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
  lea    D86(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x378,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x368(%rsp)
  xor    %eax,%eax
  movl   $0x1d,0x14(%rsp)
  movabs $0x1e0000001f,%rax
  lea    0x1d0(%rsp),%rbx
  mov    %rax,0x18(%rsp)
  mov    %rax,0x20(%rsp)
  movabs $0x1f0000001f,%rax
  mov    %rax,0x28(%rsp)
  sub    $0x1,%rax
  mov    %rax,0x30(%rsp)
  mov    %rax,0x38(%rsp)
  lea    0x40(%rsp),%rax
  mov    %rax,0x8(%rsp)
  mov    %rax,%r14
L140:
  lea    (%rbx,%r12,4),%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    %r12,%rbp
  call   <scanf@plt>
  mov    (%r14),%eax
  mov    %r12d,%r15d
  add    $0x4,%r14
  lea    0x1(%rbp),%r12
  test   %eax,%eax
  jne    L140
  xor    %r13d,%r13d
  lea    D53(%rip),%r12
  lea    D43(%rip),%r14
  test   %r15d,%r15d
  je     L8
  xchg   %ax,%ax
L80:
  mov    0x8(%rsp),%rax
  mov    (%rax,%r13,4),%edx
  cmp    $0x1,%edx
  jle    L144
  sub    $0x2,%edx
  lea    0x14(%rsp),%rax
  mov    $0x1f,%edi
  lea    (%rax,%rdx,4),%r9
  xor    %edx,%edx
  jmp    L97
  nopw   0x0(%rax,%rax,1)
L142:
  mov    (%rax),%edi
  add    $0x4,%rax
L97:
  add    %edi,%edx
  cmp    %rax,%r9
  jne    L142
L9:
  add    (%rbx,%r13,4),%edx
  movslq %edx,%rax
  mov    %edx,%edi
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%edi
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  sub    %edi,%eax
  lea    0x0(,%rax,8),%edi
  sub    %eax,%edi
  sub    %edi,%edx
  cmp    $0x6,%edx
  ja     L113
  movslq (%r12,%rdx,4),%rax
  add    %r12,%rax
  notrack jmp *%rax
  nopl   (%rax)
  lea    D58(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L113:
  add    $0x1,%r13
  cmp    %rbp,%r13
  jne    L80
L8:
  mov    0x368(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x378,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  lea    D36(%rip),%rdi
  call   <printf@plt>
  jmp    L113
  xchg   %ax,%ax
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L113
  xchg   %ax,%ax
  lea    D19(%rip),%rdi
  call   <printf@plt>
  jmp    L113
  xchg   %ax,%ax
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L113
  xchg   %ax,%ax
  lea    D54(%rip),%rdi
  call   <printf@plt>
  jmp    L113
  nopl   0x0(%rax)
  mov    %r14,%rdi
  call   <printf@plt>
  jmp    L113
  nopl   (%rax)
L144:
  xor    %edx,%edx
  jmp    L9
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D86
data_type: string
name: parse_data
arguments: data_tag: D43
data_type: string
name: parse_data
arguments: data_tag: D58
data_type: string
name: parse_data
arguments: data_tag: D36
data_type: string
name: parse_data
arguments: data_tag: D7
data_type: string
name: parse_data
arguments: data_tag: D19
data_type: string
name: parse_data
arguments: data_tag: D12
data_type: string
name: parse_data
arguments: data_tag: D54
data_type: string

tool: D86: %d %d

tool: D43: Wednesday


tool: D58: Monday


tool: D36: Sunday


tool: D7: Saturday


tool: D19: Friday


tool: D12: Tuesday


tool: D54: Thursday


assistant: ```c
int main(void) {
  int i = 0, j, k, sum, n, month[100], date[100], md[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  while (1) {
    scanf("%d %d", &month[i], &date[i]);
    if (month[i] == 0)
      break;
    i++;
  }
  for (j = 0; j < i; j++) {
    sum = 0;
    for (k = 0; k < month[j] - 1; k++)
      sum += md[k];
    sum += date[j];
    n = sum % 7;
    switch (n) {
    case 1:
      printf("Thursday\n");
      break;
    case 2:
      printf("Friday\n");
      break;
    case 3:
      printf("Saturday\n");
      break;
    case 4:
      printf("Sunday\n");
      break;
    case 5:
      printf("Monday\n");
      break;
    case 6:
      printf("Tuesday\n");
      break;
    case 0:
      printf("Wednesday\n");
      break;
    }
  }
  return 0;
}
```