user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mergeAndCount>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %ecx,%eax
  sub    %edx,%eax
  movl   $0x0,0x4(%rsp)
  cmp    $0x1,%eax
  jle    L43
  mov    %rdi,%r12
  mov    %rsi,%r15
  mov    %edx,%edi
  mov    %ecx,%ebp
  mov    %r8d,%r14d
  mov    %r9d,%r13d
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,0xc(%rsp)
  mov    %eax,%ebx
  add    %edi,%ebx
  mov    %ebx,%ecx
  mov    %edi,0x8(%rsp)
  mov    %edi,%edx
  mov    %r12,%rdi
  call   <mergeAndCount>
  mov    %eax,0x4(%rsp)
  mov    %r13d,%r9d
  mov    %r14d,%r8d
  mov    %ebp,%ecx
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <mergeAndCount>
  mov    %eax,%ecx
  add    0x4(%rsp),%ecx
  mov    %ecx,0x4(%rsp)
  mov    0x8(%rsp),%edi
  cmp    %ebx,%edi
  jge    L133
  movslq %edi,%rdx
  lea    (%r12,%rdx,8),%r8
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%eax
  add    %rdx,%rax
  lea    0x8(%r12,%rax,8),%r11
  mov    %ebx,%edi
  mov    %ebx,%ecx
  mov    $0x0,%edx
  movslq %r14d,%r14
  movslq %r13d,%r13
  jmp    L69
L87:
  cmp    %edi,%ebp
  jle    L101
  movslq %edi,%rax
  lea    0x0(%rbp,%rdx,1),%r9d
  sub    %edi,%r9d
  movslq %edx,%rsi
  sub    %rax,%rsi
  lea    (%r15,%rsi,8),%r10
L34:
  mov    %eax,%edi
  mov    (%r12,%rax,8),%rsi
  cmp    (%r8),%rsi
  jge    L101
  lea    0x1(%rax),%edi
  add    $0x1,%edx
  mov    %rsi,(%r10,%rax,8)
  add    $0x1,%rax
  cmp    %r9d,%edx
  jne    L34
  mov    %r9d,%edx
L101:
  mov    %ecx,%eax
  sub    %ebx,%eax
  add    %eax,0x4(%rsp)
  mov    (%r8),%rsi
  movslq %edx,%rax
  mov    %rsi,(%r15,%rax,8)
  add    $0x1,%edx
  add    $0x8,%r8
  cmp    %r11,%r8
  je     L59
L69:
  cmp    %ebx,%ebp
  jle    L55
  mov    (%r8),%r9
  movslq %ebx,%rax
  lea    (%r12,%rax,8),%rsi
L91:
  mov    (%rsi),%rax
  sub    %r9,%rax
  cmp    %r14,%rax
  jge    L55
  add    $0x1,%ebx
  add    $0x8,%rsi
  cmp    %ebx,%ebp
  jne    L91
  mov    %ebp,%ebx
L55:
  cmp    %ecx,%ebp
  jle    L87
  mov    (%r8),%r9
  movslq %ecx,%rax
  lea    (%r12,%rax,8),%rsi
L75:
  mov    (%rsi),%rax
  sub    %r9,%rax
  cmp    %r13,%rax
  jg     L87
  add    $0x1,%ecx
  add    $0x8,%rsi
  cmp    %ecx,%ebp
  jne    L75
  mov    %ebp,%ecx
  jmp    L87
L133:
  mov    %ebx,%edi
L59:
  mov    0x8(%rsp),%eax
  sub    %eax,%edi
  lea    0x0(,%rdi,8),%edx
  cltq
  lea    (%r12,%rax,8),%rdi
  mov    %r15,%rsi
  call   <memcpy@plt>
L43:
  mov    0x4(%rsp),%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int mergeAndCount(long *sums, long *buff, int start, int end, int lower, int upper) {
  if (end - start <= 1)
    return 0;
  int mid = start + (end - start) / 2;
  int cnt = mergeAndCount(sums, buff, start, mid, lower, upper) + mergeAndCount(sums, buff, mid, end, lower, upper);
  int lowerIdx = mid, upperIdx = mid, tempIdx = mid;
  for (int i = start, bi = 0; i < mid; i++, bi++) {
    while (lowerIdx < end && sums[lowerIdx] - sums[i] < lower)
      lowerIdx++;
    while (upperIdx < end && sums[upperIdx] - sums[i] <= upper)
      upperIdx++;
    while (tempIdx < end && sums[tempIdx] < sums[i])
      buff[bi++] = sums[tempIdx++];
    cnt += upperIdx - lowerIdx;
    buff[bi] = sums[i];
  }
  memcpy(sums + start, buff, (tempIdx - start) * sizeof(long));
  return cnt;
}
```