user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x41,%esi
  lea    D2(%rip),%rdi
  lea    D12(%rip),%r13
  push   %r12
  lea    D73(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x1c8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1b8(%rsp)
  xor    %eax,%eax
  lea    0x1a0(%rsp),%rbp
  lea    0x8(%rsp),%rbx
  call   <printf@plt>
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movl   $0xf,0x8(%rsp)
  mov    $0xf,%eax
  lea    D82(%rip),%rbp
  movl   $0x18,0xc(%rsp)
L40:
  cmp    $0x9,%eax
  jle    L48
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmpl   $0x9,0x8(%rsp)
  jg     L53
L48:
  lea    D15(%rip),%r12
  lea    0xc(%rsp),%rbx
  lea    D12(%rip),%r13
L31:
  movslq 0xc(%rsp),%rdx
  cmp    $0x9,%edx
  jle    L11
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movslq 0xc(%rsp),%rdx
  cmp    $0x9,%edx
  jg     L43
L11:
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    $0x32,%ecx
  rep stos %rax,%es:(%rdi)
  movslq 0x8(%rsp),%rax
  lea    0x38(%rsp),%r12
  lea    0x1c8(%rsp),%r13
  lea    (%rax,%rax,4),%rax
  lea    (%rdx,%rax,2),%rax
  movl   $0x1,0x10(%rsp,%rax,4)
  nopl   0x0(%rax)
L82:
  lea    -0x28(%r12),%rbx
  nopl   (%rax)
L8:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L8
  mov    $0xa,%edi
  lea    0x28(%rbx),%r12
  call   <putchar@plt>
  cmp    %r13,%r12
  jne    L82
  mov    0x1b8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x1c8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L53:
  mov    %r13,%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%eax
  jmp    L40
L43:
  mov    %r13,%rdi
  call   <puts@plt>
  jmp    L31
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
data_type: string
name: unpack_data
arguments: data_label: D12
data_type: string
name: unpack_data
arguments: data_label: D73
data_type: string
name: unpack_data
arguments: data_label: D19
data_type: string
name: unpack_data
arguments: data_label: D13
data_type: string
name: unpack_data
arguments: data_label: D4
data_type: string
name: unpack_data
arguments: data_label: D82
data_type: string
name: unpack_data
arguments: data_label: D15
data_type: string

tool: D2: %c


tool: D12: You can not chose more than 10

tool: D73: Chose a row 

tool: D19: Write a name:

tool: D13: %s

tool: D4: Hello %s


tool: D82: %i

tool: D15: Chose a col

assistant: ```c
int main() {
  char letter = 'A';
  printf("%c\n", letter);
  char str[20];
  printf("Write a name:\n");
  scanf("%s", str);
  printf("Hello %s\n", str);
  int number = 15, number2 = 24;
  while (number >= 10) {
    printf("Chose a row ");
    scanf("%i", &number);
    if (number >= 10) {
      printf("You can not chose more than 10\n");
    }
  }
  while (number2 >= 10) {
    printf("Chose a col");
    scanf("%i", &number2);
    if (number2 >= 10) {
      printf("You can not chose more than 10\n");
    }
  }
  int matrix[10][10] = {0};
  matrix[number][number2] = 1;
  for (int row = 0; row < 10; row++) {
    for (int col = 0; col < 10; col++) {
      printf("%i", matrix[row][col]);
    }
    printf("\n");
  }
  return 0;
}
```