user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r14
  lea    D62(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  lea    D27(%rip),%rbp
  push   %rbx
  sub    $0xcc0,%rsp
  movdqa D75(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0xcb8(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x4dc(%rsp)
  movabs $0x5756555453525150,%rax
  lea    0x4e0(%rsp),%rbx
  movaps %xmm0,0x4c0(%rsp)
  mov    %rax,0x4d0(%rsp)
  movl   $0x2a5a5958,0x4d8(%rsp)
  call   <puts@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D67(%rip),%rdi
  call   <puts@plt>
  mov    %rbx,%rsi
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    0x10(%rsp),%rax
  lea    0x4c0(%rsp),%rdx
  xchg   %ax,%ax
L20:
  movq   $0x0,(%rax)
  add    $0x18,%rax
  movq   $0x0,-0xc(%rax)
  cmp    %rax,%rdx
  jne    L20
  movslq 0x8(%rsp),%rax
  mov    0x20(%rsp),%r10d
  movl   $0x1c,0x1c(%rsp)
  mov    %rax,%rbp
  imul   $0xffffffff92492493,%rax,%rax
  mov    %ebp,%edx
  sar    $0x1f,%edx
  shr    $0x20,%rax
  add    %ebp,%eax
  sar    $0x4,%eax
  mov    %eax,%ecx
  sub    %edx,%eax
  sub    %edx,%ecx
  imul   %r10d,%eax
  imul   $0x1c,%ecx,%esi
  mov    %ebp,%ecx
  sub    %esi,%ecx
  mov    $0x1,%esi
  sub    %eax,%esi
  movd   %ecx,%xmm0
  movd   %esi,%xmm2
  punpckldq %xmm2,%xmm0
  movq   %xmm0,0x28(%rsp)
  test   %ecx,%ecx
  je     L2
  lea    0x34(%rsp),%r8
  mov    $0x1c,%eax
  mov    $0x2,%edi
  jmp    L97
  nopl   (%rax)
L86:
  mov    %edx,%ecx
  mov    %r9d,%esi
L97:
  cltd
  movslq %edi,%r11
  add    $0xc,%r8
  add    $0x1,%edi
  idiv   %ecx
  imul   %esi,%eax
  movd   %edx,%xmm0
  sub    %eax,%r10d
  mov    %ecx,%eax
  movd   %r10d,%xmm1
  mov    %r10d,%r9d
  mov    %esi,%r10d
  punpckldq %xmm1,%xmm0
  movq   %xmm0,-0xc(%r8)
  test   %edx,%edx
  jne    L86
L26:
  lea    (%r11,%r11,2),%rax
  mov    %ebp,%esi
  lea    D19(%rip),%rdi
  xor    %ebp,%ebp
  movslq 0x14(%rsp,%rax,4),%r12
  xor    %eax,%eax
  movslq %ebp,%r14
  mov    %r12d,%edx
  mov    %r12,%r13
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  mov    %r12d,%ecx
  mov    %r12d,%edx
  imul   $0xffffffff92492493,%r12,%r12
  lea    D55(%rip),%rdi
  xor    %eax,%eax
  imul   %esi,%ecx
  shr    $0x20,%r12
  call   <printf@plt>
  add    %r13d,%r12d
  mov    %r13d,%eax
  mov    %rbx,%rdi
  sar    $0x1f,%eax
  sar    $0x4,%r12d
  sub    %eax,%r12d
  imul   $0x1c,%r12d,%eax
  sub    %eax,%r13d
  mov    %r13d,%r12d
  lea    0x8d0(%rsp),%r13
  call   <strlen@plt>
  cmp    %ebp,%eax
  jle    L56
  nopl   0x0(%rax,%rax,1)
L63:
  movzbl (%rbx,%rbp,1),%ecx
  xor    %eax,%eax
  cmp    $0x41,%cl
  je     L74
  mov    $0x1,%edx
  lea    0x4bf(%rsp),%rsi
  nopl   0x0(%rax,%rax,1)
L59:
  mov    %edx,%eax
  add    $0x1,%rdx
  cmp    %cl,(%rsi,%rdx,1)
  jne    L59
L74:
  sub    0xc(%rsp),%eax
  mov    %rbx,%rdi
  imul   %r12d,%eax
  add    $0xaf0,%eax
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0xffffffff92492493,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x4,%edx
  sub    %ecx,%edx
  imul   $0x1c,%edx,%edx
  sub    %edx,%eax
  cltq
  movzbl 0x4c0(%rsp,%rax,1),%eax
  mov    %al,0x0(%r13,%rbp,1)
  add    $0x1,%rbp
  movslq %ebp,%r14
  call   <strlen@plt>
  cmp    %ebp,%eax
  jg     L63
L56:
  mov    %r13,%rsi
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  movb   $0x0,0x8d0(%rsp,%r14,1)
  call   <printf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0xcb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0xcc0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  mov    $0x1,%r11d
  jmp    L26
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D62
data_type: string
name: parse_data
arguments: tag: D27
data_type: string
name: parse_data
arguments: tag: D75
data_type: byte[16]
name: parse_data
arguments: tag: D67
data_type: string
name: parse_data
arguments: tag: D20
data_type: string
name: parse_data
arguments: tag: D19
data_type: string
name: parse_data
arguments: tag: D55
data_type: string
name: parse_data
arguments: tag: D41
data_type: string

tool: D62: Introduce los dos numeros del criptosistema afin: 

tool: D27: %d

tool: {"D75": [0x41,0x42,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x4b,0x4c,0x4d,0x4e,0x6e,0x4f]}

tool: D67: Introduce el mensaje a decodificar: 

tool: D20: %s

tool: D19: El inverso de %d modulo 28 es %d


tool: D55: Comprobamos: %d * %d = %d


tool: D41: Mensaje decodificado: 
%s

assistant: ```c
int main(void) {
  int a, b;
  int a_inv;
  int D;
  char clave[] = "ABCDEFGHIJKLMNnOPQRSTUVWXYZ*";
  char mensaje[1000];
  char decodificado[1000];
  int codif;
  int matriz[100][3];
  int i = 2;
  int cociente;
  int j, k;
  printf("Introduce los dos numeros del criptosistema afin: \n");
  scanf("%d", &a);
  scanf("%d", &b);
  printf("Introduce el mensaje a decodificar: \n");
  scanf("%s", mensaje);
  for (j = 0; j < 100; j++) {
    for (k = 0; k < 3; k++) {
      matriz[j][k] = 0;
    }
  }
  matriz[0][1] = 1;
  matriz[1][2] = 1;
  matriz[0][0] = a;
  matriz[1][0] = 28;
  while (1) {
    cociente = (matriz[i - 2][0] / matriz[i - 1][0]);
    matriz[i][0] = (matriz[i - 2][0] % matriz[i - 1][0]);
    matriz[i][1] = (matriz[i - 2][1] - (cociente * matriz[i - 1][1]));
    matriz[i][2] = (matriz[i - 2][2] - (cociente * matriz[i - 1][2]));
    if (matriz[i][0] == 0)
      break;
    i++;
  }
  printf("El inverso de %d modulo 28 es %d\n", a, matriz[i - 1][1]);
  printf("Comprobamos: %d * %d = %d\n", a, matriz[i - 1][1], matriz[i - 1][1] * a);
  a_inv = (matriz[i - 1][1] % 28);
  for (i = 0; i < strlen(mensaje); i++) {
    codif = 0;
    j = 0;
    while (codif == 0) {
      if (mensaje[i] == clave[j]) {
        decodificado[i] = clave[((a_inv * (j - b)) + 2800) % 28];
        codif = 1;
      }
      j++;
    }
  }
  decodificado[i] = '\0';
  printf("Mensaje decodificado: \n%s", decodificado);
  getchar();
  getchar();
}
```