# -*- coding: utf-8 -*-
"""用真机 collect 密文的**长度与填充**判决「四段加密」结构。

不需要密钥：段结构在 base64 的长度与 '=' 分布上留下指纹。

判决依据（docs/tdc-案例.md §5.1）:
  chunk1/trajectory/chunk2  各 pad 到 24 字节 -> 密文长度是 24 的倍数 -> base64 无 '='
  sd                        不 pad，只被 XTEA 补到 8 的倍数 -> 长度%3 != 0 -> base64 带 '='

用法:
    .venv/bin/python scripts/verify/verify_seglen.py <collect 的 base64>
"""
import base64
import re
import sys
from pathlib import Path

if len(sys.argv) < 2:
    print("用法: verify_seglen.py <collect 的 base64（文件路径或直接给串）>")
    raise SystemExit(2)
_arg = sys.argv[1]
_p = Path(_arg)
b64 = (_p.read_text(encoding="ascii") if _p.exists() else _arg).strip()
print("真机 collect: %d 字符, 含 '=' : %s" % (len(b64), "=" in b64))
print()

raw = base64.b64decode(b64)
print("解码后密文: %d 字节" % len(raw))
print()

# 真机明文（另一份同会话样本）给出的各段长度
segs = [("chunk1", 355, 5), ("trajectory", 62, 10), ("chunk2", 288, 24), ("sd", 33, None)]
print("%-11s %8s %6s %10s %12s %10s" % ("段", "内容", "pad", "密文字节", "base64字符", "含'='"))
print("-" * 66)
total = 0
total_b64 = 0
for name, content, pad in segs:
    if pad is not None:
        n = content + pad                 # 空格 pad 到 24
        assert n % 24 == 0
    else:
        n = content + ((8 - content % 8) % 8)   # sd: 只被 XTEA 补到 8 的倍数
    enc = base64.b64encode(b"\x00" * n).decode()
    total += n
    total_b64 += len(enc)
    print("%-11s %8d %6s %10d %12d %10s"
          % (name, content, pad if pad is not None else "(xtea)",
             n, len(enc), "=" in enc))
print("-" * 66)
print("%-11s %8s %6s %10d %12d" % ("合计", "", "", total, total_b64))
print()
print(">>> 推算总密文字节 %d  == 实际 %d : %s" % (total, len(raw), total == len(raw)))
print(">>> 推算 base64 字符 %d == 实际 %d : %s" % (total_b64, len(b64), total_b64 == len(b64)))
print()

# 反过来看：把整串按 24 字节切，前三段应能整除
print("[反向校验] 把密文按 24 字节切分:")
off = 0
for name, content, pad in segs:
    if pad is None:
        print("    %-11s 剩余 %3d 字节 (sd, 非 24 倍数)" % (name, len(raw) - off))
        break
    n = content + pad
    print("    %-11s [%4d:%4d] %3d 字节  24倍数=%s"
          % (name, off, off + n, n, n % 24 == 0))
    off += n
print()

# 关键指纹: '=' 只应出现在最后(来自 sd 段)
idx = b64.find("=")
print("'=' 首次出现位置: %d / 总长 %d  -> %s"
      % (idx, len(b64), "在末尾区域（符合 sd 段不 pad）"
         if idx >= len(b64) - 4 else "⚠️ 在中间（与四段结构矛盾）"))
if idx >= 0:
    print("    尾部填充: %r" % b64[idx:])