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
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xa8,%rsp
  lea    D3(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x1098(%rsp)
  xor    %eax,%eax
  lea    0x1c(%rsp),%rsi
  lea    0x860(%rsp),%r15
  call   <scanf@plt>
  lea    0x20(%rsp),%rax
  mov    %rax,0x8(%rsp)
  mov    0x1c(%rsp),%eax
  test   %eax,%eax
  je     L14
  nopl   0x0(%rax,%rax,1)
L135:
  mov    0x8(%rsp),%r14
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  mov    %r14,%rsi
  call   <scanf@plt>
  mov    %r15,%rsi
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r15,%rdi
  mov    %eax,%r14d
  call   <strlen@plt>
  movslq %eax,%r12
  lea    0x1(%r12),%ebp
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  js     L43
  mov    %rax,%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L134:
  mov    %eax,%ecx
  mov    %eax,(%rdx)
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %ecx,%r12d
  jne    L134
  test   %r14d,%r14d
  jle    L115
L10:
  lea    -0x1(%r12),%eax
  movslq %ebp,%rbp
  mov    0x8(%rsp),%r10
  add    $0x1,%r14d
  lea    0x4(%r13,%rax,4),%r11
  lea    -0x4(%r13,%rbp,4),%rax
  mov    $0x1,%ebp
  mov    %rax,(%rsp)
  nopl   (%rax)
L105:
  mov    %r15,%rdi
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %ebp,%eax
  test   %r12d,%r12d
  jg     L75
  jmp    L69
  nopl   0x0(%rax)
L45:
  cmp    %r8d,%esi
  lea    0x1(%r8),%ebx
  mov    %esi,(%r9)
  cmovge %ebx,%ecx
  cmp    %eax,%ecx
  cmovle %ecx,%eax
  add    $0x1,%rdi
  cmp    %r11,%rdx
  je     L72
L0:
  mov    %eax,%esi
L75:
  mov    (%rdx),%r8d
  mov    %rdx,%r9
  mov    0x4(%rdx),%eax
  add    $0x4,%rdx
  movzbl (%rdi),%ebx
  lea    0x1(%rsi),%ecx
  add    $0x1,%eax
  cmp    %bl,(%r10)
  jne    L45
  cmp    %ecx,%r8d
  mov    %esi,(%r9)
  cmovg  %ecx,%r8d
  cmp    %eax,%r8d
  cmovle %r8d,%eax
  add    $0x1,%rdi
  cmp    %r11,%rdx
  jne    L0
L72:
  mov    (%rsp),%rdx
L69:
  add    $0x1,%ebp
  mov    %eax,(%rdx)
  add    $0x1,%r10
  cmp    %ebp,%r14d
  jne    L105
  mov    %eax,%ebx
  mov    %rdx,%rax
L70:
  mov    %ebx,(%rax)
  mov    0x0(%r13,%r12,4),%esi
  lea    D30(%rip),%rdi
  call   <printf@plt>
  subl   $0x1,0x1c(%rsp)
  jne    L135
L14:
  mov    0x1098(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x10a8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L43:
  xor    %eax,%eax
  test   %r14d,%r14d
  jg     L10
L115:
  cltq
  lea    -0x4(%r13,%rax,4),%rax
  jmp    L70
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D3
type_guessed: string
name: unpack_data
arguments: label: D37
type_guessed: string
name: unpack_data
arguments: label: D30
type_guessed: string

tool: D3: %d

tool: D37: %s

tool: D30: %d


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;
}
```