user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x2,%edi
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <log10@plt>
  imul   $0x3e8,%eax,%edi
  call   <ceil@plt>
  mov    %rsp,%rcx
  movslq %eax,%r13
  mov    %rax,%rbx
  shl    $0x2,%r13
  lea    0xf(%r13),%rdx
  mov    %rdx,%rax
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rcx
  and    $0xfffffffffffffff0,%rax
  cmp    %rcx,%rsp
  je     L106
L93:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L93
L106:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L116
L148:
  mov    %rsp,%rdi
  test   %ebx,%ebx
  jle    L91
  lea    -0x1(%rbx),%r14d
  xor    %esi,%esi
  lea    0x4(,%r14,4),%rdx
  mov    %r14,%r12
  call   <memset@plt>
  mov    $0x3e7,%r8d
  mov    %rax,%rdi
  movslq %r14d,%rax
  shl    $0x2,%r14
  movl   $0x2,(%rdi,%rax,4)
  lea    -0x4(%rdi,%r13,1),%rsi
  lea    (%rdi,%r13,1),%r9
  sub    %r14,%rsi
  nopl   0x0(%rax,%rax,1)
L53:
  mov    %r9,%rcx
  xor    %eax,%eax
  nopl   (%rax)
L44:
  mov    -0x4(%rcx),%edx
  sub    $0x4,%rcx
  lea    (%rax,%rdx,2),%edx
  movslq %edx,%rax
  mov    %edx,%r10d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%r10d
  sar    $0x22,%rax
  sub    %r10d,%eax
  lea    (%rax,%rax,4),%r10d
  add    %r10d,%r10d
  sub    %r10d,%edx
  mov    %edx,(%rcx)
  cmp    %rsi,%rcx
  jne    L44
  sub    $0x1,%r8d
  jne    L53
  test   %ebx,%ebx
  jle    L91
  cmp    $0x2,%r12d
  jbe    L38
  mov    %ebx,%edx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
L71:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rdx,%rax
  jne    L71
  movdqa %xmm0,%xmm1
  mov    %ebx,%eax
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%eax
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%esi
  test   $0x3,%bl
  je     L102
L25:
  movslq %eax,%rdx
  lea    0x1(%rax),%ecx
  lea    (%rdi,%rdx,4),%rdx
  add    (%rdx),%esi
  cmp    %ecx,%ebx
  jle    L102
  add    $0x2,%eax
  add    0x4(%rdx),%esi
  cmp    %eax,%ebx
  jle    L102
  add    0x8(%rdx),%esi
L102:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  lea    -0x20(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L91:
  xor    %esi,%esi
  jmp    L102
L116:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L148
L38:
  xor    %esi,%esi
  xor    %eax,%eax
  jmp    L25
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D4
data_type: string

tool: D4: %d


assistant: ```c
int main(void) {
  int i, j, ca = 0, n = (int)ceil(1000 * log10(2)), a[n];
  for (i = 0; i < n; i++)
    a[i] = 0;
  a[n - 1] = 2;
  for (i = 0; i < 1000 - 1; i++) {
    for (j = n - 1; j >= 0; j--) {
      ca = 2 * a[j] + ca;
      a[j] = ca % 10;
      ca /= 10;
    }
    ca = 0;
  }
  ca = 0;
  for (i = 0; i < n; i++)
    ca += a[i];
  printf("%d\n", ca);
  return 0;
}
```