user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rdx
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x28(%rbp),%eax
  cmp    $0x1,%eax
  jne    L113
  mov    -0x24(%rbp),%eax
  test   %eax,%eax
  jne    L24
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  jmp    L68
L24:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x1,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L91
L153:
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  cmp    %eax,%edx
  jne    L140
  addl   $0x1,-0x18(%rbp)
  jmp    L86
L140:
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x20(%rbp)
L86:
  addl   $0x1,-0x14(%rbp)
L91:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L153
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L48
L113:
  mov    -0x28(%rbp),%eax
  cmp    $0x2,%eax
  jne    L48
  movl   $0x0,-0x10(%rbp)
  jmp    L6
L147:
  lea    -0x1c(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L161
L114:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x10(%rbp)
  jne    L168
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jne    L168
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L62
L168:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L62:
  addl   $0x1,-0xc(%rbp)
L161:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L114
  addl   $0x1,-0x10(%rbp)
L6:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L147
L48:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
L68:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L84
  call   <__stack_chk_fail@plt>
L84:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D59
type: string
name: unpack_data
arguments: label: D26
type: string
name: unpack_data
arguments: label: D1
type: string
name: unpack_data
arguments: label: D30
type: string

tool: D59: %d %d

tool: D26: %d

tool: D1: %d %d 

tool: D30: %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;
}
```