user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  lea    D128(%rip),%rdi
  push   %rbp
  push   %rbx
  call   <puts@plt>
  movss  D47(%rip),%xmm5
  xor    %esi,%esi
  movss  D95(%rip),%xmm2
  movsd  D39(%rip),%xmm1
  movss  D81(%rip),%xmm3
  pxor   %xmm0,%xmm0
  movss  D119(%rip),%xmm4
L147:
  addss  %xmm5,%xmm0
  lea    0x1(%rsi),%edx
  mov    $0x2,%eax
  subss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
L7:
  comiss %xmm0,%xmm3
  jbe    L91
  lea    0x1(%rdx),%esi
  cmp    $0x5,%eax
  jne    L11
  subss  %xmm4,%xmm0
  subss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  comiss %xmm0,%xmm3
  jbe    L15
  subss  %xmm2,%xmm0
  lea    0x2(%rdx),%esi
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  comiss %xmm0,%xmm3
  jbe    L15
  subss  %xmm2,%xmm0
  lea    0x3(%rdx),%esi
  mov    $0x8,%eax
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
L24:
  mov    %esi,%edx
  jmp    L7
  nopl   0x0(%rax)
L11:
  subss  %xmm2,%xmm0
  add    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  cmp    $0x1f,%eax
  jne    L24
  comiss %xmm0,%xmm3
  ja     L147
L15:
  mov    %esi,%edx
  mov    %esi,%ecx
  lea    D44(%rip),%rdi
  mov    $0x14dc,%ebx
  imul   $0x6719f361,%rdx,%rax
  lea    D51(%rip),%rbp
  shr    $0x20,%rax
  sub    %eax,%ecx
  shr    %ecx
  add    %eax,%ecx
  mov    $0x88888889,%eax
  imul   %rax,%rdx
  shr    $0x8,%ecx
  xor    %eax,%eax
  shr    $0x24,%rdx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x1c41,%edx
  mov    $0x14dc,%esi
  xor    %eax,%eax
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    $0x1c41,%eax
  jmp    L90
  nopw   0x0(%rax,%rax,1)
L94:
  sub    %eax,%ebx
  cmp    %eax,%ebx
  je     L115
L90:
  cmp    %eax,%ebx
  jg     L94
  sub    %ebx,%eax
  cmp    %eax,%ebx
  jne    L90
L115:
  lea    D99(%rip),%r12
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  cmp    $0x1,%ebx
  je     L96
L54:
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x7,%esi
  lea    D107(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x10e5,%esi
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%edi
  xor    %esi,%esi
  mov    $0x10e5,%edx
L105:
  movslq %edx,%rax
  mov    %edx,%ecx
  lea    (%rsi,%rsi,4),%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rsi,2),%esi
  mov    %eax,%edx
  sub    $0x1,%edi
  jne    L105
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L91:
  mov    %edx,%esi
  jmp    L15
L96:
  lea    D121(%rip),%rdi
  call   <puts@plt>
  jmp    L54
```

tool_calls: name: extract_symbolic_data
arguments: tag: D128
data_type: string
name: extract_symbolic_data
arguments: tag: D47
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D95
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D39
data_type: f64
name: extract_symbolic_data
arguments: tag: D81
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D119
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D44
data_type: string
name: extract_symbolic_data
arguments: tag: D51
data_type: string
name: extract_symbolic_data
arguments: tag: D27
data_type: string
name: extract_symbolic_data
arguments: tag: D99
data_type: string
name: extract_symbolic_data
arguments: tag: D107
data_type: string
name: extract_symbolic_data
arguments: tag: D58
data_type: string
name: extract_symbolic_data
arguments: tag: D121
data_type: string

tool: D128: Problema do milionario:

tool: {"D47": [0x00,0x00,0xfa,0x43]}

tool: {"D95": [0x00,0x00,0x20,0x41]}

tool: D39: 1.001

tool: {"D81": [0x00,0x24,0x74,0x49]}

tool: {"D119": [0x00,0x00,0x16,0x43]}

tool: D44: Em %d dias (%d meses, %d anos) ficarei milionario!


tool: D51: ----

tool: D27: MDC(%d, %d)=

tool: D99: %d


tool: D107: %d eh primo!


tool: D58: Vou inverter %d: 

tool: D121: Primos entre si!

assistant: ```c
int main() {
  printf("Problema do milionario:\n");
  int contar_dias = 0;
  int dia = 1;
  float conta = 0;
  while (conta < 1000000) {
    if (dia == 1) {
      conta = conta + 500;
    }
    if (dia == 5) {
      conta = conta - 150;
    }
    conta = conta - 10;
    conta = conta * 1.001;
    contar_dias = contar_dias + 1;
    dia = dia + 1;
    if (dia == 31) {
      dia = 1;
    }
  }
  printf("Em %d dias (%d meses, %d anos) ficarei milionario!\n", contar_dias, contar_dias / 30, contar_dias / 365);
  printf("----\n");
  int A, B;
  A = 5340;
  B = 7233;
  printf("MDC(%d, %d)=", A, B);
  while (A != B) {
    if (A > B)
      A = A - B;
    else
      B = B - A;
  }
  printf("%d\n", A);
  if (A == 1)
    printf("Primos entre si!\n");
  printf("----\n");
  A = 7;
  B = 2;
  while (((A % B) != 0) && (B < A)) {
    B = B + 1;
  }
  if (B == A) {
    printf("%d eh primo!\n", A);
  } else {
    printf("%d eh divisor de %d\n", B, A);
  }
  printf("----\n");
  A = 4325;
  B = 0;
  printf("Vou inverter %d: ", A);
  while (A > 0) {
    B = (B * 10) + (A % 10);
    A = A / 10;
  }
  printf("%d\n", B);
  printf("----\n");
  return 0;
}
```