摘要：
一个神奇的算法，在题目中看到的，单独拿出来整理枚举一个集合S（用一个二进制表示）的所有子集S0：for(int S0 = S; S0; S0 = (S0-1)&S){}还有不包括S本身的： for(int S0 = (S-1)&S; S0; S0 = (S0-1)&S){} 阅读全文
posted @ 2016-04-20 21:20
FuTaimeng
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问题描述： This is a two player game. Initially there are n integer numbers in an array and players A and B get chance to take them alternatively. Each pla 阅读全文
posted @ 2016-04-20 21:04
FuTaimeng
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http://www.cnblogs.com/acm-bingzi/p/3272898.html Hackers’ Crackdown Miracle Corporations has a number of system services running in a distributed comp 阅读全文
posted @ 2016-04-20 17:59
FuTaimeng
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