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
  push   %rbx
  sub    $0x78,%rsp
  mov    %edi,-0x64(%rbp)
  mov    %rsi,-0x70(%rbp)
  movl   $0x64,-0x54(%rbp)
  movsd  D13(%rip),%xmm0
  movsd  %xmm0,-0x40(%rbp)
  movsd  D32(%rip),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  D12(%rip),%xmm0
  movsd  %xmm0,-0x48(%rbp)
  movsd  -0x38(%rbp),%xmm0
  subsd  -0x40(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x54(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  -0x40(%rbp),%xmm0
  movsd  %xmm0,-0x50(%rbp)
  movl   $0x1,-0x58(%rbp)
  jmp    L13
L1:
  mov    -0x48(%rbp),%rax
  movq   %rax,%xmm0
  call   <sin@plt>
  movsd  %xmm0,-0x78(%rbp)
  mov    -0x50(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  addsd  -0x78(%rbp),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  mov    -0x48(%rbp),%rax
  movq   %rax,%xmm0
  call   <sin@plt>
  movq   %xmm0,%rbx
  mov    -0x50(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  movq   %rbx,%xmm1
  subsd  %xmm0,%xmm1
  movsd  -0x28(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x48(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm3
  mulsd  -0x20(%rbp),%xmm3
  movsd  %xmm3,-0x78(%rbp)
  movsd  -0x28(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  %xmm0,-0x80(%rbp)
  mov    -0x48(%rbp),%rax
  movq   %rax,%xmm0
  call   <sin@plt>
  mulsd  -0x80(%rbp),%xmm0
  movsd  -0x78(%rbp),%xmm3
  subsd  %xmm0,%xmm3
  movsd  %xmm3,-0x78(%rbp)
  mov    -0x50(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  movapd %xmm0,%xmm1
  movsd  -0x78(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x30(%rbp),%xmm1
  pxor   %xmm0,%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x30(%rbp),%xmm2
  pxor   %xmm0,%xmm0
  mulsd  %xmm2,%xmm0
  mulsd  -0x18(%rbp),%xmm0
  addsd  -0x20(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  addsd  -0x28(%rbp),%xmm0
  mulsd  -0x30(%rbp),%xmm0
  movsd  -0x48(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x48(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x58(%rbp),%xmm0
  mulsd  -0x30(%rbp),%xmm0
  movsd  -0x40(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x50(%rbp)
  movsd  -0x48(%rbp),%xmm0
  mov    -0x50(%rbp),%rdx
  mov    -0x58(%rbp),%eax
  movapd %xmm0,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x58(%rbp)
L13:
  mov    -0x58(%rbp),%eax
  cmp    -0x54(%rbp),%eax
  jle    L1
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
datatype: f64
name: extract_symbolic_data
arguments: tag: D32
datatype: f64
name: extract_symbolic_data
arguments: tag: D12
datatype: f64
name: extract_symbolic_data
arguments: tag: D24
datatype: string

tool: D13: 2

tool: D32: 5

tool: D12: 0.32

tool: D24: no. of step size %d the value of t is %f solution of Taylor series is %f


assistant: ```c
int main(int argc, const char *argv[]) {
  int k;
  double h, t, x1, x2, x3;
  int n = 100;
  double a = 2, b = 5, x = 0.32;
  h = (b - a) / n;
  t = a;
  for (k = 1; k <= n; k++) {
    x1 = sin(x) + cos(t);
    x2 = x1 * (sin(x) - cos(t));
    x3 = x2 * cos(x) - x1 * x1 * sin(x) - cos(t);
    x = x + h * (x1 + (1 / 2) * h * (x2 + (1 / 3) * h * (x3)));
    t = a + k * h;
    printf("no. of step size %d the value of t is %f solution of Taylor series is %f\n", k, t, x);
  }
  return 0;
}
```