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    -0x40(%rbp),%rsi
  push   %r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbx
  mov    %r12,%rdi
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  lea    -0x3c(%rbp),%rax
  movl   $0x0,-0x44(%rbp)
  mov    %rax,-0x60(%rbp)
  mov    -0x40(%rbp),%eax
  test   %eax,%eax
  jle    L210
  nopl   0x0(%rax)
L29:
  mov    -0x60(%rbp),%rsi
  mov    %r12,%rdi
  mov    %rsp,-0x50(%rbp)
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  mov    %rsp,%rcx
  lea    0x1(%rax),%edi
  movslq %edi,%rax
  lea    0xf(,%rax,8),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L47
L107:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L107
L47:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L197
  orq    $0x0,-0x8(%rsp,%rdx,1)
L197:
  shl    $0x2,%edi
  mov    %rsp,%r14
  call   <malloc@plt>
  movl   $0x0,(%rax)
  mov    -0x3c(%rbp),%r9d
  mov    %rax,-0x58(%rbp)
  test   %r9d,%r9d
  jle    L124
  lea    0x4(%rax),%r15
  mov    $0x1,%ebx
  xor    %r13d,%r13d
  nopl   0x0(%rax,%rax,1)
L162:
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    (%r15),%eax
  mov    -0x3c(%rbp),%r9d
  cmp    %eax,%r13d
  cmovl  %eax,%r13d
  add    $0x1,%ebx
  add    $0x4,%r15
  cmp    %ebx,%r9d
  jge    L162
  lea    0x1(%r13),%eax
  mov    %eax,%edi
  test   %r9d,%r9d
  js     L188
L198:
  movslq %r9d,%r15
  mov    %r14,%rax
  lea    0x8(%r14,%r15,8),%rdx
  nop
L34:
  movl   $0x0,(%rax)
  add    $0x8,%rax
  movl   $0x0,-0x4(%rax)
  cmp    %rdx,%rax
  jne    L34
L54:
  mov    -0x58(%rbp),%rsi
  lea    -0x2(%r9),%eax
  mov    $0x2,%r11d
  lea    0x8(%rsi,%rax,4),%r10
  cmp    $0x1,%r9d
  jle    L99
L51:
  lea    0x4(%rsi),%rcx
  mov    %r14,%rax
  mov    %edi,%r8d
  jmp    L221
  nop
L234:
  cmp    %edx,%ebx
  cmovl  %edx,%ebx
L121:
  cmp    %r13d,%ebx
  jge    L190
  mov    %r13d,0x14(%rax)
L190:
  add    $0x8,%rax
  cmp    %rcx,%r10
  je     L11
L221:
  mov    (%rcx),%edx
  sub    (%rax),%edx
  cmp    %edx,%r8d
  mov    0x14(%rax),%ebx
  mov    0xc(%rax),%r13d
  cmovg  %edx,%r8d
  mov    0x4(%rcx),%edx
  add    $0x4,%rcx
  sub    %r8d,%edx
  cmp    %edx,%ebx
  jge    L121
  cmp    %r13d,%edx
  jl     L234
  mov    %edx,0x14(%rax)
  add    $0x8,%rax
  cmp    %rcx,%r10
  jne    L221
L11:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L218:
  mov    0x4(%r14,%rax,8),%edx
  mov    %edx,(%r14,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%r9d
  jge    L218
  cmp    $0x1,%r11d
  je     L135
L38:
  mov    $0x1,%r11d
  cmp    $0x1,%r9d
  jg     L51
L99:
  test   %r9d,%r9d
  jns    L11
  cmp    $0x1,%r11d
  jne    L38
  nopl   0x0(%rax,%rax,1)
L135:
  mov    0x4(%r14,%r15,8),%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  addl   $0x1,-0x44(%rbp)
  mov    -0x50(%rbp),%rsp
  mov    -0x44(%rbp),%eax
  cmp    %eax,-0x40(%rbp)
  jg     L29
L210:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L124:
  mov    $0x1,%edi
  test   %r9d,%r9d
  jns    L198
L188:
  movslq %r9d,%r15
  jmp    L54
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D10
datatype: string
name: extract_symbolic_data
arguments: data_label: D14
datatype: string

tool: D10: %d

tool: D14: %d


assistant: ```c
int main() {
  int T, t;
  scanf("%d", &T);
  for (t = 0; t < T; t++) {
    int *stock, max = 0, min, n, k = 2, i, j, x;
    scanf("%d", &n);
    int maxprofit[n + 1][2];
    stock = (int *)malloc(sizeof(int) * (n + 1));
    stock[0] = 0;
    for (i = 1; i < n + 1; i++) {
      scanf("%d", &stock[i]);
      if (stock[i] > max)
        max = stock[i];
    }
    for (i = 0; i < n + 1; i++) {
      maxprofit[i][0] = 0;
      maxprofit[i][1] = 0;
    }
    for (i = 1; i < k + 1; i++) {
      min = max + 1;
      for (j = 2; j < n + 1; j++) {
        if ((stock[j - 1] - maxprofit[j - 2][0]) < min)
          min = stock[j - 1] - maxprofit[j - 2][0];
        x = stock[j] - min;
        if ((x > maxprofit[j][1]) && (x >= maxprofit[j - 1][1]))
          maxprofit[j][1] = x;
        if ((maxprofit[j - 1][1] > maxprofit[j][1]) && (maxprofit[j - 1][1] > x))
          maxprofit[j][1] = maxprofit[j - 1][1];
      }
      for (j = 0; j < n + 1; j++)
        maxprofit[j][0] = maxprofit[j][1];
    }
    printf("%d\n", maxprofit[n][1]);
  }
  return 0;
}
```