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
  lea    D14(%rip),%r14
  push   %r13
  mov    %r14,%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%ebp
  cmp    $0x1,%ebp
  je     L95
  cmp    $0x2,%ebp
  je     L26
L122:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L116
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  cmpl   $0x0,0xc(%rsp)
  jle    L122
  xor    %ebx,%ebx
  lea    0x10(%rsp),%r14
  lea    0x14(%rsp),%r12
  lea    D13(%rip),%r13
  lea    D1(%rip),%rbp
  nopl   (%rax)
L73:
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%edx
  xor    %r15d,%r15d
  test   %edx,%edx
  jle    L69
  nopl   0x0(%rax,%rax,1)
L142:
  mov    0xc(%rsp),%eax
  mov    0x14(%rsp),%esi
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jne    L135
  sub    $0x1,%edx
  cmp    %r15d,%edx
  je     L8
L135:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L168:
  mov    0x10(%rsp),%edx
  add    $0x1,%r15d
  cmp    %r15d,%edx
  jg     L142
L69:
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jg     L73
  jmp    L122
L95:
  cmpl   $0x0,0xc(%rsp)
  je     L122
  lea    D8(%rip),%r13
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x1,0xc(%rsp)
  mov    0x10(%rsp),%edx
  jle    L123
  mov    $0x1,%r15d
  xor    %ebx,%ebx
  lea    0x14(%rsp),%r12
  nopl   0x0(%rax)
L56:
  mov    %r12,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%edx
  cmp    0x14(%rsp),%edx
  jne    L110
  mov    0xc(%rsp),%eax
  add    $0x1,%ebx
  add    $0x1,%r15d
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jg     L56
  mov    %r15d,%ebp
L123:
  mov    %ebp,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L122
  nopl   (%rax)
L8:
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L168
  nopl   0x0(%rax,%rax,1)
L110:
  mov    %r15d,%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  mov    0x14(%rsp),%edx
  sub    $0x1,%eax
  mov    %edx,0x10(%rsp)
  cmp    %ebx,%eax
  jle    L123
  mov    $0x1,%r15d
  jmp    L56
L116:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D14
type: string
name: read_data
arguments: data_tag: D13
type: string
name: read_data
arguments: data_tag: D1
type: string
name: read_data
arguments: data_tag: D8
type: string

tool: D14: %d %d

tool: D13: %d %d 

tool: D1: %d 

tool: D8: %d

assistant: ```c
int main() {
  int mode, numInputs;
  scanf("%d %d", &mode, &numInputs);
  if (mode == 1) {
    if (numInputs == 0) {
      printf("\n");
      return 0;
    }
    int prevNum;
    scanf("%d", &prevNum);
    int prevNumCount = 1;
    for (int i = 0; i < numInputs - 1; i++) {
      int currentNum;
      scanf("%d", &currentNum);
      if (prevNum == currentNum) {
        prevNumCount++;
      } else {
        printf("%d %d ", prevNumCount, prevNum);
        prevNumCount = 1;
        prevNum = currentNum;
      }
    }
    printf("%d %d", prevNumCount, prevNum);
  } else if (mode == 2) {
    for (int i = 0; i < numInputs; i++) {
      int numCount;
      int currentNum;
      scanf("%d %d ", &numCount, &currentNum);
      for (int j = 0; j < numCount; j++) {
        if (i == numInputs - 1 && j == numCount - 1) {
          printf("%d", currentNum);
        } else {
          printf("%d ", currentNum);
        }
      }
    }
  }
  printf("\n");
  return 0;
}
```