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    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x44(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x3c(%rbp)
  jmp    L52
L30:
  mov    -0x3c(%rbp),%eax
  add    $0x5,%eax
  mov    %eax,%esi
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x3c(%rbp),%eax
  cltq
  movsd  D82(%rip),%xmm0
  movsd  %xmm0,-0x30(%rbp,%rax,8)
  addl   $0x1,-0x3c(%rbp)
L52:
  cmpl   $0x4,-0x3c(%rbp)
  jle    L30
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x1,-0x38(%rbp)
  jmp    L34
L15:
  mov    -0x38(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L40
L22:
  mov    -0x44(%rbp),%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  add    $0x1,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mov    -0x34(%rbp),%eax
  cltq
  movsd  -0x30(%rbp,%rax,8),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x34(%rbp),%eax
  cltq
  movsd  %xmm0,-0x30(%rbp,%rax,8)
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,8),%rax
  movq   %rax,%xmm0
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L40:
  cmpl   $0x4,-0x34(%rbp)
  jle    L22
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x38(%rbp)
L34:
  mov    -0x40(%rbp),%eax
  cmp    %eax,-0x38(%rbp)
  jle    L15
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L38
  call   <__stack_chk_fail@plt>
L38:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D58
type_infered: string
name: retrieve_data
arguments: label: D49
type_infered: string
name: retrieve_data
arguments: label: D79
type_infered: string
name: retrieve_data
arguments: label: D34
type_infered: string
name: retrieve_data
arguments: label: D56
type_infered: string
name: retrieve_data
arguments: label: D82
type_infered: byte[8]
name: retrieve_data
arguments: label: D29
type_infered: string
name: retrieve_data
arguments: label: D20
type_infered: string

tool: D58: enter interest rate:

tool: D49: %d

tool: D79: enter number of years:

tool: D34: 
Years

tool: D56: %6d%

tool: {"D82": [0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

tool: D29: %3d	

tool: D20: %7.2f


assistant: ```c
int main(void) {
  int start_rate, num_years;
  double value[5];
  printf("enter interest rate:\n");
  scanf("%d", &start_rate);
  printf("enter number of years:\n");
  scanf("%d", &num_years);
  printf("\nYears");
  for (int i = 0; i < ((int)(sizeof(value) / sizeof(value[0]))); i++) {
    printf("%6d%", ((int)(sizeof(value) / sizeof(value[0]))) + i);
    value[i] = 100;
  }
  printf("\n");
  for (int i = 1; i < num_years + 1; i++) {
    printf("%3d\t", i);
    for (int i = 0; i < ((int)(sizeof(value) / sizeof(value[0]))); i++) {
      value[i] = (1 + (start_rate + i) / 100) * value[i];
      printf("%7.2f\n", value[i]);
    }
    printf("\n");
  }
  return 0;
}
```