user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf70,%rsp
  mov    %edi,-0x3f64(%rbp)
  mov    %rsi,-0x3f70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3f44(%rbp)
  jmp    L48
L13:
  movl   $0x0,-0x3f40(%rbp)
  jmp    L108
L118:
  mov    -0x3f40(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x3f44(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  movl   $0x0,-0x3f30(%rbp,%rax,4)
  addl   $0x1,-0x3f40(%rbp)
L108:
  cmpl   $0x1,-0x3f40(%rbp)
  jle    L118
  addl   $0x1,-0x3f44(%rbp)
L48:
  cmpl   $0x7e3,-0x3f44(%rbp)
  jle    L13
  lea    -0x3f54(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x3f3c(%rbp)
  jmp    L3
L23:
  lea    -0x3f4c(%rbp),%rdx
  lea    -0x3f50(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x3f4c(%rbp),%eax
  mov    -0x3f50(%rbp),%edx
  sub    %edx,%eax
  add    $0x3e8,%eax
  movslq %eax,%rdx
  mov    -0x3f30(%rbp,%rdx,8),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x3f30(%rbp,%rax,8)
  mov    -0x3f4c(%rbp),%edx
  mov    $0x3e9,%eax
  sub    %edx,%eax
  mov    %eax,-0x3f4c(%rbp)
  mov    -0x3f4c(%rbp),%eax
  mov    -0x3f50(%rbp),%edx
  sub    %edx,%eax
  add    $0x3e8,%eax
  movslq %eax,%rdx
  mov    -0x3f2c(%rbp,%rdx,8),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x3f2c(%rbp,%rax,8)
  addl   $0x1,-0x3f3c(%rbp)
L3:
  mov    -0x3f54(%rbp),%eax
  cmp    %eax,-0x3f3c(%rbp)
  jl     L23
  movl   $0x0,-0x3f38(%rbp)
  jmp    L11
L87:
  movl   $0x0,-0x3f34(%rbp)
  jmp    L59
L101:
  mov    -0x3f34(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x3f38(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  mov    -0x3f30(%rbp,%rax,4),%eax
  cmp    $0x1,%eax
  jle    L107
  mov    -0x3f34(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x3f38(%rbp),%eax
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  mov    -0x3f30(%rbp,%rax,4),%eax
  mov    -0x3f34(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x3f38(%rbp),%edx
  movslq %edx,%rdx
  add    %rdx,%rdx
  add    %rcx,%rdx
  mov    -0x3f30(%rbp,%rdx,4),%edx
  sub    $0x1,%edx
  imul   %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  add    %eax,-0x3f48(%rbp)
L107:
  addl   $0x1,-0x3f34(%rbp)
L59:
  cmpl   $0x1,-0x3f34(%rbp)
  jle    L101
  addl   $0x1,-0x3f38(%rbp)
L11:
  cmpl   $0x7e3,-0x3f38(%rbp)
  jle    L87
  mov    -0x3f48(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
type_guessed: string
name: read_data
arguments: data_label: D21
type_guessed: string
name: read_data
arguments: data_label: D18
type_guessed: string

tool: D9: %d

tool: D21: %d %d

tool: D18: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int n, x, y, ans;
  int mark[2020][2];
  for (int i = 0; i < 2020; i++)
    for (int j = 0; j < 2; j++)
      mark[i][j] = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &x, &y);
    mark[y - x + 1000][0]++;
    y = 1000 - y + 1;
    mark[y - x + 1000][1]++;
  }
  for (int i = 0; i < 2020; i++)
    for (int j = 0; j < 2; j++)
      if (mark[i][j] > 1)
        ans += mark[i][j] * (mark[i][j] - 1) / 2;
  printf("%d\n", ans);
  return 0;
}
```