user: 请从给定的汇编指令推导出C源代码:

```asm
<test_replace_solution>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbx
  lea    (%rdi,%rdi,1),%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    0x5(%rbx),%eax
  mov    %rsp,%rcx
  cltq
  lea    0xf(,%rax,8),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L52
L6:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L6
L52:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L24
L20:
  lea    D97(%rip),%rdi
  lea    0x7(%rsp),%r12
  call   <printf@plt>
  lea    -0x2c(%rbp),%rsi
  lea    D13(%rip),%rdi
  mov    %r12,%r14
  call   <scanf@plt>
  mov    -0x2c(%rbp),%esi
  mov    %r13d,%edi
  shr    $0x3,%r14
  and    $0xfffffffffffffff8,%r12
  call   <retrieve_and_write@plt>
  lea    D72(%rip),%rdi
  call   <printf@plt>
  mov    D69(%rip),%rax
  mov    D58(%rip),%rsi
  mov    %rax,0x0(,%r14,8)
  mov    D2(%rip),%rax
  mov    %rax,0x8(,%r14,8)
  lea    0x2(%rbx),%eax
  cltq
  mov    %rsi,(%r12,%rax,8)
  mov    D23(%rip),%rsi
  lea    0x3(%rbx),%eax
  cltq
  mov    %rsi,(%r12,%rax,8)
  mov    D39(%rip),%rsi
  lea    0x4(%rbx),%eax
  cltq
  mov    %rsi,(%r12,%rax,8)
  test   %ebx,%ebx
  jle    L37
  movslq %ebx,%rdx
  xor    %eax,%eax
  xchg   %ax,%ax
L22:
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,0x10(%r12,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L22
L37:
  mov    -0x2c(%rbp),%esi
  mov    %r12,%rcx
  mov    $0x3039,%edx
  mov    %r13d,%edi
  call   <solcon_replace_standard_solution@plt>
  cmp    $0x1,%eax
  je     L26
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    -0x2c(%rbp),%esi
  mov    %r13d,%edi
  call   <retrieve_and_write@plt>
L81:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L80
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L81
  cs nopw 0x0(%rax,%rax,1)
L24:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L20
L80:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D97
type: string
name: parse_data
arguments: tag: D13
type: string
name: parse_data
arguments: tag: D72
type: string
name: parse_data
arguments: tag: D69
type: f64
name: parse_data
arguments: tag: D58
type: f64
name: parse_data
arguments: tag: D2
type: f64
name: parse_data
arguments: tag: D23
type: f64
name: parse_data
arguments: tag: D39
type: f64
name: parse_data
arguments: tag: D30
type: string
name: parse_data
arguments: tag: D7
type: string

tool: D97: Give k : 

tool: D13: %d

tool: D72: Setting solution vector to 1,2,3,...


tool: D69: 1.111

tool: D58: 3.333

tool: D2: 2.222

tool: D23: 4.444

tool: D39: 5.555

tool: D30: The changed solution retrieved again :


tool: D7: some failure occurred...


assistant: ```c
void test_replace_solution(int n) {
  int i, k, m, fail;
  double sol[2 * n + 5];
  printf("Give k : ");
  scanf("%d", &k);
  retrieve_and_write(n, k);
  printf("Setting solution vector to 1,2,3,...\n");
  sol[0] = 1.111;
  sol[1] = 2.222;
  m = 12345;
  sol[2 * n + 2] = 3.333;
  sol[2 * n + 3] = 4.444;
  sol[2 * n + 4] = 5.555;
  for (i = 0; i < 2 * n; i++)
    sol[2 + i] = i;
  fail = solcon_replace_standard_solution(n, k, m, sol);
  if (fail == 1)
    printf("some failure occurred...\n");
  else {
    printf("The changed solution retrieved again :\n");
    retrieve_and_write(n, k);
  }
}
```