user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  push   %r14
  push   %r13
  lea    D125(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%r12
  lea    0x70(%rsp),%r15
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x10,%esi
  lea    D103(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x6c(%rsp),%rax
  mov    %rax,0x8(%rsp)
  lea    0x2c(%rsp),%rax
  mov    %rax,0x10(%rsp)
  lea    0x34(%rsp),%rax
  mov    %rax,0x18(%rsp)
L67:
  mov    %r12,%rbx
  xor    %ebp,%ebp
  mov    $0x1,%r14d
L113:
  xor    %eax,%eax
  call   <rand@plt>
  cltd
  shr    $0x1c,%edx
  lea    (%rax,%rdx,1),%ecx
  mov    %r14d,%eax
  and    $0xf,%ecx
  sub    %edx,%ecx
  add    $0x1,%ecx
  shl    %cl,%eax
  mov    %ecx,(%rbx)
  cltq
  test   %rbp,%rax
  jne    L113
  add    $0x4,%rbx
  or     %rax,%rbp
  cmp    %r15,%rbx
  jne    L113
  lea    D84(%rip),%rdi
  xor    %ebp,%ebp
  call   <puts@plt>
L57:
  mov    $0x9,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%r14
  mov    %r14,%rbx
L178:
  mov    (%r14),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r14,%rax
  sub    $0x4,%r14
  cmp    %r12,%rax
  jne    L178
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    0x2c(%rsp),%esi
  lea    -0x2(%rsi),%eax
  cmp    $0xe,%eax
  jbe    L110
  lea    D51(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x2c(%rsp),%esi
L188:
  mov    0x30(%rsp),%edx
  mov    0x18(%rsp),%rax
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %edx,%ecx
  mov    (%rax),%edx
  cmp    %ecx,%edx
  jge    L23
  add    $0x4,%rax
  cmp    %r15,%rax
  jne    L6
  test   %esi,%esi
  je     L125
  mov    $0x9,%edi
  call   <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L186:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rax
  sub    $0x4,%rbx
  cmp    %r12,%rax
  jne    L186
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D73(%rip),%rdi
  call   <printf@plt>
  cmp    $0x1d,%ebp
  jg     L30
  cmp    $0x10,%ebp
  jle    L98
L125:
  lea    D122(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  nopl   0x0(%rax,%rax,1)
L171:
  xor    %eax,%eax
  call   <getchar@plt>
  movsbl %al,%edi
  mov    %edi,%ebx
  call   <isalpha@plt>
  test   %eax,%eax
  je     L171
  cmp    $0x79,%bl
  je     L67
  lea    D50(%rip),%rdi
  call   <puts@plt>
  lea    D93(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  lea    D36(%rip),%rdi
  call   <puts@plt>
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L23:
  test   %esi,%esi
  jne    L57
  jmp    L125
  nopl   (%rax)
L110:
  movslq %esi,%rax
  mov    %r12,%rdx
  lea    -0x4(%r12,%rax,4),%rcx
  mov    %esi,%eax
  sar    %eax
  sub    $0x1,%eax
  lea    0x4(%r12,%rax,4),%rdi
  nopw   0x0(%rax,%rax,1)
L174:
  mov    (%rdx),%eax
  xor    (%rcx),%eax
  sub    $0x4,%rcx
  mov    %eax,(%rdx)
  xor    0x4(%rcx),%eax
  mov    %eax,0x4(%rcx)
  xor    %eax,(%rdx)
  add    $0x4,%rdx
  cmp    %rdi,%rdx
  jne    L174
  add    $0x1,%ebp
  jmp    L188
L98:
  lea    D60(%rip),%rdi
  call   <puts@plt>
  jmp    L125
L30:
  mov    $0x1d,%esi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L125
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D125
type_infered: string
name: retrieve_data
arguments: data_tag: D103
type_infered: string
name: retrieve_data
arguments: data_tag: D84
type_infered: string
name: retrieve_data
arguments: data_tag: D13
type_infered: string
name: retrieve_data
arguments: data_tag: D4
type_infered: string
name: retrieve_data
arguments: data_tag: D51
type_infered: string
name: retrieve_data
arguments: data_tag: D73
type_infered: string
name: retrieve_data
arguments: data_tag: D122
type_infered: string
name: retrieve_data
arguments: data_tag: D50
type_infered: string
name: retrieve_data
arguments: data_tag: D93
type_infered: string
name: retrieve_data
arguments: data_tag: D36
type_infered: string
name: retrieve_data
arguments: data_tag: D60
type_infered: string
name: retrieve_data
arguments: data_tag: D23
type_infered: string

tool: D125: %2d 

tool: D103: Reverse
-------

This is the game of Reverse. To win all you have to do is arrange a list
of numbers (1 through %d) in numerical order left to right.
To move, you input how many numbers from the right you want to reverse.
You will no doubt like this game but, if you want to quit just input '0'
as your move.



tool: D84: Here we go...


tool: D13: 	? 

tool: D4: %d

tool: D51: You can not reverse %d. Try again.


tool: D73: 

You ordered the list in %d moves.


tool: D122: 
Would you like to try another? (y/n) : 

tool: D50: 







	Cymon's Games

	http://WWW.CYMONSGAMES.COM


   This game and its code is a 2008 Cymon's Games game.
   If you have enjoyed this game or would just like to have a new game
   each week please visit:
   http://WWW.CYMONSGAMES.COM for C/C++ programming resources and programs
   for everyone, beginners and advanced users alike.



tool: D93: 
	Press ENTER to continue...






tool: D36: Good bye for now then!

tool: D60: Wow, you are either extremely good, or extremely lucky.

tool: D23: But it shouldn't have taken more than %d moves.


assistant: ```c
int main(void) {
  int list[16], c, d, turns;
  long mask;
  char yn;
  srand(time(NULL));
  printf("Reverse\n-------\n\n"
         "This is the game of Reverse. To win all you have to do is arrange a list\n"
         "of numbers (1 through %d) in numerical order left to right.\n"
         "To move, you input how many numbers from the right you want to reverse.\n"
         "You will no doubt like this game but, if you want to quit just input '0'\n"
         "as your move.\n\n",
         16);
  do {
    mask = 0;
    for (c = 0; c < 16; c++) {
      while (1 << (list[c] = rand() % 16 + 1) & mask)
        ;
      mask |= 1 << list[c];
    }
    turns = 0;
    puts("Here we go...\n");
    do {
      putchar('\t');
      for (c = 16; c > 0;)
        printf("%2d ", list[--c]);
      printf("\t? ");
      scanf("%d", &d);
      if (d > 1 && d <= 16) {
        for (c = 0; c < (d / 2); c++) {
          list[c] ^= list[d - c - 1];
          list[d - c - 1] ^= list[c];
          list[c] ^= list[d - c - 1];
        }
        turns++;
      } else
        printf("You can not reverse %d. Try again.\n", d);
      for (c = 0; c < 16 - 1 && list[c] > list[c + 1]; c++)
        ;
    } while (d && c < 16 - 1);
    if (d) {
      putchar('\t');
      for (c = 16; c > 0;)
        printf("%2d ", list[--c]);
      printf("\n\nYou ordered the list in %d moves.\n", turns);
      if (turns > 16 * 2 - 3)
        printf("But it shouldn't have taken more than %d moves.\n", 16 * 2 - 3);
      if (turns <= 16)
        puts("Wow, you are either extremely good, or extremely lucky.");
    }
    printf("\nWould you like to try another? (y/n) : ");
    while (!isalpha(yn = getchar()))
      ;
  } while (yn == 'y');
  puts("\n\n\n\n\n\n\n\n\tCymon's Games\n\n\thttp://WWW.CYMONSGAMES.COM\n\n\n"
       "   This game and its code is a 2008 Cymon's Games game.\n"
       "   If you have enjoyed this game or would just like to have a new game\n"
       "   each week please visit:\n"
       "   http://WWW.CYMONSGAMES.COM for C/C++ programming resources and programs\n"
       "   for everyone, beginners and advanced users alike.\n\n");
  printf("\n\tPress ENTER to continue...\n\n\n\n\n\n");
  getchar();
  getchar();
  puts("Good bye for now then!");
  exit(0);
}
```