user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x58,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%r13
  lea    0x60(%rax),%rbp
  mov    %rax,%rbx
  nopw   0x0(%rax,%rax,1)
L20:
  mov    $0x60,%edi
  add    $0x8,%rbx
  call   <malloc@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %rbp,%rbx
  jne    L20
  mov    D49(%rip),%rax
  movsd  D2(%rip),%xmm1
  mov    $0xc,%ecx
  mov    %r12,%rdi
  mov    $0xb,%edx
  mov    $0xa,%esi
  lea    D57(%rip),%r14
  mov    $0xc9,%ebx
  movq   %rax,%xmm0
  mov    $0x2,%eax
  lea    D67(%rip),%rbp
  call   <iniMatriz@plt>
  mov    %r14,%rdi
  call   <puts@plt>
  mov    D49(%rip),%rax
  movq   %rax,%xmm0
  xchg   %ax,%ax
L2:
  movapd %xmm0,%xmm1
  mov    %rbp,%rdi
  andpd  D50(%rip),%xmm1
  mov    $0x2,%eax
  mulsd  D54(%rip),%xmm1
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  addsd  D0(%rip),%xmm0
  sub    $0x1,%ebx
  jne    L2
  mov    %r14,%rdi
  mov    $0xc9,%ebx
  call   <puts@plt>
  mov    $0xc,%edx
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    $0xb,%esi
  call   <f1@plt>
  mov    D49(%rip),%rax
  movq   %rax,%xmm0
  nopw   0x0(%rax,%rax,1)
L19:
  mov    $0xc,%edx
  mov    $0xb,%esi
  mov    %r12,%rdi
  mov    $0x1,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <Px@plt>
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  cvtsi2sd %eax,%xmm1
  mov    $0x2,%eax
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  addsd  D0(%rip),%xmm0
  sub    $0x1,%ebx
  jne    L19
  lea    0x58(%r12),%rbx
L1:
  mov    0x0(%r13),%rdi
  add    $0x8,%r13
  call   <free@plt>
  cmp    %rbx,%r13
  jne    L1
  mov    %r12,%rdi
  call   <free@plt>
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: label: D49
data_type: byte[8]
name: retrieve_data
arguments: label: D2
data_type: f64
name: retrieve_data
arguments: label: D57
data_type: string
name: retrieve_data
arguments: label: D67
data_type: string
name: retrieve_data
arguments: label: D50
data_type: qword[2]
name: retrieve_data
arguments: label: D54
data_type: byte[8]
name: retrieve_data
arguments: label: D0
data_type: f64

tool: {"D49": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: D2: 1

tool: D57: 


tool: D67: %lf %lf 


tool: {"D50": [0x7fffffffffffffff,0x0000000000000000]}

tool: {"D54": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40]}

tool: D0: 0.01

assistant: ```c
int main() {
  double **matriz;
  double x = 0.91;
  double a = -1;
  double b = 1;
  double sol;
  int npartes;
  int FIL;
  int COL;
  int j;
  double i;
  npartes = 10;
  FIL = npartes + 1;
  COL = npartes + 2;
  matriz = (double **)malloc(FIL * sizeof(double *));
  for (j = 0; j < COL; j++)
    matriz[j] = (double *)malloc(COL * sizeof(double));
  iniMatriz(matriz, a, b, npartes, FIL, COL);
  printf("\n\n");
  for (i = a; i < b + 0.01; i += 0.01)
    printf("%lf %lf \n", i, (10 * fabs(i)));
  printf("\n\n");
  f1(matriz, FIL, COL);
  for (i = a; i < b + 0.01; i += 0.01) {
    sol = Px(matriz, i, FIL, COL);
    printf("%lf %lf \n", i, sol);
  }
  for (j = 0; j < FIL; j++)
    free(matriz[j]);
  free(matriz);
  return 0;
}
```