user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x350,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x350(%rbp)
  jmp    L19
L136:
  mov    -0x350(%rbp),%eax
  cltq
  movl   $0xffffffff,-0x330(%rbp,%rax,4)
  addl   $0x1,-0x350(%rbp)
L19:
  cmpl   $0xc7,-0x350(%rbp)
  jle    L136
  movl   $0xffffffff,-0x34c(%rbp)
  movl   $0xf4240,-0x348(%rbp)
  movl   $0x63,-0x338(%rbp)
  movl   $0x0,-0x350(%rbp)
  movl   $0x0,-0x344(%rbp)
  jmp    L20
L71:
  mov    $0x0,%eax
  call   <random_number@plt>
  mov    -0x338(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x334(%rbp)
  mov    -0x334(%rbp),%eax
  cmp    -0x34c(%rbp),%eax
  jle    L125
  mov    -0x334(%rbp),%eax
  mov    %eax,-0x34c(%rbp)
L125:
  mov    -0x334(%rbp),%eax
  cmp    -0x348(%rbp),%eax
  jge    L70
  mov    -0x334(%rbp),%eax
  mov    %eax,-0x348(%rbp)
L70:
  movl   $0x0,-0x33c(%rbp)
  movl   $0x0,-0x340(%rbp)
  jmp    L84
L89:
  mov    -0x340(%rbp),%eax
  cltq
  mov    -0x330(%rbp,%rax,4),%eax
  cmp    %eax,-0x334(%rbp)
  jne    L48
  addl   $0x1,-0x33c(%rbp)
L48:
  addl   $0x1,-0x340(%rbp)
L84:
  mov    -0x340(%rbp),%eax
  cmp    -0x350(%rbp),%eax
  jl     L89
  cmpl   $0x0,-0x33c(%rbp)
  jne    L26
  mov    -0x350(%rbp),%eax
  cltq
  mov    -0x334(%rbp),%edx
  mov    %edx,-0x330(%rbp,%rax,4)
  mov    -0x344(%rbp),%edx
  mov    -0x350(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x350(%rbp)
  jmp    L65
L26:
  cmpl   $0x1,-0x33c(%rbp)
  jle    L65
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L65:
  addl   $0x1,-0x344(%rbp)
L20:
  mov    -0x350(%rbp),%eax
  cmp    -0x338(%rbp),%eax
  jne    L71
  mov    -0x348(%rbp),%edx
  mov    -0x34c(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x344(%rbp),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x350(%rbp)
  jmp    L116
L14:
  mov    -0x350(%rbp),%eax
  cltq
  mov    -0x330(%rbp,%rax,4),%edx
  mov    -0x350(%rbp),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x350(%rbp)
L116:
  mov    -0x350(%rbp),%eax
  cmp    -0x338(%rbp),%eax
  jl     L14
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L128
  call   <__stack_chk_fail@plt>
L128:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D6
type_infered: string
name: retrieve_data
arguments: data_label: D34
type_infered: string
name: retrieve_data
arguments: data_label: D7
type_infered: string
name: retrieve_data
arguments: data_label: D41
type_infered: string
name: retrieve_data
arguments: data_label: D39
type_infered: string

tool: D6: k =%3d c =%3d


tool: D34: 

on a enregistre plusieurs fois le meme

tool: D7: max %d 	 min %d 


tool: D41: nbre de tirage : %d


tool: D39: %3d %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]);
}
```