user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x108,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf8(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L198
  mov    $0x0,%ebp
  lea    0x50(%rsp),%rbx
  lea    D1(%rip),%r14
  lea    D32(%rip),%r13
  lea    D0(%rip),%r12
  jmp    L63
L11:
  jle    L55
  lea    (%rax,%rax,1),%edx
  mov    $0xffffffff,%esi
  mov    $0x0,%ecx
  mov    $0x0,%eax
  test   %edx,%edx
  jle    L141
L132:
  mov    $0x0,%eax
L91:
  movzbl 0x10(%rsp,%rax,1),%ecx
  mov    %cl,0x90(%rsp,%rax,2)
  movzbl (%rax,%rbx,1),%ecx
  mov    %cl,0x91(%rsp,%rax,2)
  add    $0x1,%rax
  lea    (%rax,%rax,1),%ecx
  cmp    %ecx,%edx
  jg     L91
  lea    -0x1(%rdx),%eax
  shr    %eax
  add    $0x1,%eax
  test   %edx,%edx
  mov    $0x1,%ecx
  cmovg  %eax,%ecx
  add    %eax,%eax
  test   %edx,%edx
  mov    $0x2,%edx
  cmovle %edx,%eax
  lea    0x1(%rsi),%edx
  cmp    $0x1,%edx
  jbe    L48
L202:
  cltq
  movb   $0x0,0x90(%rsp,%rax,1)
  lea    0x90(%rsp),%r15
  mov    %r15,%rsi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r15,%rdi
  call   <strcpy@plt>
  add    $0x1,%ebp
  cmp    %ebp,0xc(%rsp)
  jle    L198
L63:
  lea    0x10(%rsp),%r15
  mov    %rbx,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %eax,%r15d
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    %eax,%r15d
  jge    L11
  lea    (%r15,%r15,1),%edx
  mov    $0x0,%esi
  mov    $0x0,%ecx
  mov    $0x0,%eax
  test   %edx,%edx
  jg     L132
  jmp    L97
L55:
  lea    (%rax,%rax,1),%edx
  test   %edx,%edx
  jle    L89
  mov    $0x1,%esi
  jmp    L132
L48:
  cmp    $0xffffffff,%esi
  je     L141
L97:
  movslq %ecx,%rdx
  movzbl 0x50(%rsp,%rdx,1),%edi
  lea    0x1(%rax),%edx
  movslq %edx,%rdx
  test   %dil,%dil
  je     L202
  lea    0x8f(%rsp),%rsi
  movslq %ecx,%rcx
  cltq
  sub    %rax,%rcx
  add    %rbx,%rcx
L147:
  mov    %dil,(%rsi,%rdx,1)
  mov    %edx,%eax
  add    $0x1,%rdx
  movzbl -0x1(%rcx,%rdx,1),%edi
  test   %dil,%dil
  jne    L147
  jmp    L202
L141:
  movslq %ecx,%rdx
  movzbl 0x10(%rsp,%rdx,1),%esi
  test   %sil,%sil
  je     L202
  lea    0x1(%rax),%edx
  movslq %edx,%rdx
  lea    0x8f(%rsp),%rdi
  movslq %ecx,%rcx
  cltq
  sub    %rax,%rcx
  lea    0x10(%rsp,%rcx,1),%rcx
L148:
  mov    %sil,(%rdi,%rdx,1)
  mov    %edx,%eax
  add    $0x1,%rdx
  movzbl -0x1(%rcx,%rdx,1),%esi
  test   %sil,%sil
  jne    L148
  jmp    L202
L89:
  mov    $0x0,%eax
  jmp    L202
L198:
  mov    0xf8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L126
  mov    $0x0,%eax
  add    $0x108,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L126:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D8
type_guessed: string
name: unpack_data
arguments: data_label: D1
type_guessed: string
name: unpack_data
arguments: data_label: D32
type_guessed: string
name: unpack_data
arguments: data_label: D0
type_guessed: string

tool: D8: %d

tool: D1: %s%s

tool: D32: %s


tool: 

assistant: ```c
int main() {
  int N, Cont, I, J, TamResult, TamStr1, TamStr2, Bolean;
  char string1[53], string2[53], Result[102];
  scanf("%d", &N);
  for (Cont = 0; Cont < N; Cont++) {
    scanf("%s%s", string1, string2);
    TamStr1 = strlen(string1);
    TamStr2 = strlen(string2);
    Bolean = 0;
    if (TamStr1 < TamStr2)
      TamResult = TamStr1 * 2;
    else {
      if (TamStr1 > TamStr2) {
        TamResult = TamStr2 * 2;
        Bolean = -1;
      } else {
        TamResult = TamStr2 * 2;
        Bolean = 1;
      }
    }
    J = 0;
    for (I = 0; I < TamResult; I = I + 2) {
      Result[I] = string1[J];
      Result[I + 1] = string2[J];
      J++;
    }
    if (Bolean == 0 || Bolean == -1) {
      if (Bolean == -1)
        for (J = J; string1[J] != '\0'; J++) {
          Result[I] = string1[J];
          I++;
        }
      else
        for (J = J; string2[J] != '\0'; J++) {
          Result[I] = string2[J];
          I++;
        }
    }
    Result[I] = '\0';
    printf("%s\n", Result);
    strcpy(Result, "");
  }
  return 0;
}
```