user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xeb0,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x3ea8(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  cmpq   $0x0,0x8(%rsp)
  jle    L50
  xor    %ebx,%ebx
  lea    0x18(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    D5(%rip),%r12
  lea    D24(%rip),%rbp
  nopl   0x0(%rax)
L90:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %r14,%rdx
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  xor    %r9d,%r9d
  xor    %edi,%edi
  mov    $0x2,%ecx
  cmp    $0x1,%rsi
  jle    L203
  nopl   0x0(%rax)
L83:
  mov    %rsi,%rax
  lea    0x20(%rsp),%r8
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L170
  nopw   0x0(%rax,%rax,1)
L167:
  mov    %rsi,%rax
  mov    %rcx,(%r8,%rdi,8)
  add    $0x1,%rdi
  cqto
  idiv   %rcx
  cqto
  mov    %rax,%rsi
  idiv   %rcx
  test   %rdx,%rdx
  je     L167
  mov    $0x1,%r9d
L170:
  cmp    %rcx,%rsi
  jl     L137
  add    $0x1,%rcx
  jmp    L83
  xchg   %ax,%ax
L137:
  cmp    $0x1,%rsi
  jle    L188
  mov    $0x2,%ecx
  jmp    L83
  nopl   (%rax)
L188:
  test   %r9b,%r9b
  je     L172
  mov    %rsi,0x10(%rsp)
L172:
  mov    0x18(%rsp),%rsi
  cmp    $0x1,%rsi
  jle    L194
L32:
  xor    %r10d,%r10d
  xor    %r8d,%r8d
  mov    $0x2,%ecx
L204:
  mov    %rsi,%rax
  lea    0x1f60(%rsp),%r9
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L69
  nopl   (%rax)
L1:
  mov    %rsi,%rax
  mov    %rcx,(%r9,%r8,8)
  add    $0x1,%r8
  cqto
  idiv   %rcx
  cqto
  mov    %rax,%rsi
  idiv   %rcx
  test   %rdx,%rdx
  je     L1
  mov    $0x1,%r10d
L69:
  cmp    %rcx,%rsi
  jl     L20
  add    $0x1,%rcx
  jmp    L204
  cs nopw 0x0(%rax,%rax,1)
L20:
  cmp    $0x1,%rsi
  jle    L64
  mov    $0x2,%ecx
  jmp    L204
  nopl   (%rax)
L64:
  test   %r10b,%r10b
  je     L194
  mov    %rsi,0x18(%rsp)
L194:
  test   %rdi,%rdi
  je     L225
  xor    %eax,%eax
  mov    $0x1,%esi
  lea    0x20(%rsp),%r8
  lea    0x1f60(%rsp),%r9
  nopw   0x0(%rax,%rax,1)
L155:
  mov    (%r8,%rax,8),%rdx
  cmp    %rdx,(%r9,%rax,8)
  jne    L81
  imul   %rdx,%rsi
L81:
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L155
L231:
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    %rbx,0x8(%rsp)
  jg     L90
L50:
  mov    0x3ea8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L110
  add    $0x3eb0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L203:
  mov    0x18(%rsp),%rsi
  cmp    $0x1,%rsi
  jg     L32
L225:
  mov    $0x1,%esi
  jmp    L231
L110:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D7
type: string
name: read_data
arguments: data_tag: D5
type: string
name: read_data
arguments: data_tag: D24
type: string

tool: D7: %lld

tool: D5: %lld%lld

tool: D24: 
%lld

assistant: ```c
int main() {
  long long int n, i, j = 2, k = 2, l = 0, m = 0, a[1000], b[1000], p = 1, c, d, o;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld%lld", &c, &d);
    while (c > 1) {
      while (c % j == 0) {
        if (c % j == 0) {
          a[l] = j;
          c = c / j;
          l++;
        }
      }
      if (c < j)
        j = 1;
      j++;
    }
    while (d > 1) {
      while (d % k == 0) {
        if (d % k == 0) {
          b[m] = k;
          d = d / k;
          m++;
        }
      }
      if (d < k)
        k = 1;
      k++;
    }
    for (o = 0; o < l; o++) {
      if (a[o] == b[o])
        p = p * a[o];
    }
    printf("\n%lld", p);
    printf("\n");
    j = 2;
    k = 2;
    l = 0;
    m = 0;
    p = 1;
  }
  return 0;
}
```