user: 下面的汇编指令对应的C源代码是什么:

```asm
<main1>:
  endbr64
  push   %r12
  mov    $0x1,%ecx
  mov    $0x64,%edx
  push   %rbp
  push   %rbx
  sub    $0xe0,%rsp
  movss  D10(%rip),%xmm0
  movss  D19(%rip),%xmm1
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  lea    -0x1(%rsp),%r8
L83:
  divss  %xmm1,%xmm0
  movslq %edx,%rax
  mov    %edx,%esi
  movslq %ecx,%rdi
  imul   $0x55555556,%rax,%rax
  sar    $0x1f,%esi
  shr    $0x20,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,2),%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,(%r8,%rcx,1)
  add    $0x1,%rcx
  mov    %eax,%edx
  cmp    $0x6,%rcx
  jne    L83
  pxor   %xmm2,%xmm2
  movb   $0x0,(%rsp,%rdi,1)
  lea    0x70(%rsp),%rbp
  comiss %xmm2,%xmm0
  jbe    L49
  movss  D36(%rip),%xmm1
  lea    0x70(%rsp),%rbp
  xor    %esi,%esi
  xor    %r8d,%r8d
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %rbp,%rcx
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  add    $0x30,%edx
  mov    %dl,0x70(%rsp)
  jmp    L34
  nopl   (%rax)
L24:
  xor    %r8d,%r8d
L52:
  add    $0x1,%rcx
  movslq %edi,%rsi
L34:
  divss  %xmm1,%xmm0
  lea    0x1(%rsi),%edi
  comiss %xmm2,%xmm0
  jbe    L49
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%eax
  movslq %eax,%rdx
  mov    %eax,%r9d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r9d
  sar    $0x22,%rdx
  sub    %r9d,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  add    $0x30,%eax
  mov    %al,0x1(%rcx)
  cmp    (%rcx),%al
  jne    L24
  add    $0x1,%r8d
  cmp    $0x6,%r8d
  jne    L52
  lea    -0x1(%rsi),%eax
  cltq
  movb   $0x29,0x70(%rsp,%rax,1)
  lea    -0x3(%rsi),%eax
  cltq
  movb   $0x28,0x70(%rsp,%rax,1)
  movb   $0x0,0x70(%rsp,%rsi,1)
  nopw   0x0(%rax,%rax,1)
L49:
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  js     L12
  movslq %eax,%rbx
  nopl   0x0(%rax,%rax,1)
L0:
  movsbl (%r12,%rbx,1),%edi
  sub    $0x1,%rbx
  call   <putchar@plt>
  test   %ebx,%ebx
  jns    L0
L12:
  mov    $0x2e,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  add    $0xe0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D10
type: f32
name: disassemble_data
arguments: data_tag: D19
type: byte[4]
name: disassemble_data
arguments: data_tag: D36
type: byte[4]
name: disassemble_data
arguments: data_tag: D17
type: string

tool: D10: 100

tool: {"D19": [0x00,0x00,0x40,0x40]}

tool: {"D36": [0x00,0x00,0x20,0x41]}

tool: D17: %s

assistant: ```c
void main1() {
  float a = 100;
  int b = 3;
  int c = a;
  char str1[100];
  char str2[100];
  int i = 0;
  while (c > 0) {
    str1[i++] = '0' + (c % b);
    c = c / b;
    a = a / b;
  }
  str1[i] = '\0';
  i = 0;
  int count = 0;
  while (a > 0) {
    a = a * 10;
    c = a;
    str2[i] = '0' + (c % 10);
    if (i > 0 && str2[i - 1] == str2[i]) {
      count++;
    } else {
      count = 0;
    }
    if (count > 5) {
      str2[i - 2] = ')';
      str2[i - 4] = '(';
      str2[i - 1] = '\0';
      break;
    }
    a = a / 10;
    i++;
  }
  int len = strlen(str1);
  for (i = len - 1; i >= 0; i--) {
    printf("%c", str1[i]);
  }
  printf(".");
  printf("%s", str2);
}
```