user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<normalize>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rsi),%r13d
  lea    -0x1(%r13),%ebx
  test   %ebx,%ebx
  jle    L25
  mov    %rdi,%rax
  mov    $0x9,%r9d
  xor    %edi,%edi
  mov    $0xcccccccd,%r8d
  jmp    L125
  cs nopw 0x0(%rax,%rax,1)
L87:
  movslq %ecx,%rdx
  mov    %ecx,%r10d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r10d
  sar    $0x22,%rdx
  sub    %r10d,%edx
  add    %edx,%esi
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  mov    %esi,0x4(%rax)
  sub    %edx,%ecx
  mov    %ecx,(%rax)
L100:
  mov    (%r12),%r13d
  add    $0x1,%edi
  add    $0x4,%rax
  lea    -0x1(%r13),%ebx
  cmp    %edi,%ebx
  jle    L25
L125:
  mov    (%rax),%ecx
  mov    0x4(%rax),%esi
  test   %ecx,%ecx
  jns    L87
  mov    %r9d,%edx
  sub    %ecx,%edx
  imul   %r8,%rdx
  shr    $0x23,%rdx
  sub    %edx,%esi
  lea    (%rdx,%rdx,4),%edx
  lea    (%rcx,%rdx,2),%edx
  mov    %esi,0x4(%rax)
  mov    %edx,(%rax)
  jmp    L100
  xchg   %ax,%ax
L25:
  test   %ebx,%ebx
  js     L183
  movslq %ebx,%rbx
  jmp    L85
  nopl   (%rax)
L57:
  sub    $0x1,%rbx
  test   %ebx,%ebx
  js     L183
L85:
  mov    0x0(%rbp,%rbx,4),%eax
  mov    %ebx,%r13d
  test   %eax,%eax
  je     L57
  lea    0x1(%rbx),%r15d
  lea    0x0(,%r15,4),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %ebx,%ebx
  js     L98
  lea    0x4(%rbp),%rdx
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L3
  cmp    $0x2,%ebx
  jbe    L3
  mov    %r15d,%edx
  mov    %rbp,%rcx
  mov    %r14,%rax
  shr    $0x2,%edx
  sub    %r14,%rcx
  shl    $0x4,%rdx
  add    %r14,%rdx
  nopl   0x0(%rax,%rax,1)
L44:
  movdqu (%rax,%rcx,1),%xmm0
  add    $0x10,%rax
  movups %xmm0,-0x10(%rax)
  cmp    %rax,%rdx
  jne    L44
  mov    %r15d,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%r15b
  je     L98
  movslq %eax,%rcx
  mov    0x0(%rbp,%rcx,4),%esi
  lea    0x0(,%rcx,4),%rdx
  mov    %esi,(%r14,%rcx,4)
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r13d
  jl     L98
  mov    0x4(%rbp,%rdx,1),%ecx
  add    $0x2,%eax
  mov    %ecx,0x4(%r14,%rdx,1)
  cmp    %r13d,%eax
  jg     L98
  mov    0x8(%rbp,%rdx,1),%eax
  mov    %eax,0x8(%r14,%rdx,1)
L98:
  mov    %r15d,(%r12)
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L3:
  mov    %ebx,%ebx
  mov    %rbp,%rcx
  mov    %r14,%rax
  lea    0x4(%r14,%rbx,4),%rsi
  sub    %r14,%rcx
L29:
  mov    (%rcx,%rax,1),%edx
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  cmp    %rax,%rsi
  jne    L29
  jmp    L98
  nopl   0x0(%rax)
L183:
  lea    0x0(,%r13,4),%edi
  mov    %r13d,%r15d
  call   <malloc@plt>
  mov    %rax,%r14
  jmp    L98
```

assistant: ```c
int *normalize(int *a, int *sizeA) {
  int i = 0, j = 0;
  for (i = 0; i < (*sizeA) - 1; i++) {
    if (a[i] < 0) {
      int shift = (abs(a[i]) + 9) / 10;
      a[i + 1] -= shift;
      a[i] += shift * 10;
    } else {
      a[i + 1] += a[i] / 10;
      a[i] %= 10;
    }
  }
  for (i = (*sizeA) - 1; i >= 0; i--) {
    if (a[i] != 0)
      break;
  }
  int *result = (int *)malloc(sizeof(int) * (i + 1));
  for (j = 0; j <= i; j++) {
    result[j] = a[j];
  }
  *sizeA = i + 1;
  free(a);
  return result;
}
```