As we all know, Matt is an outstanding contestant in ACM-ICPC. Graph problems are his favorite.Once, he came up with a simple algorithm for finding the maximal independent set in trees by mistake.A tree is a connected undirected graph without cycles,…

Just A Mistake Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 168    Accepted Submission(s): 41 Problem Description As we all know, Matt is an outstanding contestant in ACM-ICPC. Graph proble…

作业 poj 1091 跳蚤 容斥原理. 考虑能否跳到旁边就是卡牌的\(gcd\)是否是1,可以根据裴蜀定理证明. 考虑正着做十分的麻烦,所以倒着做,也就是用\(M^N - (不合法)\)即可. 不合法显然就是\(gcd\)不为1的情况,那么我们考虑枚举\(gcd\),\((1 \leq gcd \leq 15)\),第\(n + 1\)个数一直是\(m\)所以不用理它. 考虑每个\(gcd\)的贡献,如果所有的都能被整除,那么产生的方案数就是: \(({m \over gcd})^n\) 表示…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=4293 Problem Description After the regional contest, all the ACMers are walking alone a very long avenue to the dining hall in groups. Groups can vary in size for kinds of reasons, which means, sev…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3007 Each person had do something foolish along with his or her growth.But,when he or she did this that time,they could not predict that this thing is a mistake and they will want this thing would rather…

Clarke and chemistry Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 63    Accepted Submission(s): 33 Problem Description Clarke is a patient with multiple personality disorder. One day, Clarke…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…

http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; int jc[100003]; int p; int ipow(int x, int b) { ll t = 1, w = x;…

http://acm.hdu.edu.cn/showproblem.php?pid=4859 题目大意: 在一个矩形周围都是海,这个矩形中有陆地,深海和浅海.浅海是可以填成陆地的. 求最多有多少条方格线满足两侧分别是海洋和陆地 这道题很神 首先考虑一下,什么情况下能够对答案做出贡献 就是相邻的两块不一样的时候 这样我们可以建立最小割模型,可是都说是最小割了 无法求出最大的不相同的东西 所以我们考虑转化,用总的配对数目 - 最小的相同的对数 至于最小的相同的对数怎么算呢? 我们考虑这样的构造方法:…

Special equations Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4569 Description Let f(x) = a nx n +...+ a 1x +a 0, in which a i (0 <= i <= n) are all known integers. We call f(x) 0 (mod…