user: 请从给定的汇编指令推导出C源代码：

```asm
<crypto_onetimeauth>:
  endbr64
  push   %r15
  pxor   %xmm0,%xmm0
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x108,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf8(%rsp)
  xor    %eax,%eax
  movzbl (%rcx),%eax
  lea    0x60(%rsp),%r15
  lea    0xb0(%rsp),%rbp
  mov    %eax,0x10(%rsp)
  movzbl 0x1(%rcx),%eax
  mov    %eax,0x14(%rsp)
  movzbl 0x2(%rcx),%eax
  mov    %eax,0x18(%rsp)
  movzbl 0x3(%rcx),%eax
  and    $0xf,%eax
  mov    %eax,0x1c(%rsp)
  movzbl 0x4(%rcx),%eax
  and    $0xfc,%eax
  mov    %eax,0x20(%rsp)
  movzbl 0x5(%rcx),%eax
  mov    %eax,0x24(%rsp)
  movzbl 0x6(%rcx),%eax
  mov    %eax,0x28(%rsp)
  movzbl 0x7(%rcx),%eax
  and    $0xf,%eax
  mov    %eax,0x2c(%rsp)
  movzbl 0x8(%rcx),%eax
  and    $0xfc,%eax
  mov    %eax,0x30(%rsp)
  movzbl 0x9(%rcx),%eax
  mov    %eax,0x34(%rsp)
  movzbl 0xa(%rcx),%eax
  mov    %eax,0x38(%rsp)
  movzbl 0xb(%rcx),%eax
  and    $0xf,%eax
  mov    %eax,0x3c(%rsp)
  movzbl 0xc(%rcx),%eax
  movl   $0x0,0x50(%rsp)
  and    $0xfc,%eax
  movaps %xmm0,0x60(%rsp)
  mov    %eax,0x40(%rsp)
  movzbl 0xd(%rcx),%eax
  movl   $0x0,0xa0(%rsp)
  mov    %eax,0x44(%rsp)
  movzbl 0xe(%rcx),%eax
  movaps %xmm0,0x70(%rsp)
  mov    %eax,0x48(%rsp)
  movzbl 0xf(%rcx),%eax
  movaps %xmm0,0x80(%rsp)
  and    $0xf,%eax
  movaps %xmm0,0x90(%rsp)
  mov    %eax,0x4c(%rsp)
  lea    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
  test   %rdx,%rdx
  je     L27
  nopl   0x0(%rax,%rax,1)
L30:
  movl   $0x0,0x40(%rbp)
  pxor   %xmm0,%xmm0
  xor    %eax,%eax
  movaps %xmm0,0x0(%rbp)
  movaps %xmm0,0x10(%rbp)
  movaps %xmm0,0x20(%rbp)
  movaps %xmm0,0x30(%rbp)
  nopl   (%rax)
L31:
  mov    %eax,%edx
  cmp    %rax,%rbx
  je     L7
  movzbl (%r14,%rax,1),%edx
  mov    %edx,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    $0x10,%rax
  jne    L31
  mov    %rbp,%rsi
  mov    %r15,%rdi
  sub    $0x10,%rbx
  add    $0x10,%r14
  movl   $0x1,0xf0(%rsp)
  call   <add@plt>
  mov    0x8(%rsp),%rsi
  mov    %r15,%rdi
  call   <mulmod@plt>
  test   %rbx,%rbx
  jne    L30
L27:
  mov    %r15,%rdi
  call   <freeze@plt>
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L2:
  movzbl 0x10(%r13,%rax,1),%edx
  mov    %edx,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    $0x10,%rax
  jne    L2
  mov    %rbp,%rsi
  mov    %r15,%rdi
  movl   $0x0,0xf0(%rsp)
  call   <add@plt>
  xor    %eax,%eax
  nopl   0x0(%rax)
L79:
  mov    (%r15,%rax,4),%edx
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%rax
  cmp    $0x10,%rax
  jne    L79
  mov    0xf8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0x108,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L7:
  mov    %rbp,%rsi
  mov    %r15,%rdi
  movl   $0x1,0xb0(%rsp,%rdx,4)
  call   <add@plt>
  mov    0x8(%rsp),%rsi
  mov    %r15,%rdi
  call   <mulmod@plt>
  jmp    L27
L64:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int crypto_onetimeauth(unsigned char *out, const unsigned char *in, unsigned long long inlen, const unsigned char *k) {
  unsigned int j;
  unsigned int r[17];
  unsigned int h[17];
  unsigned int c[17];
  r[0] = k[0];
  r[1] = k[1];
  r[2] = k[2];
  r[3] = k[3] & 15;
  r[4] = k[4] & 252;
  r[5] = k[5];
  r[6] = k[6];
  r[7] = k[7] & 15;
  r[8] = k[8] & 252;
  r[9] = k[9];
  r[10] = k[10];
  r[11] = k[11] & 15;
  r[12] = k[12] & 252;
  r[13] = k[13];
  r[14] = k[14];
  r[15] = k[15] & 15;
  r[16] = 0;
  for (j = 0; j < 17; ++j)
    h[j] = 0;
  while (inlen > 0) {
    for (j = 0; j < 17; ++j)
      c[j] = 0;
    for (j = 0; (j < 16) && (j < inlen); ++j)
      c[j] = in[j];
    c[j] = 1;
    in += j;
    inlen -= j;
    add(h, c);
    mulmod(h, r);
  }
  freeze(h);
  for (j = 0; j < 16; ++j)
    c[j] = k[j + 16];
  c[16] = 0;
  add(h, c);
  for (j = 0; j < 16; ++j)
    out[j] = h[j];
  return 0;
}
```