user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x64,%ecx
  xor    %r15d,%r15d
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0xf4240,%r12d
  push   %rbp
  mov    $0xffffffff,%ebp
  push   %rbx
  sub    $0x338,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0xffffffffffffffff,%rax
  lea    0x4(%rsp),%r14
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  nopl   0x0(%rax,%rax,1)
L64:
  xor    %eax,%eax
  call   <random_number@plt>
  imul   $0x63,%eax,%eax
  cmp    %eax,%ebp
  cmovl  %eax,%ebp
  cmp    %eax,%r12d
  cmovg  %eax,%r12d
  test   %r15d,%r15d
  je     L14
  lea    -0x1(%r15),%ecx
  mov    %rbx,%rdx
  lea    (%r14,%rcx,4),%rdi
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L11:
  xor    %esi,%esi
  cmp    %eax,(%rdx)
  sete   %sil
  add    $0x4,%rdx
  add    %esi,%ecx
  cmp    %rdx,%rdi
  jne    L11
  test   %ecx,%ecx
  je     L14
  cmp    $0x1,%ecx
  je     L51
  lea    D35(%rip),%rdi
  call   <puts@plt>
  xor    %edi,%edi
  call   <exit@plt>
L51:
  add    $0x1,%r13d
  cmp    $0x63,%r15d
  jne    L64
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13d,%esi
  xor    %eax,%eax
  xor    %ebp,%ebp
  lea    D22(%rip),%rdi
  lea    D40(%rip),%r12
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L39:
  mov    (%rbx,%rbp,4),%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0x63,%rbp
  jne    L39
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  add    $0x338,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  movslq %r15d,%rdx
  mov    %r15d,%esi
  lea    D31(%rip),%rdi
  add    $0x1,%r15d
  mov    %eax,(%rsp,%rdx,4)
  mov    %r13d,%edx
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L51
L46:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D35
datatype: string
name: extract_symbolic_data
arguments: tag: D13
datatype: string
name: extract_symbolic_data
arguments: tag: D22
datatype: string
name: extract_symbolic_data
arguments: tag: D40
datatype: string
name: extract_symbolic_data
arguments: tag: D31
datatype: string

tool: D35: 

on a enregistre plusieurs fois le meme

tool: D13: max %d 	 min %d 


tool: D22: nbre de tirage : %d


tool: D40: %3d %3d


tool: D31: k =%3d c =%3d


assistant: ```c
void main(void) {
  int k, nvois, ivois, max, min, c, i, dejavu;
  int liste[200];
  for (k = 0; k < 200; k++)
    liste[k] = -1;
  max = -1;
  min = 1000000;
  nvois = 99;
  k = 0;
  c = 0;
  while (k != nvois) {
    ivois = (int)(nvois * random_number());
    if (ivois > max)
      max = ivois;
    if (ivois < min)
      min = ivois;
    dejavu = 0;
    for (i = 0; i < k; i++) {
      if (liste[i] == ivois)
        dejavu++;
    }
    if (dejavu == 0) {
      liste[k] = ivois;
      printf("k =%3d c =%3d\n", k, c);
      k++;
    } else if (dejavu > 1) {
      printf("\n\non a enregistre plusieurs fois le meme\n");
      exit(0);
    }
    c++;
  }
  printf("max %d \t min %d \n", max, min);
  printf("nbre de tirage : %d\n", c);
  for (k = 0; k < nvois; k++)
    printf("%3d %3d\n", k, liste[k]);
}
```