user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Knapsack>:
  endbr64
  push   %r15
  pxor   %xmm0,%xmm0
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x1e8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1d8(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x10(%rsp)
  movq   $0x0,0x30(%rsp)
  movaps %xmm0,0x20(%rsp)
  test   %edi,%edi
  js     L102
  movslq %edi,%rdi
  lea    0x40(%rsp),%rax
  mov    %rdi,0x8(%rsp)
  lea    (%rdi,%rdi,4),%rdx
  lea    0x68(%rsp,%rdx,8),%rdx
  nopl   0x0(%rax,%rax,1)
L106:
  movl   $0x0,(%rax)
  add    $0x28,%rax
  cmp    %rdx,%rax
  jne    L106
L7:
  movslq %ebp,%r12
  test   %ebp,%ebp
  js     L13
  lea    0x4(,%r12,4),%rdx
  lea    0x40(%rsp),%rdi
  xor    %esi,%esi
  call   <memset@plt>
L13:
  test   %r13d,%r13d
  jle    L107
  lea    0x68(%rsp),%r10
  mov    $0x1,%r8d
  mov    $0x1,%r15d
  lea    0x1(%rbp),%r11d
  nopl   0x0(%rax,%rax,1)
L12:
  test   %ebp,%ebp
  jle    L131
  mov    %r15d,%edx
  sub    (%r14,%r8,4),%edx
  mov    %r11d,%edi
  mov    %r10,%rax
  sub    (%r14,%r8,4),%edi
  movslq %edx,%r9
  nopl   0x0(%rax,%rax,1)
L26:
  mov    -0x24(%rax),%ecx
  test   %edx,%edx
  js     L20
  mov    -0x28(%rax,%r9,4),%esi
  add    (%rbx,%r8,4),%esi
  cmp    %esi,%ecx
  cmovl  %esi,%ecx
L20:
  add    $0x1,%edx
  mov    %ecx,0x4(%rax)
  add    $0x4,%rax
  cmp    %edi,%edx
  jne    L26
L131:
  add    $0x1,%r8
  add    $0x28,%r10
  cmp    %r8d,%r13d
  jge    L12
  mov    0x8(%rsp),%rax
  lea    D34(%rip),%rdi
  sub    $0x1,%r13d
  lea    (%rax,%rax,4),%rax
  lea    (%r12,%rax,2),%rax
  mov    0x40(%rsp,%rax,4),%ebx
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  lea    D36(%rip),%rdi
  call   <puts@plt>
  movslq %r13d,%rax
  jmp    L61
  cs nopw 0x0(%rax,%rax,1)
L90:
  movslq %ebp,%r12
  lea    (%rdx,%rdx,4),%rdx
  sub    $0x1,%rax
  lea    (%r12,%rdx,2),%rdx
  mov    0x40(%rsp,%rdx,4),%ebx
L61:
  movslq %eax,%rcx
  movslq %eax,%rdx
  lea    (%rcx,%rcx,4),%rcx
  lea    (%r12,%rcx,2),%rcx
  cmp    %ebx,0x40(%rsp,%rcx,4)
  je     L135
  movl   $0x1,0x14(%rsp,%rax,4)
  sub    0x4(%r14,%rax,4),%ebp
L135:
  test   %eax,%eax
  jg     L90
  lea    0x14(%rsp),%rbx
  lea    0x18(%rsp,%r13,4),%r12
  lea    D30(%rip),%rbp
  nopl   0x0(%rax)
L57:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L57
L32:
  mov    0x1d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L68
  add    $0x1e8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <putchar@plt>
L102:
  movslq %edi,%rax
  mov    %rax,0x8(%rsp)
  jmp    L7
L107:
  mov    0x8(%rsp),%rax
  lea    D34(%rip),%rdi
  lea    (%rax,%rax,4),%rax
  lea    (%r12,%rax,2),%rax
  mov    0x40(%rsp,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D36(%rip),%rdi
  call   <puts@plt>
  jmp    L32
L68:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D34
type_guessed: string
name: read_data
arguments: label: D36
type_guessed: string
name: read_data
arguments: label: D30
type_guessed: string

tool: D34: Max Profit : %d 


tool: D36: Edges are :

tool: D30: %d	

assistant: ```c
void Knapsack(int no, int wt, int pt[10], int weight[10]) {
  int knap[10][10], x[10] = {0};
  int i, j;
  for (i = 0; i <= no; i++)
    knap[i][0] = 0;
  for (i = 0; i <= wt; i++)
    knap[0][i] = 0;
  for (i = 1; i <= no; i++)
    for (j = 1; j <= wt; j++) {
      if ((j - weight[i]) < 0)
        knap[i][j] = knap[i - 1][j];
      else
        knap[i][j] = knap[i - 1][j] > pt[i] + knap[i - 1][j - weight[i]] ? knap[i - 1][j] : pt[i] + knap[i - 1][j - weight[i]];
    }
  printf("Max Profit : %d \n", knap[no][wt]);
  printf("Edges are :\n");
  for (i = no; i >= 1; i--)
    if (knap[i][wt] != knap[i - 1][wt]) {
      x[i] = 1;
      wt -= weight[i];
    }
  for (i = 1; i <= no; i++)
    printf("%d\t", x[i]);
  printf("\n");
}
```