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    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D125(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D82(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D82(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm1
  movsd  D25(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  D123(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  lea    D66(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D90(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x2c(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  mov    $0x1,%edi
  call   <balanceCalculator@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  mov    $0x1,%esi
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movl   $0x1,-0x24(%rbp)
  jmp    L44
L18:
  mov    -0x2c(%rbp),%eax
  imul   -0x24(%rbp),%eax
  cmp    $0x1,%eax
  je     L9
  movsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  imul   -0x24(%rbp),%eax
  movapd %xmm0,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <balanceCalculator@plt>
  movq   %xmm0,%rax
  mov    -0x2c(%rbp),%edx
  imul   -0x24(%rbp),%edx
  movq   %rax,%xmm0
  mov    %edx,%esi
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L6
L9:
  nop
L6:
  addl   $0x1,-0x24(%rbp)
L44:
  mov    -0x28(%rbp),%eax
  mov    -0x2c(%rbp),%ecx
  cltd
  idiv   %ecx
  cmp    %eax,-0x24(%rbp)
  jle    L18
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D125
type: string
name: unpack_data
arguments: data_label: D75
type: string
name: unpack_data
arguments: data_label: D82
type: string
name: unpack_data
arguments: data_label: D43
type: string
name: unpack_data
arguments: data_label: D25
type: f64
name: unpack_data
arguments: data_label: D123
type: byte[8]
name: unpack_data
arguments: data_label: D66
type: string
name: unpack_data
arguments: data_label: D4
type: string
name: unpack_data
arguments: data_label: D90
type: string
name: unpack_data
arguments: data_label: D55
type: string

tool: D125: Interest calculation program.

tool: D75: Starting balance? 

tool: D82: %lf

tool: D43: Annual interest rate percentage? 

tool: D25: 0.01

tool: {"D123": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: D66: Calculate for ? years: 

tool: D4: %d

tool: D90: print every ? years balance: 

tool: D55: Year %-3d  |  balance %10g


assistant: ```c
int main(int argc, const char *argv[]) {
  int N, toNYearLater;
  double start, rate, yearRate;
  printf("Interest calculation program.\n");
  printf("Starting balance? ");
  scanf("%lf", &start);
  printf("Annual interest rate percentage? ");
  scanf("%lf", &rate);
  yearRate = (1 + 0.01 * rate);
  printf("Calculate for ? years: ");
  scanf("%d", &toNYearLater);
  printf("print every ? years balance: ");
  scanf("%d", &N);
  printf("Year %-3d  |  balance %10g\n", 1, balanceCalculator(1, start, rate));
  for (int i = 1; i <= toNYearLater / N; i++) {
    if ((i * N) == 1)
      continue;
    printf("Year %-3d  |  balance %10g\n", i * N, balanceCalculator(i * N, start, rate));
  }
  return 0;
}
```