user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r14
  lea    D0(%rip),%rdi
  push   %r13
  lea    D64(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  lea    D40(%rip),%rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%r12
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D8(%rip),%rdi
  add    0x8(%rsp),%esi
  call   <printf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  js     L26
  nopl   (%rax)
L140:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  cmp    %ebp,%eax
  je     L116
  add    $0x1,%ebp
  add    $0x4,%r12
  cmp    %ebp,%eax
  jge    L140
L26:
  mov    0xc(%rsp),%esi
  lea    0x50(%rsp),%r12
  xor    %ebp,%ebp
  lea    D40(%rip),%rbx
  lea    D64(%rip),%r13
  test   %esi,%esi
  js     L8
  nopl   0x0(%rax)
L115:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    %ebp,%esi
  je     L100
  add    $0x1,%ebp
  add    $0x4,%r12
  cmp    %ebp,%esi
  jge    L115
L8:
  mov    0x8(%rsp),%edi
  mov    %edi,%eax
  add    %esi,%eax
  js     L0
  lea    0x10(%rsp),%r14
  mov    $0x1,%r12d
  xor    %ebp,%ebp
  lea    D40(%rip),%rbx
  lea    0x50(%rsp),%r13
  nop
L86:
  mov    %r14,%rdx
  xor    %eax,%eax
  xor    %r8d,%r8d
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %ebp,%ecx
  sub    %eax,%ecx
  cmp    %ecx,%edi
  jl     L88
  cmp    %eax,%esi
  jl     L88
  mov    (%rdx),%ecx
  imul   0x0(%r13,%rax,4),%ecx
  add    %ecx,%r8d
L88:
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %rax,%r12
  jne    L15
  mov    %r8d,%esi
  mov    %rbx,%rdi
  add    $0x1,%ebp
  add    $0x1,%r12
  call   <printf@plt>
  mov    0x8(%rsp),%edi
  mov    0xc(%rsp),%esi
  add    $0x4,%r14
  lea    (%rdi,%rsi,1),%eax
  cmp    %ebp,%eax
  jge    L86
L0:
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L148
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L116:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  add    $0x4,%r12
  call   <scanf@plt>
  cmp    0x8(%rsp),%ebp
  jle    L140
  jmp    L26
L100:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  add    $0x4,%r12
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    %esi,%ebp
  jle    L115
  jmp    L8
L148:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type_infered: string
name: disassemble_data
arguments: data_label: D64
type_infered: string
name: disassemble_data
arguments: data_label: D40
type_infered: string
name: disassemble_data
arguments: data_label: D8
type_infered: string

tool: D0: %d %d

tool: D64: 


tool: D40: %d 

tool: D8: %d


assistant: ```c
int main() {
  int a[15], b[15];
  int n1, n2, p, j, i, c, n;
  scanf("%d %d", &n1, &n2);
  printf("%d\n", n1 + n2);
  for (p = 0; p <= n1; p++) {
    scanf("%d ", &a[p]);
    if (p == n1) {
      scanf("\n");
    } else {
      continue;
    }
  }
  for (j = 0; j <= n2; j++) {
    scanf("%d ", &b[j]);
    if (j == n2) {
      scanf("\n");
    } else {
      continue;
    }
  }
  for (i = 0; i <= n1 + n2; i++) {
    c = 0;
    for (n = 0; n <= i; n++) {
      if (i - n <= n1 && n <= n2) {
        c = a[i - n] * b[n] + c;
      } else {
        continue;
      }
    }
    printf("%d ", c);
  }
  return 0;
}
```