user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L73:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L73
  sub    $0x338,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf4328(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r15
  jmp    L196
L101:
  test   %r8w,%r8w
  js     L8
  movswq %r8w,%r11
  movslq %ecx,%rcx
  imul   $0x3e8,%rcx,%rcx
  add    %r15,%rcx
L127:
  movswq %dx,%rbx
  movzbl 0xf4250(%rsp,%rbx,1),%ebx
  mov    %bl,(%rcx,%r11,1)
  add    $0x1,%edx
  sub    $0x1,%r11
  test   %r11w,%r11w
  jns    L127
  mov    %r10d,%r11d
L140:
  add    $0x1,%esi
  movswl %si,%ecx
  cmp    %eax,%ecx
  jge    L169
L149:
  test   %r11b,%r11b
  jne    L101
  test   %di,%di
  jle    L185
  movslq %ecx,%rcx
  imul   $0x3e8,%rcx,%rcx
  add    %r15,%rcx
  lea    (%rdx,%rdi,1),%ebx
L164:
  movswq %dx,%r11
  movzbl 0xf4250(%rsp,%r11,1),%r11d
  mov    %r11b,(%rcx)
  add    $0x1,%edx
  add    $0x1,%rcx
  cmp    %bx,%dx
  jne    L164
  mov    %ebx,%edx
  mov    %r9d,%r11d
  jmp    L140
L185:
  mov    %r9d,%r11d
  jmp    L140
L8:
  mov    %r10d,%r11d
  jmp    L140
L169:
  mov    $0x0,%r13d
  mov    $0x0,%r14d
  test   %di,%di
  jg     L14
L133:
  mov    $0xa,%edi
  call   <putchar@plt>
L196:
  lea    0xe(%rsp),%rsi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmpw   $0x0,0xe(%rsp)
  je     L65
  lea    0xf4250(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  movswl %ax,%ebp
  movzwl 0xe(%rsp),%edi
  movswl %di,%ecx
  mov    %ebp,%eax
  cltd
  idiv   %ecx
  test   %eax,%eax
  jle    L169
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%r11d
  mov    $0x0,%ecx
  lea    -0x1(%rdi),%r8d
  mov    $0x0,%r10d
  mov    $0x1,%r9d
  jmp    L149
L113:
  movslq %ecx,%rcx
  imul   $0x3e8,%rcx,%rcx
  lea    0xf4330(%rcx),%rax
  lea    (%rsp,%rax,1),%rcx
  movsbl -0xf4320(%r12,%rcx,1),%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  movswl %bx,%ecx
  movswl 0xe(%rsp),%esi
  mov    %ebp,%eax
  cltd
  idiv   %esi
  cmp    %eax,%ecx
  jl     L113
L153:
  add    $0x1,%r13d
  movsbw %r13b,%ax
  cmp    0xe(%rsp),%ax
  jge    L133
L14:
  movswl 0xe(%rsp),%ecx
  mov    %ebp,%eax
  cltd
  idiv   %ecx
  mov    $0x0,%ebx
  mov    %r14d,%ecx
  movsbq %r13b,%r12
  test   %eax,%eax
  jg     L113
  jmp    L153
L65:
  mov    0xf4328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L86
  mov    $0x0,%eax
  add    $0xf4338,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L86:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
type: string
name: extract_symbolic_data
arguments: label: D1
type: string

tool: D15: %hd

tool: D1: %s

assistant: ```c
int main() {
  char i;
  char s[201];
  short n, len, j, q, k;
  char a[1000][1000];
  while (1) {
    scanf("%hd", &n);
    if (!n) {
      break;
    }
    scanf("%s", s);
    i = 0;
    len = strlen(s);
    for (q = k = 0; q < len / n; ++q) {
      if (i == 0) {
        for (j = 0; j < n; ++j, ++k) {
          a[q][j] = s[k];
        }
        i = 1;
      } else {
        for (j = n - 1; j >= 0; --j, ++k) {
          a[q][j] = s[k];
        }
        i = 0;
      }
    }
    for (i = 0; i < n; ++i) {
      for (j = 0; j < len / n; ++j) {
        putchar(a[j][i]);
      }
    }
    putchar('\n');
  }
  return 0;
}
```