user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x64,%edx
  xor    %esi,%esi
  push   %r14
  push   %r13
  lea    D17(%rip),%r13
  push   %r12
  lea    D15(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x548,%rsp
  lea    0x330(%rsp),%rbx
  lea    0x3a0(%rsp),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x538(%rsp)
  xor    %eax,%eax
  mov    %rbx,%rdi
  call   <memset@plt>
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %r15,%rdi
  call   <gets@plt>
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <strtok@plt>
  mov    %rax,%rdi
  mov    %rax,%r14
  call   <atoi@plt>
  mov    %eax,%ebp
  test   %r14,%r14
  je     L127
  lea    0x14(%rsp),%r14
  nopl   0x0(%rax)
L5:
  xor    %edi,%edi
  mov    %r13,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L127
  call   <atoi@plt>
  add    $0x4,%r14
  mov    %eax,-0x4(%r14)
  jmp    L5
  nopl   0x0(%rax)
L127:
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %r15,%rdi
  call   <gets@plt>
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <strtok@plt>
  mov    %rax,%rdi
  mov    %rax,%r14
  call   <atoi@plt>
  mov    %eax,0x1a0(%rsp)
  test   %r14,%r14
  je     L35
  lea    0x1a4(%rsp),%r14
L113:
  xor    %edi,%edi
  mov    %r13,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L35
  call   <atoi@plt>
  add    $0x4,%r14
  mov    %eax,-0x4(%r14)
  jmp    L113
  nopl   0x0(%rax,%rax,1)
L35:
  mov    0xc(%rsp),%eax
  xor    %r8d,%r8d
  test   %eax,%eax
  jle    L138
  mov    0x8(%rsp),%r9d
  sub    $0x1,%eax
  lea    0x14(%rsp),%r10
  xor    %r8d,%r8d
  lea    0x331(%rsp,%rax,1),%r11
  lea    0x1a0(%rsp),%rdi
  lea    -0x1(%r9),%eax
  lea    (%r10,%rax,4),%rsi
L69:
  test   %r9d,%r9d
  jle    L147
  mov    (%rdi),%ecx
  mov    %r10,%rax
  mov    %ebp,%edx
  jmp    L122
  nopl   0x0(%rax)
L52:
  cmp    %rsi,%rax
  je     L147
L143:
  mov    (%rax),%edx
  add    $0x4,%rax
L122:
  cmp    %edx,%ecx
  jne    L52
  cmpb   $0x0,(%rbx)
  jne    L52
  movb   $0x1,(%rbx)
  add    $0x1,%r8d
  cmp    %rsi,%rax
  jne    L143
  nopl   0x0(%rax,%rax,1)
L147:
  add    $0x1,%rbx
  add    $0x4,%rdi
  cmp    %r11,%rbx
  jne    L69
L138:
  mov    %r8d,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x538(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x548,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D17
datatype: string
name: parse_data
arguments: data_tag: D15
datatype: string

tool: D17:  

tool: D15: %d


assistant: ```c
int main() {
  int S[100], T[100];
  char checked[100];
  int n, q;
  int i = 0;
  int j;
  int c = 0;
  char input[400];
  char *p;
  memset(checked, 0, sizeof(checked));
  scanf("%d\n", &n);
  gets(input);
  p = strtok(input, " ");
  S[0] = atoi(p);
  while (p != NULL) {
    p = strtok(NULL, " ");
    if (p != NULL) {
      i++;
      S[i] = atoi(p);
    }
  }
  scanf("%d\n", &q);
  gets(input);
  i = 0;
  p = strtok(input, " ");
  T[0] = atoi(p);
  while (p != NULL) {
    p = strtok(NULL, " ");
    if (p != NULL) {
      i++;
      T[i] = atoi(p);
    }
  }
  for (i = 0; i < q; i++) {
    for (j = 0; j < n; j++) {
      if (T[i] == S[j] && checked[i] == 0) {
        c++;
        checked[i] = 1;
      }
    }
  }
  printf("%d\n", c);
  return 0;
}
```