user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r14
  lea    D41(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L124
  mov    $0x1,%r14d
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D8(%rip),%rbp
  lea    D27(%rip),%rbx
  nopl   0x0(%rax,%rax,1)
L29:
  mov    %r12,%rsi
  xor    %eax,%eax
  mov    %r13,%rdx
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%ecx
  mov    0x14(%rsp),%esi
  movdqa D19(%rip),%xmm6
  movdqa D34(%rip),%xmm5
  cmp    %esi,%ecx
  jg     L61
  mov    %esi,%eax
  sub    %ecx,%eax
  lea    0x1(%rax),%edi
  cmp    $0x4,%eax
  jbe    L118
  movd   %ecx,%xmm7
  mov    %edi,%edx
  pxor   %xmm2,%xmm2
  xor    %eax,%eax
  pshufd $0x0,%xmm7,%xmm1
  paddd  D0(%rip),%xmm1
  shr    $0x2,%edx
  movdqa %xmm2,%xmm4
L58:
  movdqa %xmm1,%xmm3
  add    $0x1,%eax
  paddd  %xmm6,%xmm1
  movdqa %xmm3,%xmm0
  pand   %xmm5,%xmm0
  pcmpeqd %xmm4,%xmm0
  pandn  %xmm3,%xmm0
  paddd  %xmm0,%xmm2
  cmp    %eax,%edx
  jne    L58
  movdqa %xmm2,%xmm0
  mov    %edi,%eax
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%eax
  paddd  %xmm0,%xmm2
  add    %eax,%ecx
  movdqa %xmm2,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm2
  movd   %xmm2,%edx
  cmp    %eax,%edi
  je     L76
L36:
  lea    (%rdx,%rcx,1),%eax
  test   $0x1,%cl
  cmovne %eax,%edx
  lea    0x1(%rcx),%eax
  cmp    %esi,%eax
  jg     L76
  lea    (%rdx,%rax,1),%edi
  test   $0x1,%al
  lea    0x2(%rcx),%eax
  cmovne %edi,%edx
  cmp    %esi,%eax
  jg     L76
  lea    (%rdx,%rax,1),%edi
  test   $0x1,%al
  lea    0x3(%rcx),%eax
  cmovne %edi,%edx
  cmp    %eax,%esi
  jl     L76
  test   $0x1,%al
  lea    (%rdx,%rax,1),%edi
  cmovne %edi,%edx
  add    $0x4,%ecx
  cmp    %ecx,%esi
  jl     L76
  lea    (%rdx,%rcx,1),%eax
  and    $0x1,%ecx
  cmovne %eax,%edx
L76:
  mov    %r14d,%esi
  mov    %rbx,%rdi
  add    $0x1,%r14d
  call   <printf@plt>
  cmp    %r14d,0xc(%rsp)
  jge    L29
L124:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L97
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L61:
  xor    %edx,%edx
  jmp    L76
L118:
  xor    %edx,%edx
  jmp    L36
L97:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D41
data_type: string
name: read_data
arguments: label: D8
data_type: string
name: read_data
arguments: label: D27
data_type: string
name: read_data
arguments: label: D19
data_type: byte[16]
name: read_data
arguments: label: D34
data_type: byte[16]
name: read_data
arguments: label: D0
data_type: byte[16]

tool: D41: %d

tool: D8: %d %d

tool: D27: Case %d: %d


tool: {"D19": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D34": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D0": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int n, a, b, i, j, sum = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &a, &b);
    for (j = a; j <= b; j++) {
      if (j % 2 != 0)
        sum += j;
    }
    printf("Case %d: %d\n", i, sum);
    sum = 0;
  }
  return 0;
}
```