user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x48(%rbp),%rsi
  push   %r13
  push   %r12
  lea    -0x3c(%rbp),%r12
  push   %rbx
  lea    D37(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  mov    -0x48(%rbp),%eax
  lea    -0x44(%rbp),%rcx
  mov    %rcx,-0x60(%rbp)
  lea    -0x40(%rbp),%rcx
  lea    -0x1(%rax),%edx
  mov    %rcx,-0x68(%rbp)
  mov    %edx,-0x48(%rbp)
  test   %eax,%eax
  je     L40
  nopl   0x0(%rax)
L105:
  mov    -0x60(%rbp),%rdx
  mov    -0x68(%rbp),%rsi
  xor    %eax,%eax
  mov    %rsp,-0x58(%rbp)
  lea    D17(%rip),%rdi
  call   <scanf@plt>
  mov    -0x44(%rbp),%r15d
  mov    %rsp,%rcx
  lea    0x1(%r15),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L59
L63:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L63
L59:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L31
  orq    $0x0,-0x8(%rsp,%rdx,1)
L31:
  lea    0x3(%rsp),%rcx
  mov    %rcx,%rax
  and    $0xfffffffffffffffc,%rcx
  shr    $0x2,%rax
  mov    %rcx,%r14
  mov    %rax,%r13
  test   %r15d,%r15d
  jle    L57
  movslq %r15d,%rdx
  lea    0x4(%rcx),%rdi
  xor    %esi,%esi
  shl    $0x2,%rdx
  call   <memset@plt>
L57:
  movl   $0x1,0x0(,%r13,4)
  mov    -0x40(%rbp),%ecx
  xor    %r13d,%r13d
  test   %ecx,%ecx
  jle    L25
L47:
  mov    %r12,%rsi
  xor    %eax,%eax
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    -0x44(%rbp),%r15d
  mov    -0x3c(%rbp),%esi
  mov    %r15d,%edx
  sub    %esi,%edx
  movslq %edx,%rax
  js     L112
  xchg   %ax,%ax
L52:
  mov    (%r14,%rax,4),%edx
  test   %edx,%edx
  jle    L94
  lea    (%rsi,%rax,1),%edx
  movslq %edx,%rdx
  movl   $0x1,(%r14,%rdx,4)
L94:
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L52
L112:
  add    $0x1,%r13d
  cmp    %r13d,-0x40(%rbp)
  jg     L47
L25:
  movslq %r15d,%r8
  lea    D8(%rip),%rdi
  mov    (%r14,%r8,4),%eax
  test   %eax,%eax
  jg     L44
  lea    D10(%rip),%rdi
L44:
  call   <printf@plt>
  mov    -0x48(%rbp),%eax
  mov    -0x58(%rbp),%rsp
  lea    -0x1(%rax),%edx
  mov    %edx,-0x48(%rbp)
  test   %eax,%eax
  jne    L105
L40:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L95
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L95:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D37
type: string
name: retrieve_data
arguments: tag: D17
type: string
name: retrieve_data
arguments: tag: D8
type: string
name: retrieve_data
arguments: tag: D10
type: string

tool: D37: %d

tool: D17: %d %d

tool: D8: Yes


tool: D10: No


assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int m, n, i, j;
    scanf("%d %d", &n, &m);
    int b[m + 1];
    for (i = 1; i <= m; ++i)
      b[i] = 0;
    b[0] = 1;
    for (i = 0; i < n; ++i) {
      int k;
      scanf("%d", &k);
      for (j = m - k; j >= 0; --j) {
        if (b[j] > 0)
          b[j + k] = 1;
      }
    }
    if (b[m] > 0)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}
```