user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D7(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  lea    D8(%rip),%r12
  push   %rbp
  mov    $0x10,%ebp
  push   %rbx
  lea    D46(%rip),%rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <puts@plt>
  mov    %r12,%rdi
  call   <puts@plt>
L5:
  mov    $0x9,%edi
  lea    -0x10(%rbp),%r14
  call   <putchar@plt>
L61:
  mov    %r14d,%edi
  call   <inverse@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  mov    %eax,%edx
  lea    (%rax,%rax,1),%ecx
  shl    $0x4,%edx
  sar    $0x4,%esi
  or     %edx,%esi
  mov    %eax,%edx
  sar    $0x7,%edx
  or     %ecx,%edx
  lea    0x0(,%rax,8),%ecx
  xor    %edx,%esi
  mov    %eax,%edx
  sar    $0x5,%edx
  xor    %eax,%esi
  or     %ecx,%edx
  xor    %edx,%esi
  mov    %eax,%edx
  shl    $0x2,%eax
  sar    $0x6,%edx
  or     %edx,%eax
  xor    %eax,%esi
  xor    %eax,%eax
  xor    $0x63,%esi
  mov    %esi,0x0(%r13,%r14,4)
  add    $0x1,%r14
  call   <printf@plt>
  cmp    %rbp,%r14
  jne    L61
  mov    $0xa,%edi
  lea    0x10(%r14),%rbp
  call   <putchar@plt>
  cmp    $0x100,%r14
  jne    L5
  lea    D14(%rip),%r14
  mov    $0x10,%ebp
  mov    %r14,%rdi
  call   <puts@plt>
  lea    D37(%rip),%rdi
  call   <puts@plt>
  mov    %r12,%rdi
  call   <puts@plt>
  mov    (%rsp),%r12d
L0:
  mov    $0x9,%edi
  lea    -0x10(%rbp),%r15d
  call   <putchar@plt>
  nopw   0x0(%rax,%rax,1)
L43:
  mov    $0x1,%eax
  xor    %esi,%esi
  cmp    %r15d,%r12d
  je     L48
  nopl   0x0(%rax)
L11:
  mov    %eax,%esi
  add    $0x1,%rax
  cmp    %r15d,-0x4(%r13,%rax,4)
  jne    L11
L48:
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <printf@plt>
  cmp    %ebp,%r15d
  jne    L43
  mov    $0xa,%edi
  lea    0x10(%r15),%ebp
  call   <putchar@plt>
  cmp    $0x100,%r15d
  jne    L0
  mov    %r14,%rdi
  call   <puts@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x418,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D7
type: string
name: parse_data
arguments: data_label: D8
type: string
name: parse_data
arguments: data_label: D46
type: string
name: parse_data
arguments: data_label: D14
type: string
name: parse_data
arguments: data_label: D37
type: string

tool: D7: uint8_t SBox[] =

tool: D8: {

tool: D46: 0x%.2x, 

tool: D14: };

tool: D37: uint8_t ISBox[] =

assistant: ```c
int main(int argc, char **argv) {
  (void)argc;
  (void)argv;
  uint8_t SBox[256];
  printf("uint8_t SBox[] =\n");
  printf("{\n");
  for (uint32_t i = 0; i < 256;) {
    printf("\t");
    for (uint8_t j = 0; j < 16; j++, i++) {
      uint8_t b = inverse(i);
      b = b ^ (((b) >> 4) | ((b) << (8 - 4))) ^ (((b) >> 5) | ((b) << (8 - 5))) ^ (((b) >> 6) | ((b) << (8 - 6))) ^ (((b) >> 7) | ((b) << (8 - 7))) ^ 0x63;
      SBox[i] = b;
      printf("0x%.2x, ", b);
    }
    printf("\n");
  }
  printf("};\n");
  printf("uint8_t ISBox[] =\n");
  printf("{\n");
  for (uint32_t i = 0; i < 256;) {
    printf("\t");
    for (uint8_t j = 0; j < 16; j++, i++) {
      uint8_t k;
      for (k = 0; SBox[k] != i; k++)
        ;
      printf("0x%.2x, ", k);
    }
    printf("\n");
  }
  printf("};\n");
  return 0;
}
```