user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  movl   $0x2,0x14(%rsp)
  movl   $0x3,0x18(%rsp)
  movl   $0x4,0x1c(%rsp)
  movl   $0x5,0x20(%rsp)
  movl   $0x6,0x24(%rsp)
  movl   $0x7,0x28(%rsp)
  movl   $0x8,0x2c(%rsp)
  movl   $0x9,0x30(%rsp)
  movl   $0xa,0x34(%rsp)
  movl   $0xb,0x38(%rsp)
  movl   $0xc,0x3c(%rsp)
  movl   $0xd,0x40(%rsp)
  movl   $0xe,0x44(%rsp)
  movl   $0xf,0x48(%rsp)
  movl   $0x10,0x4c(%rsp)
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  call   <get_2_tupples@plt>
  mov    %rax,%rbp
  mov    %rbx,%rax
  lea    0x50(%rsp),%rcx
  mov    $0x0,%edx
L16:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L16
  lea    0x3(%rdx),%r12d
  test   %edx,%edx
  cmovns %edx,%r12d
  sar    $0x2,%r12d
  mov    %r12d,%esi
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    $0x2,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x2,%edx
  mov    0xc(%rsp),%esi
  mov    %rbp,%rdi
  call   <display_tupples@plt>
  mov    $0x2,%ebx
  lea    D4(%rip),%r14
L31:
  mov    %rbp,%r13
  lea    0x10(%rsp),%rdi
  lea    0xc(%rsp),%r9
  mov    %r12d,%r8d
  mov    %ebx,%ecx
  mov    0xc(%rsp),%edx
  mov    %rbp,%rsi
  call   <get_next_size_tupples@plt>
  mov    %rax,%rbp
  mov    %r13,%rdi
  call   <free@plt>
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %ebx,%edx
  mov    0xc(%rsp),%esi
  mov    %rbp,%rdi
  call   <display_tupples@plt>
  cmp    $0x4,%ebx
  jne    L31
  mov    $0x4,%ecx
  mov    $0x4,%edx
  mov    0xc(%rsp),%esi
  mov    %rbp,%rdi
  call   <get_all_solutions@plt>
  test   %rbp,%rbp
  je     L13
  mov    %rbp,%rdi
  call   <free@plt>
L13:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D22
data_type: string
name: retrieve_data
arguments: tag: D4
data_type: string

tool: D22: Sum = %d


tool: D4: Cardinality = %d


assistant: ```c
int main() {
  int seq[16] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16};
  int tupple_array_size = 0;
  int cardinality = 2;
  int *tupples_n = get_2_tupples(seq, &tupple_array_size);
  int *tupples_nplus1 = NULL;
  int *final_tupples = NULL;
  int sum = 0;
  int final_cardinality = 4;
  int num_tupples_in_sol = 4;
  int seq_sz = sizeof(seq) / sizeof(seq[0]);
  for (int i = 0; i < seq_sz; i++) {
    sum += seq[i];
  }
  sum /= final_cardinality;
  printf("Sum = %d\n", sum);
  printf("Cardinality = %d\n", cardinality);
  display_tupples(tupples_n, tupple_array_size, cardinality);
  while (cardinality != final_cardinality) {
    tupples_nplus1 = get_next_size_tupples(seq, tupples_n, tupple_array_size, cardinality, sum, &tupple_array_size);
    free(tupples_n);
    tupples_n = tupples_nplus1;
    cardinality++;
    printf("Cardinality = %d\n", cardinality);
    display_tupples(tupples_n, tupple_array_size, cardinality);
  }
  get_all_solutions(tupples_nplus1, tupple_array_size, final_cardinality, num_tupples_in_sol);
  if (tupples_nplus1) {
    free(tupples_nplus1);
  }
}
```