user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x64(%rbp)
  jmp    L7
L20:
  mov    -0x64(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x40(%rbp),%rcx
  mov    -0x64(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  lea    D67(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x40(%rbp),%rcx
  mov    -0x64(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x60(%rbp),%rdx
  mov    -0x64(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x64(%rbp)
L7:
  cmpl   $0x4,-0x64(%rbp)
  jle    L20
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x64(%rbp)
  jmp    L32
L39:
  mov    -0x64(%rbp),%eax
  cltq
  movss  -0x60(%rbp,%rax,4),%xmm0
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm0,%xmm1
  movsd  D2(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x64(%rbp),%eax
  cltq
  movss  -0x60(%rbp,%rax,4),%xmm1
  pxor   %xmm2,%xmm2
  cvtss2sd %xmm1,%xmm2
  movq   %xmm2,%rdx
  lea    -0x40(%rbp),%rsi
  mov    -0x64(%rbp),%eax
  movslq %eax,%rcx
  mov    %rcx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  add    %rax,%rax
  add    %rsi,%rax
  movapd %xmm0,%xmm1
  movq   %rdx,%xmm0
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  addl   $0x1,-0x64(%rbp)
L32:
  cmpl   $0x4,-0x64(%rbp)
  jle    L39
  lea    D86(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D33
data_type: string
name: extract_symbolic_data
arguments: tag: D67
data_type: string
name: extract_symbolic_data
arguments: tag: D22
data_type: string
name: extract_symbolic_data
arguments: tag: D7
data_type: string
name: extract_symbolic_data
arguments: tag: D58
data_type: string
name: extract_symbolic_data
arguments: tag: D35
data_type: string
name: extract_symbolic_data
arguments: tag: D2
data_type: f64
name: extract_symbolic_data
arguments: tag: D41
data_type: string
name: extract_symbolic_data
arguments: tag: D86
data_type: string

tool: D33: 
Entre com o nome do funcionario %d: 

tool: D67: %s

tool: D22: 
Entre com o salario do funcionario %s: 

tool: D7: %f

tool: D58: 
Funcionario		Sal.Bruto	Sal.Liq

tool: D35: 
------------------------------------------

tool: D2: 0.8

tool: D41: 
%s 			%.2f 		%.2f

tool: D86: 
-------------------------------------------

assistant: ```c
int main() {
  char nome[5][10];
  float salario[5];
  int i;
  for (i = 0; i <= 4; i++) {
    printf("\nEntre com o nome do funcionario %d: ", i + 1);
    scanf("%s", nome[i]);
    printf("\nEntre com o salario do funcionario %s: ", nome[i]);
    scanf("%f", &salario[i]);
  }
  printf("\nFuncionario\t\tSal.Bruto\tSal.Liq");
  printf("\n------------------------------------------");
  for (i = 0; i < 5; i++) {
    printf("\n%s \t\t\t%.2f \t\t%.2f", nome[i], salario[i], salario[i] * 0.8);
  }
  printf("\n-------------------------------------------");
  return 0;
}
```