user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xea000(%rsp),%r11
L62:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L62
  sub    $0x638,%rsp
  mov    $0xea618,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0xea628(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  mov    %rbp,%rdi
  call   <memset@plt>
  lea    0xea628(%rsp),%rdx
  nopl   0x0(%rax)
L164:
  movl   $0x1,0x4(%rax)
  add    $0x18,%rax
  movl   $0x1,-0x18(%rax)
  cmp    %rdx,%rax
  jne    L164
  mov    $0x2,%r12d
  mov    $0x9c41,%ebx
  nopw   0x0(%rax,%rax,1)
L101:
  mov    D26(%rip),%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  cmp    %eax,%r12d
  jg     L157
  mov    %r12d,%eax
  imul   %r12d,%eax
  cmp    $0x9c40,%eax
  jg     L124
  movslq %eax,%rdx
  xor    %edi,%edi
  lea    (%rdx,%rdx,2),%r8
  mov    %ebx,%edx
  sub    %eax,%edx
  shl    $0x3,%r8
  lea    (%rdx,%rdx,2),%r9
  lea    0x0(%rbp,%r8,1),%rsi
  neg    %r8
  shl    $0x3,%r9
  nopl   (%rax)
L135:
  mov    (%rsi),%eax
  mov    0x0(%rbp,%rdi,1),%ecx
  mov    $0x1,%edx
  cmp    %eax,%ecx
  cmovl  %eax,%ecx
  mov    %rsi,%rax
  mov    %ecx,(%rsi)
  test   %ecx,%ecx
  jle    L90
  nopl   0x0(%rax)
L68:
  lea    (%rax,%r8,1),%rcx
  add    $0x1,%edx
  mov    0x4(%rcx,%rdi,1),%ecx
  add    %ecx,0x4(%rax)
  add    $0x4,%rax
  mov    (%rsi),%r11d
  cmp    %edx,%r11d
  jge    L68
  test   %r11d,%r11d
  jle    L90
  mov    %rsi,%rdx
  mov    $0x1,%r10d
L28:
  mov    0x4(%rdx),%ecx
  cmp    $0x3b9ac9ff,%ecx
  jle    L122
  cmp    %r10d,%r11d
  jne    L159
  add    $0x1,%r11d
  mov    %r11d,(%rsi)
  mov    0x4(%rdx),%ecx
L159:
  movslq %ecx,%rax
  mov    %ecx,%r11d
  imul   $0x44b82fa1,%rax,%rax
  sar    $0x1f,%r11d
  sar    $0x3c,%rax
  sub    %r11d,%eax
  add    %eax,0x8(%rdx)
  imul   $0x3b9aca00,%eax,%eax
  sub    %eax,%ecx
  mov    %ecx,0x4(%rdx)
  mov    (%rsi),%r11d
L122:
  add    $0x1,%r10
  add    $0x4,%rdx
  cmp    %r10d,%r11d
  jge    L28
L90:
  add    $0x18,%rdi
  add    $0x18,%rsi
  sub    $0x18,%r8
  cmp    %r9,%rdi
  jne    L135
L124:
  add    $0x1,%r12d
  jmp    L101
L157:
  lea    D14(%rip),%r12
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  lea    0xc(%rsp),%r13
  lea    D20(%rip),%rbp
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  je     L100
  nopl   (%rax)
L32:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  movslq 0xc(%rsp),%rax
  mov    %r12,%rdi
  lea    (%rax,%rax,2),%rax
  movslq 0x10(%rsp,%rax,8),%rdx
  lea    (%rdx,%rax,2),%rax
  mov    %rdx,%rbx
  mov    0x10(%rsp,%rax,4),%esi
  sub    $0x1,%ebx
  xor    %eax,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  jle    L118
  nopl   (%rax)
L84:
  movslq 0xc(%rsp),%rax
  movslq %ebx,%rdx
  mov    $0x9,%esi
  mov    %rbp,%rdi
  lea    (%rax,%rax,2),%rax
  lea    (%rdx,%rax,2),%rax
  mov    0x10(%rsp,%rax,4),%edx
  xor    %eax,%eax
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L84
L118:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  jne    L32
L100:
  mov    0xea628(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0xea638,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D26
type: byte[8]
name: retrieve_data
arguments: data_tag: D14
type: string
name: retrieve_data
arguments: data_tag: D20
type: string

tool: {"D26": [0x00,0x00,0x00,0x00,0x00,0x88,0xe3,0x40]}

tool: D14: %d

tool: D20: %0*d

assistant: ```c
int main(void) {
  int i, j, jj, k, T, N, DP[40000 + 1][6] = {0};
  for (i = 0; i <= 40000; i++)
    DP[i][0] = DP[i][1] = 1;
  for (j = 2; j <= (int)sqrt((double)40000); j++) {
    jj = j * j;
    for (i = jj; i <= 40000; i++) {
      DP[i][0] = ((DP[i][0]) > (DP[i - jj][0]) ? (DP[i][0]) : (DP[i - jj][0]));
      for (k = 1; k <= DP[i][0]; k++) {
        DP[i][k] += DP[i - jj][k];
      }
      for (k = 1; k <= DP[i][0]; k++) {
        if (DP[i][k] >= 1000000000) {
          if (k == DP[i][0]) {
            DP[i][0]++;
          }
          DP[i][k + 1] += DP[i][k] / 1000000000;
          DP[i][k] %= 1000000000;
        }
      }
    }
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &N);
    i = DP[N][0];
    printf("%d", DP[N][i]);
    while (--i > 0)
      printf("%0*d", 9, DP[N][i]);
    putchar('\n');
  }
  return 0;
}
```