user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xc3000(%rsp),%r11
L190:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L190
  sub    $0x528,%rsp
  lea    D24(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0xc3518(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  cmpq   $0x0,0x8(%rsp)
  jle    L90
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  lea    0x10(%rsp),%rbx
  lea    D8(%rip),%r13
  xchg   %ax,%ax
L172:
  lea    (%rbx,%rbp,8),%rsi
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    (%rbx,%rbp,8),%rax
  cmp    %rax,%r12
  cmovl  %rax,%r12
  mov    0x8(%rsp),%rax
  add    $0x1,%rbp
  cmp    %rbp,%rax
  jg     L172
  mov    (%rsp),%rdx
  test   %rdx,%rdx
  jne    L166
L40:
  xor    %ebp,%ebp
  lea    0x10(%rsp),%rbx
  lea    D35(%rip),%r12
  lea    D8(%rip),%r13
  test   %rax,%rax
  jg     L21
  jmp    L29
L147:
  mov    %r12,%rdi
  add    $0x1,%rbp
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  cmp    %rbp,%rax
  jle    L29
L21:
  sub    $0x1,%rax
  mov    (%rbx,%rbp,8),%rsi
  cmp    %rbp,%rax
  jne    L147
  mov    %r13,%rdi
  add    $0x1,%rbp
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  cmp    %rbp,%rax
  jg     L21
L29:
  mov    0xc3518(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L169
  add    $0xc3528,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L166:
  and    $0x1,%dl
  je     L110
  test   %rax,%rax
  jle    L40
  lea    (%rbx,%rax,8),%rcx
  nopw   0x0(%rax,%rax,1)
L65:
  mov    %r12,%rdx
  sub    (%rbx),%rdx
  add    $0x8,%rbx
  mov    %rdx,-0x8(%rbx)
  cmp    %rcx,%rbx
  jne    L65
  jmp    L40
L110:
  test   %rax,%rax
  jle    L40
  lea    (%rbx,%rax,8),%rdi
  mov    %rbx,%rdx
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L53:
  mov    %r12,%rsi
  sub    (%rdx),%rsi
  cmp    %rsi,%rcx
  mov    %rsi,(%rdx)
  cmovl  %rsi,%rcx
  add    $0x8,%rdx
  cmp    %rdi,%rdx
  jne    L53
  nopl   0x0(%rax)
L89:
  mov    %rcx,%rdx
  sub    (%rbx),%rdx
  add    $0x8,%rbx
  mov    %rdx,-0x8(%rbx)
  cmp    %rdi,%rbx
  jne    L89
  jmp    L40
L90:
  cmpq   $0x0,(%rsp)
  je     L29
  mov    0x8(%rsp),%rax
  jmp    L40
L169:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D24
type_infered: string
name: read_data
arguments: label: D8
type_infered: string
name: read_data
arguments: label: D35
type_infered: string

tool: D24: %ld %lld

tool: D8: %lld

tool: D35: %lld 

assistant: ```c
int main() {
  long long int A[100000], K, largest = 0, new_largest = 0;
  long N, i;
  int z = 0;
  scanf("%ld %lld", &N, &K);
  for (i = 0; i < N; i++) {
    scanf("%lld", &A[i]);
    if (largest < A[i])
      largest = A[i];
  }
  if (K == 0) {
  } else if (K % 2 == 0) {
    for (i = 0; i < N; i++) {
      A[i] = largest - A[i];
      if (new_largest < A[i])
        new_largest = A[i];
    }
    largest = new_largest;
    for (i = 0; i < N; i++) {
      A[i] = largest - A[i];
    }
  } else {
    for (i = 0; i < N; i++) {
      A[i] = largest - A[i];
    }
  }
  for (i = 0; i < N; i++) {
    if (i == (N - 1)) {
      printf("%lld", A[i]);
    } else
      printf("%lld ", A[i]);
  }
  return 0;
}
```