;
  size_t size = sample * sizeof(int);
  int *restrict a1, *restrict a2, *restrict a3;
  a1 = (int *)malloc(size);
  a2 = (int *)malloc(size);
  a3 = (int *)malloc(size);
  int sum = 0;
  int i;
#pragma acc data create(i, a1 [0:sample], a2 [0:sample], a3 [0:sample]), copyin(sample), copy(sum)
  {
#pragma acc parallel loop
    for (i = 0; i < sample; ++i) {
      a1[i] = a2[i] = 1;
    }
#pragma acc parallel loop
    for (i = 0; i < sample; ++i) {
      a3[i] = a1[i] + a2[i];
    }
#pragma acc parallel loop reduction(+ : sum)
    for (i = 0; i < sample; ++i) {
      sum += a3[i];
    }
  }
  free(a1);
  free(a2);
  free(a3);
  printf("Sample size: %d | Sum: %d\n", sample, sum);
}
```