user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D18(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x168,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x158(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L122
  lea    0x10(%rsp),%rbp
  mov    $0x1,%ebx
  lea    D43(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L49:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  add    $0x4,%rbp
  cmp    %ebx,0x8(%rsp)
  jge    L49
L122:
  lea    D56(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L51
  lea    0x60(%rsp),%rbp
  mov    $0x1,%ebx
  lea    D22(%rip),%r12
  nopl   0x0(%rax)
L79:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  add    $0x4,%rbp
  cmp    %ebx,%esi
  jge    L79
L51:
  add    0x8(%rsp),%esi
  lea    D49(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%r13d
  mov    0xc(%rsp),%ebx
  lea    0x0(%r13,%rbx,1),%r12d
  test   %r12d,%r12d
  jle    L56
  movslq %r12d,%rdx
  lea    0xb0(%rsp),%rbp
  xor    %esi,%esi
  shl    $0x2,%rdx
  mov    %rbp,%rdi
  call   <memset@plt>
  lea    -0x1(%r13),%r10d
  test   %r10d,%r10d
  jle    L127
L0:
  lea    -0x1(%rbx),%eax
  mov    %rbp,%r9
  lea    0x60(%rsp),%r11
  xor    %r8d,%r8d
  lea    0x64(%rsp,%rax,4),%rdi
  nopl   0x0(%rax,%rax,1)
L62:
  test   %ebx,%ebx
  jle    L11
  mov    0x10(%rsp,%r8,4),%esi
  mov    %r11,%rdx
  mov    %r9,%rax
  nop
L138:
  mov    (%rdx),%ecx
  add    $0x4,%rdx
  imul   %esi,%ecx
  add    %ecx,(%rax)
  add    $0x4,%rax
  cmp    %rdi,%rdx
  jne    L138
L11:
  add    $0x1,%r8
  add    $0x4,%r9
  cmp    %r8d,%r10d
  jg     L62
  test   %r12d,%r12d
  jle    L99
L127:
  xor    %ebx,%ebx
  lea    D43(%rip),%r12
  nop
L27:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  add    0x8(%rsp),%eax
  cmp    %ebx,%eax
  jg     L27
L99:
  mov    0x158(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x168,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L56:
  lea    -0x1(%r13),%r10d
  lea    0xb0(%rsp),%rbp
  test   %r10d,%r10d
  jg     L0
  jmp    L99
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D18
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D43
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D56
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D22
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D49
type_infered: string

tool: D18: %d %d


tool: D43: %d 

tool: D56: 


tool: D22: %d

tool: D49: %d


assistant: ```c
int main() {
  int n1, n2, a[20], b[20];
  scanf("%d %d\n", &n1, &n2);
  for (int i = 1; i < n1 + 1; i++) {
    scanf("%d ", &a[i - 1]);
  }
  scanf("\n");
  for (int j = 1; j < n2 + 1; j++) {
    scanf("%d", &b[j - 1]);
  }
  printf("%d\n", n1 + n2);
  int m[40];
  for (int q = 0; q < n1 + n2; q++) {
    m[q] = 0;
  }
  for (int k = 0; k < n1 - 1; k++) {
    for (int l = 0; l <= n2 - 1; l++) {
      m[k + l] = m[k + l] + (a[k] * b[l]);
    }
  }
  for (int z = 0; z < n1 + n2; z++) {
    printf("%d ", m[z]);
  }
  return 0;
}
```