user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x84(%rbp)
  movl   $0x0,-0x80(%rbp)
  movl   $0x0,-0x7c(%rbp)
  movl   $0x0,-0x78(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x74(%rbp)
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <getline_my@plt>
  mov    %eax,-0x80(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x7c(%rbp)
  mov    -0x7c(%rbp),%eax
  mov    %eax,%esi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <clear_line@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <getline_my@plt>
  mov    %eax,-0x80(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x84(%rbp)
  mov    -0x84(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <clear_line@plt>
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <getline_my@plt>
  mov    %eax,-0x80(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x78(%rbp)
  mov    -0x78(%rbp),%eax
  mov    %eax,%esi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <clear_line@plt>
  mov    -0x7c(%rbp),%eax
  imul   -0x84(%rbp),%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
  imul   -0x78(%rbp),%eax
  imul   $0x1e,%eax,%eax
  movslq %eax,%rdx
  imul   $0xffffffffb38cf9b1,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x8,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x74(%rbp)
  pxor   %xmm1,%xmm1
  cvtss2sd -0x74(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
datatype: string
name: extract_symbolic_data
arguments: data_tag: D62
datatype: string
name: extract_symbolic_data
arguments: data_tag: D7
datatype: string
name: extract_symbolic_data
arguments: data_tag: D25
datatype: string
name: extract_symbolic_data
arguments: data_tag: D65
datatype: string
name: extract_symbolic_data
arguments: data_tag: D37
datatype: string
name: extract_symbolic_data
arguments: data_tag: D68
datatype: string

tool: D5: Input please amount of money of deposit:

tool: D62: Deposit: %d;


tool: D7: Input please amount of procents of deposit:

tool: D25: Amount of procents: %d;


tool: D65: Input please duration of deposit (in monthes):

tool: D37: Duration: %d;


tool: D68: in the end we have: %f.
 

assistant: ```c
int main() {
  int proc = 0;
  int len = 0;
  int dep = 0;
  int dur = 0;
  float sum = 0.0;
  char line[100];
  printf("Input please amount of money of deposit:");
  len = getline_my(line);
  dep = atoi(line);
  printf("Deposit: %d;\n", dep);
  clear_line(line);
  printf("Input please amount of procents of deposit:");
  len = getline_my(line);
  proc = atoi(line);
  printf("Amount of procents: %d;\n", proc);
  clear_line(line);
  printf("Input please duration of deposit (in monthes):");
  len = getline_my(line);
  dur = atoi(line);
  printf("Duration: %d;\n", dur);
  clear_line(line);
  sum = (dep * proc) / 100 * 30 * dur / 365;
  printf("in the end we have: %f.\n ", sum);
}
```