user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  mov    $0x2d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x298,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x288(%rsp)
  xor    %eax,%eax
  lea    0x1f(%rsp),%r13
  jmp    L105
  xchg   %ax,%ax
L28:
  cmp    $0x20,%al
  cmove  %r14d,%eax
  mov    %al,0x1b0(%rsp,%rbx,1)
  add    $0x1,%rbx
L105:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movzbl 0x1f(%rsp),%eax
  cmp    $0xa,%al
  jne    L28
  lea    0x20(%rsp),%rdi
  mov    $0x32,%ecx
  xor    %eax,%eax
  mov    %rdi,0x8(%rsp)
  rep stos %rax,%es:(%rdi)
  test   %ebx,%ebx
  je     L51
  movslq %ebx,%rdi
  lea    -0x1(%rbx),%ecx
  lea    0x220(%rsp),%rax
  lea    0x1af(%rsp,%rdi,1),%rdx
  lea    0x221(%rsp,%rcx,1),%rsi
  nop
L61:
  movzbl (%rdx),%ecx
  add    $0x1,%rax
  sub    $0x1,%rdx
  mov    %cl,-0x1(%rax)
  cmp    %rsi,%rax
  jne    L61
L69:
  mov    0x8(%rsp),%rbp
  movb   $0x2a,0x220(%rsp,%rdi,1)
  xor    %r12d,%r12d
  lea    0x21f(%rsp),%r13
  movsbl 0x220(%rsp),%edi
  mov    %rbp,%r14
  cs nopw 0x0(%rax,%rax,1)
L82:
  cmp    $0x2d,%dil
  je     L86
  lea    0x1(%r12),%r15d
  xor    %ebx,%ebx
  movslq %r15d,%r15
  jmp    L73
  cs nopw 0x0(%rax,%rax,1)
L88:
  sub    $0x30,%edi
  call   <expo@plt>
  add    %eax,%ebx
  mov    %r15d,%eax
  add    $0x1,%r15
  movsbl 0x0(%r13,%r15,1),%edi
  cmp    $0x2d,%dil
  je     L20
L73:
  mov    %r15d,%esi
  lea    -0x1(%r15),%eax
  sub    %r12d,%esi
  sub    $0x1,%esi
  cmp    $0x2a,%dil
  jne    L88
  mov    %eax,%r12d
L17:
  movslq %r12d,%rax
  mov    %ebx,0x0(%rbp)
  add    $0x4,%rbp
  movsbl 0x220(%rsp,%rax,1),%edi
  cmp    $0x2a,%dil
  jne    L82
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rbp
  mov    %rbp,%rdi
  add    $0x190,%rbp
  call   <stack@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %esi,%esi
  nopl   (%rax)
L100:
  add    (%r14),%esi
  add    $0x4,%r14
  cmp    %rbp,%r14
  jne    L100
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x288(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L66
  add    $0x298,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L86:
  mov    %r12d,%eax
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L20:
  lea    0x1(%rax),%r12d
  jmp    L17
L51:
  xor    %edi,%edi
  jmp    L69
L66:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D5
type_infered: string
name: read_data
arguments: tag: D13
type_infered: string
name: read_data
arguments: tag: D11
type_infered: string

tool: D5: %c

tool: D13: Your list: 

tool: D11: The sum of the integers is: %d


assistant: ```c
int main(void) {
  char list[100];
  int counter = 0;
  char temp;
  do {
    scanf("%c", &temp);
    if (temp == '\n') {
      break;
    }
    if (temp == ' ') {
      list[counter] = '-';
      counter++;
      continue;
    }
    list[counter] = temp;
    counter++;
  } while (1);
  int intlist[100] = {0};
  char revlist[100];
  int i;
  for (i = 0; i < counter; i++) {
    revlist[i] = list[counter - 1 - i];
  }
  revlist[counter] = '*';
  int tet = 0, sum, exp, cot = 0, temp1;
  do {
    exp = 0;
    sum = 0;
    while (revlist[tet] != '-') {
      if (revlist[tet] == '*') {
        break;
      }
      temp1 = revlist[tet] - '0';
      sum += expo(temp1, exp);
      exp++;
      tet++;
    }
    intlist[cot] = sum;
    cot++;
    if (revlist[tet] == '-') {
      tet++;
      continue;
    }
  } while (revlist[tet] != '*');
  printf("Your list: ");
  stack(intlist);
  printf("\n");
  int sum2 = 0;
  for (i = 0; i < 100; i++) {
    sum2 += intlist[i];
  }
  printf("The sum of the integers is: %d\n", sum2);
  return 0;
}
```