user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Problem7>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1f5c(%rbp)
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1f68(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x2,-0x1f64(%rbp)
  jmp    L55
L155:
  movl   $0x2,-0x1f60(%rbp)
  jmp    L138
L139:
  mov    -0x1f64(%rbp),%eax
  cltd
  idivl  -0x1f60(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  je     L158
  addl   $0x1,-0x1f60(%rbp)
L138:
  mov    -0x1f64(%rbp),%eax
  cltd
  idivl  -0x1f60(%rbp)
  cmp    %eax,-0x1f60(%rbp)
  jle    L139
  jmp    L140
L158:
  nop
L140:
  mov    -0x1f64(%rbp),%eax
  cltd
  idivl  -0x1f60(%rbp)
  cmp    %eax,-0x1f60(%rbp)
  jle    L3
  mov    -0x1f68(%rbp),%eax
  cltd
  idivl  -0x1f64(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L3
  mov    -0x1f5c(%rbp),%eax
  cltq
  mov    -0x1f64(%rbp),%edx
  mov    %edx,-0x1f50(%rbp,%rax,4)
  addl   $0x1,-0x1f5c(%rbp)
L3:
  addl   $0x1,-0x1f64(%rbp)
L55:
  cmpl   $0x1869f,-0x1f64(%rbp)
  jle    L155
  mov    -0x1f5c(%rbp),%eax
  mov    %eax,-0x1f54(%rbp)
  movl   $0x2,-0x1f58(%rbp)
  movl   $0x0,-0x1f64(%rbp)
  jmp    L78
L76:
  mov    -0x1f64(%rbp),%eax
  cltq
  mov    -0x1f58(%rbp),%edx
  mov    %edx,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0x1f58(%rbp)
  addl   $0x1,-0x1f64(%rbp)
L78:
  mov    -0x1f68(%rbp),%eax
  sub    $0x2,%eax
  cmp    %eax,-0x1f64(%rbp)
  jl     L76
  mov    -0x1f68(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x1f64(%rbp)
  jmp    L6
L48:
  movl   $0x0,-0x1f5c(%rbp)
  jmp    L134
L31:
  mov    -0x1f64(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    -0x1f5c(%rbp),%edx
  movslq %edx,%rdx
  mov    -0x1f50(%rbp,%rdx,4),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  je     L62
  mov    -0x1f5c(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x1f54(%rbp)
  jne    L35
  mov    -0x1f68(%rbp),%edx
  mov    -0x1f64(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L35:
  addl   $0x1,-0x1f5c(%rbp)
L134:
  mov    -0x1f5c(%rbp),%eax
  cmp    -0x1f54(%rbp),%eax
  jl     L31
  jmp    L83
L62:
  nop
L83:
  addl   $0x1,-0x1f64(%rbp)
L6:
  mov    -0x1f68(%rbp),%eax
  sub    $0x2,%eax
  cmp    %eax,-0x1f64(%rbp)
  jl     L48
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L97
  call   <__stack_chk_fail@plt>
L97:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D43
data_type: string
name: disassemble_data
arguments: data_label: D20
data_type: string
name: disassemble_data
arguments: data_label: D49
data_type: string
name: disassemble_data
arguments: data_label: D27
data_type: string
name: disassemble_data
arguments: data_label: D33
data_type: string

tool: D43: Problem 7: Given a number n, all the fractions less than 1, such that n is the denominator, is printed (in lowest terms).

tool: D20: Enter n: 

tool: D49: %d

tool: D27: The output is: 1/%d

tool: D33: , %d/%d

assistant: ```c
void Problem7() {
  int n, temp, i, j, arr[1000], k = 0, ix, jx, tempo[1000], tempor;
  printf("Problem 7: Given a number n, all the fractions less than 1, such that n is the denominator, is printed (in lowest terms).\n");
  printf("Enter n: ");
  scanf("%d", &n);
  for (i = 2; i < 100000; i++) {
    for (j = 2; j <= (i / j); j++) {
      if (!(i % j)) {
        break;
      }
    }
    if (j > (i / j)) {
      if (n % i == 0) {
        arr[k] = i;
        k++;
      }
    }
  }
  temp = k;
  tempor = 2;
  for (i = 0; i < n - 2; i++) {
    tempo[i] = tempor;
    tempor++;
  }
  printf("The output is: 1/%d", n);
  for (i = 0; i < n - 2; i++) {
    for (k = 0; k < temp; k++) {
      if (tempo[i] % arr[k] == 0) {
        break;
      }
      if (k + 1 == temp) {
        printf(", %d/%d", tempo[i], n);
      }
    }
  }
  printf("\n");
}
```