user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%r12
  mov    0x10(%rsi),%rbp
  mov    %r12,%rsi
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D45(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L54
  lea    -0x1(%rax),%eax
  mov    %rbp,%r15
  lea    D33(%rip),%r13
  lea    0x1(%rbp,%rax,1),%r14
  jmp    L82
  nopl   0x0(%rax)
L85:
  sub    $0x61,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  movsbl %sil,%esi
  call   <printf@plt>
  cmp    %r14,%r15
  je     L54
L82:
  movsbl (%r15),%esi
  cmp    $0x60,%sil
  jg     L85
  sub    $0x41,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r14,%r15
  jne    L82
L54:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  xor    %ecx,%ecx
  movslq %eax,%rsi
  test   %eax,%eax
  jle    L59
  nopl   (%rax)
L98:
  mov    %ecx,%eax
  cltd
  idiv   %ebx
  movslq %edx,%rdx
  movsbl 0x0(%rbp,%rdx,1),%eax
  lea    -0x41(%rax),%edx
  cmp    $0x60,%al
  jle    L2
  sub    $0x61,%eax
  movsbl %al,%edx
L2:
  movsbl (%r12,%rcx,1),%eax
  mov    %eax,%edi
  add    %edx,%eax
  cmp    $0x60,%dil
  jle    L107
  sub    $0x61,%eax
  movslq %eax,%rdx
  mov    %eax,%edi
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%edi
  sar    $0x23,%rdx
  sub    %edi,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x61,%eax
  mov    %al,(%r12,%rcx,1)
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  jne    L98
L59:
  mov    %r12,%rsi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L107:
  sub    $0x41,%eax
  movslq %eax,%rdx
  mov    %eax,%edi
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%edi
  sar    $0x23,%rdx
  sub    %edi,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x41,%eax
  mov    %al,(%r12,%rcx,1)
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  jne    L98
  jmp    L59
```

tool_calls: name: disassemble_data
arguments: tag: D28
datatype: string
name: disassemble_data
arguments: tag: D45
datatype: string
name: disassemble_data
arguments: tag: D14
datatype: string
name: disassemble_data
arguments: tag: D33
datatype: string
name: disassemble_data
arguments: tag: D22
datatype: string

tool: D28: Plaintext is %s


tool: D45: Vigenere key is %s


tool: D14: Equivalent k is : 

tool: D33: %i 

tool: D22: Encyphered text is %s


assistant: ```c
int main(int argc, char **argv) {
  char *key = argv[2];
  char *s = argv[1];
  printf("Plaintext is %s\n", s);
  printf("Vigenere key is %s\n", key);
  printf("Equivalent k is : ");
  int l = strlen(key);
  for (int c = 0; c < l; c++) {
    if (key[c] >= 97) {
      printf("%i ", key[c] - 'a');
    } else {
      printf("%i ", key[c] - 'A');
    }
  }
  printf("\n");
  int s_l = strlen(s);
  int k;
  for (int c = 0; c < s_l; c++) {
    if (key[c % l] >= 97) {
      k = key[c % l] - 'a';
    } else {
      k = key[c % l] - 'A';
    }
    if (s[c] >= 97) {
      s[c] = (s[c] + k - 'a') % 26 + 'a';
    } else {
      s[c] = (s[c] + k - 'A') % 26 + 'A';
    }
  }
  printf("Encyphered text is %s\n", s);
  return 0;
}
```