user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D24(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x3c(%rbp),%rsi
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdi
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rdi
  and    $0xfffffffffffffff0,%rdx
  sub    %rdi,%rsi
  cmp    %rsi,%rsp
  je     L21
L129:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L129
L21:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L29
L183:
  mov    %rax,%rdx
  mov    %rsp,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rbx
  and    $0xfffffffffffffff0,%rdx
  sub    %rax,%rsi
L7:
  cmp    %rsi,%rsp
  je     L48
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L7
L29:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L183
L48:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L142
L37:
  mov    %rsp,%r12
  test   %ecx,%ecx
  jle    L66
  mov    %rbx,%r14
  mov    %r12,%r15
  xor    %r13d,%r13d
  nopl   (%rax)
L186:
  mov    %r15,%rdx
  mov    %r14,%rsi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x3c(%rbp),%r9
  add    $0x1,%r13d
  add    $0x4,%r14
  add    $0x4,%r15
  cmp    %r13d,%r9d
  jg     L186
  test   %r9d,%r9d
  jle    L66
  xor    %edx,%edx
  xor    %r10d,%r10d
  mov    $0x1,%r11d
  nopl   (%rax)
L76:
  xor    %eax,%eax
  jmp    L56
  nopl   0x0(%rax)
L141:
  mov    (%r12,%rdx,4),%esi
  mov    (%r12,%rax,4),%r8d
  cmp    %r8d,%esi
  je     L14
  sub    %edi,%ecx
  sub    %r8d,%esi
  cmp    %esi,%ecx
  cmove  %r11d,%r10d
L151:
  add    $0x1,%rax
  cmp    %r9,%rax
  je     L17
L56:
  cmp    %eax,%edx
  je     L151
  mov    (%rbx,%rdx,4),%ecx
  mov    (%rbx,%rax,4),%edi
  cmp    %edi,%ecx
  jne    L141
L14:
  add    $0x1,%rax
  mov    $0x1,%r10d
  cmp    %r9,%rax
  jne    L56
L17:
  add    $0x1,%rdx
  cmp    %rdx,%rax
  jne    L76
  test   %r10d,%r10d
  je     L66
  lea    D26(%rip),%rdi
  call   <printf@plt>
L144:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L182
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L66:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L144
L142:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L37
L182:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D24
type_guessed: string
name: read_data
arguments: data_label: D8
type_guessed: string
name: read_data
arguments: data_label: D26
type_guessed: string
name: read_data
arguments: data_label: D10
type_guessed: string

tool: D24: %d

tool: D8: %d%d

tool: D26: no

tool: D10: yes

assistant: ```c
int main() {
  int i, N, j;
  scanf("%d", &N);
  int r[N], c[N];
  for (i = 0; i < N; i++) {
    scanf("%d%d", &r[i], &c[i]);
  }
  int p = 0;
  for (i = 0; i < N; i++) {
    for (j = 0; j < N; j++) {
      if (i != j && (r[i] == r[j] || c[i] == c[j] || (r[i] - r[j]) == (c[i] - c[j]))) {
        p = 1;
      } else
        continue;
    }
  }
  if (p == 0) {
    printf("yes");
  } else {
    printf("no");
  }
  return 0;
}
```