user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1088(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jne    L58
L41:
  mov    0x1088(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L113
  mov    $0x0,%eax
  add    $0x1098,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L121:
  mov    $0x0,%r12d
  jmp    L75
L51:
  lea    0x1(%rdx),%ebx
  mov    (%rax),%r12d
  cmp    %r12d,%ebx
  cmovg  %r12d,%ebx
  mov    (%rsi),%r15d
  lea    0x1(%r15),%r12d
  cmp    %r12d,%ebx
  cmovg  %r12d,%ebx
L103:
  mov    %edx,(%rax)
  add    $0x1,%rcx
  add    $0x4,%rax
  add    $0x4,%rsi
  cmp    %r8,%rax
  je     L59
  mov    %ebx,%edx
L114:
  movzbl (%rcx),%ebx
  cmp    %bl,(%rdi)
  je     L51
  mov    (%rax),%ebx
  lea    0x1(%rbx),%r15d
  lea    0x1(%rdx),%r12d
  cmp    %edx,%ebx
  cmovle %r15d,%r12d
  mov    (%rsi),%ebx
  add    $0x1,%ebx
  cmp    %r12d,%ebx
  cmovg  %r12d,%ebx
  jmp    L103
L59:
  mov    %r14d,%r12d
L122:
  movslq %r12d,%rax
  mov    %ebx,-0x4(%r10,%rax,4)
  add    $0x1,%r9d
  add    $0x1,%rdi
  cmp    %r11d,%r9d
  je     L85
L83:
  mov    %r9d,%ebx
  mov    $0x1,%r12d
  test   %ebp,%ebp
  jle    L122
  lea    0x850(%rsp),%rcx
  mov    %r13,%rsi
  mov    %r10,%rax
  mov    %r9d,%edx
  jmp    L114
L85:
  movslq %r12d,%r12
  mov    %ebx,-0x4(%r10,%r12,4)
  movslq %ebp,%rbp
  mov    (%r10,%rbp,4),%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0xc(%rsp)
  je     L41
L58:
  lea    0x10(%rsp),%r12
  mov    %r12,%rsi
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    0x850(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  lea    0x1(%rax),%r12d
  lea    0x0(,%r12,4),%edi
  call   <malloc@plt>
  mov    %rax,%r10
  test   %ebp,%ebp
  js     L121
  mov    %rax,%rdx
  mov    %r12d,%ecx
  mov    $0x0,%eax
L80:
  mov    %eax,(%rdx)
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %ecx,%eax
  jne    L80
L75:
  test   %r13d,%r13d
  jle    L85
  lea    0x10(%rsp),%rdi
  lea    0x1(%r13),%r11d
  lea    0x4(%r10),%r13
  lea    -0x1(%rbp),%eax
  lea    0x0(%r13,%rax,4),%r8
  mov    $0x1,%r9d
  lea    0x1(%rbp),%r14d
  jmp    L83
L113:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D28
data_type: string
name: extract_symbolic_data
arguments: label: D14
data_type: string
name: extract_symbolic_data
arguments: label: D22
data_type: string

tool: D28: %d

tool: D14: %d


tool: D22: %s

assistant: ```c
int main() {
  int T, i, j;
  char s1[2100], s2[2100];
  int *Val;
  scanf("%d", &T);
  while (T != 0) {
    scanf("%s", s1);
    scanf("%s", s2);
    int l1 = strlen(s1);
    int l2 = strlen(s2);
    Val = (int *)malloc((l2 + 1) * sizeof(int));
    for (j = 0; j <= l2; j++) {
      Val[j] = j;
    }
    int curr;
    for (i = 1; i <= l1; i++) {
      curr = i;
      int curr2;
      for (j = 1; j <= l2; j++) {
        if (s1[i - 1] == s2[j - 1]) {
          int min = Val[j - 1];
          if (min > (curr + 1)) {
            min = curr + 1;
          }
          if (min > (Val[j] + 1)) {
            min = Val[j] + 1;
          }
          curr2 = min;
        } else {
          int min = Val[j - 1] + 1;
          if (min > (curr + 1)) {
            min = curr + 1;
          }
          if (min > (Val[j] + 1)) {
            min = Val[j] + 1;
          }
          curr2 = min;
        }
        Val[j - 1] = curr;
        curr = curr2;
      }
      Val[j - 1] = curr;
    }
    Val[j - 1] = curr;
    printf("%d\n", Val[l2]);
    T--;
  }
  return 0;
}
```