429 纪念日：只给 SHA-256 的口令红包

#!/usr/bin/env python3
"""穷举 8 位纯数字，撞出给定 SHA-256 的原像。

8 位纯数字只有 10^8 个候选。SHA-256 的「不可反推」挡得住原像攻击，
挡不住候选空间这么小的穷举。
实测：6 进程、纯 Python hashlib（无 GPU、无 SIMD 技巧），
跑完 0~99999999 全空间约 34 秒。

用法：
    python brute_sha8.py 3c126f9a0270bd6bbec46d5c1b8774b95407ef8145c1a0a16b5d87a511b9b9b0
    python brute_sha8.py <hash1> <hash2> -p 8
"""
import argparse
import hashlib
import multiprocessing as mp
import time

N = 100_000_000


def work(pack):
    lo, hi, targets = pack
    sha = hashlib.sha256
    found = []
    for i in range(lo, hi):
        d = sha(b"%08d" % i).digest()
        if d in targets:
            found.append((i, targets[d]))
    return found


def main():
    ap = argparse.ArgumentParser()
    ap.add_argument("hashes", nargs="+", help="一个或多个 SHA-256（十六进制）")
    ap.add_argument("-p", "--processes", type=int, default=6)
    args = ap.parse_args()

    targets = {}
    for h in args.hashes:
        h = h.strip().lower()
        if len(h) != 64:
            raise SystemExit(f"✗ 不是 SHA-256（长度 {len(h)}）：{h}")
        targets[bytes.fromhex(h)] = h

    procs = max(1, args.processes)
    step = N // procs
    packs = [(k * step, N if k == procs - 1 else (k + 1) * step, targets) for k in range(procs)]

    t0 = time.time()
    hits = []
    with mp.Pool(procs) as pool:
        for part in pool.imap_unordered(work, packs):
            hits.extend(part)
    dt = time.time() - t0

    print(f"0~{N - 1} 全空间 · {procs} 进程 · 耗时 {dt:.1f}s")
    hits.sort()
    for i, h in hits:
        print(f"  {h[:16]}… -> {i:08d}")
    for h in targets.values():
        if not any(x[1] == h for x in hits):
            print(f"  {h[:16]}… -> 未命中（候选不在 8 位纯数字里）")


if __name__ == "__main__":
    main()