将\(m\)种随机映射到\(7\)种,有\(\frac{7!}{7^7}\approx 0.06\)的可能得到正确结果
现在考虑\(7\)种,折半,分成\((4,3)\),然后暴力枚举
用\(2^7-2\)去判断分组,使得暴力枚举的花费最小
结论:暴力枚举花费最小为\(O(10^5)\)
令分好组后花费分别为\(cnt_{min},cnt_{max}\),最优则有\(cnt_{min}\times N\ge cnt_{max}\),\(cnt_{min}\times cnt_{max}\le (\frac{n}{7})^7\)
则\(cnt_{max}\le \sqrt{n\times (\frac{n}{7})^7}\approx 1.3\times 10^5\)
原文:https://www.cnblogs.com/Grice/p/12830479.html