user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  lea    D63(%rip),%rdi
  lea    D68(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  movsd  D54(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movsd  %xmm0,0x10(%rsp)
  movsd  %xmm0,0x18(%rsp)
  movsd  %xmm0,0x20(%rsp)
  movsd  %xmm0,0x28(%rsp)
  movsd  %xmm0,0x30(%rsp)
  call   <puts@plt>
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D67(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x20(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D85(%rip),%rdi
  xor    %eax,%eax
  cvttsd2si 0x20(%rsp),%ebp
  call   <printf@plt>
  lea    0x28(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x30(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  movsd  0x30(%rsp),%xmm3
  mov    %ebp,%esi
  movsd  0x18(%rsp),%xmm2
  movsd  0x28(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm0
  mov    $0x4,%eax
  lea    D75(%rip),%rdi
  call   <printf@plt>
  test   %ebp,%ebp
  js     L4
  xor    %ebx,%ebx
  pxor   %xmm1,%xmm1
  nopl   0x0(%rax,%rax,1)
L25:
  mov    %ebx,%esi
  mov    %ebp,%edi
  movsd  %xmm1,0x8(%rsp)
  call   <combinaciones@plt>
  mov    %ebp,%edi
  movsd  %xmm0,(%rsp)
  movsd  0x10(%rsp),%xmm0
  sub    %ebx,%edi
  mulsd  0x28(%rsp),%xmm0
  call   <potencia@plt>
  movsd  (%rsp),%xmm4
  mov    %ebx,%edi
  add    $0x1,%ebx
  mulsd  %xmm0,%xmm4
  movsd  0x18(%rsp),%xmm0
  mulsd  0x30(%rsp),%xmm0
  movsd  %xmm4,(%rsp)
  call   <potencia@plt>
  mulsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm0,%xmm1
  cmp    %ebx,%ebp
  jge    L25
L38:
  movapd %xmm1,%xmm0
  lea    D70(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  pxor   %xmm1,%xmm1
  jmp    L38
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D63
data_type: string
name: extract_symbolic_data
arguments: tag: D68
data_type: string
name: extract_symbolic_data
arguments: tag: D54
data_type: f64
name: extract_symbolic_data
arguments: tag: D15
data_type: string
name: extract_symbolic_data
arguments: tag: D0
data_type: string
name: extract_symbolic_data
arguments: tag: D67
data_type: string
name: extract_symbolic_data
arguments: tag: D85
data_type: string
name: extract_symbolic_data
arguments: tag: D9
data_type: string
name: extract_symbolic_data
arguments: tag: D75
data_type: string
name: extract_symbolic_data
arguments: tag: D70
data_type: string

tool: D63: Binomio (ax + by)^n:

tool: D68: %lf

tool: D54: 1

tool: D15: a = 

tool: D0: b = 

tool: D67: n = 

tool: D85: x = 

tool: D9: y = 

tool: D75: Calcular (%g*%g + %g*%g)^%d:


tool: D70: Resultado: %g.


assistant: ```c
main() {
  double a = 1, b = 1, m = 1, x = 1, y = 1, result = 0;
  int n = 1, k = 0;
  printf("Binomio (ax + by)^n:\n");
  printf("a = ");
  scanf("%lf", &a);
  printf("b = ");
  scanf("%lf", &b);
  printf("n = ");
  scanf("%lf", &m);
  n = (int)m;
  printf("x = ");
  scanf("%lf", &x);
  printf("y = ");
  scanf("%lf", &y);
  printf("Calcular (%g*%g + %g*%g)^%d:\n", a, x, b, y, n);
  for (k = 0, result = 0; k <= n; k++)
    result += combinaciones(n, k) * potencia(a * x, n - k) * potencia(b * y, k);
  printf("Resultado: %g.\n", result);
}
```