Yxter - 博客园

HDU 4549 M斐波那契数列（矩阵快速幂+费马小定理）

摘要： M斐波那契数列 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description M斐波那契数列F[n]是一种整数数列，它的定义如下：F[0] = aF[1] = bF[n] = F[n-1] * F[n-2] 阅读全文

posted @ 2017-07-10 16:06 Yxter 阅读(733) 评论(0) 推荐(1)

Poj 3318 Matrix Multiplication( 矩阵压缩)

摘要： Matrix Multiplication Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 18928 Accepted: 4074 Description You are given three n × n matrices A 阅读全文

posted @ 2017-07-10 15:57 Yxter 阅读(321) 评论(0) 推荐(1)

HDU 4704 Sum (隔板原理 + 费马小定理)

摘要： Sum Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Sample Input 2 Sample Output 2 Hint 1. For N = 2, S(1) = S(2) = 1.2. 阅读全文

posted @ 2017-07-10 10:16 Yxter 阅读(353) 评论(0) 推荐(0)

HDU 3923 Invoker(polya定理+乘法逆元（扩展欧几里德+费马小定理）)

摘要： Invoker Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description On of Vance's favourite hero is Invoker, Kael. As many people kno 阅读全文

posted @ 2017-07-06 17:31 Yxter 阅读(299) 评论(1) 推荐(0)

Polya定理

摘要： Polya定理：设G={π1，π2，π3........πn}是X={a1，a2，a3.......an}上一个置换群，用m中颜色对X中的元素进行涂色，那么不同的涂色方案数为：1/|G|*(mC(π1)+mC(π2)+mC(π3)+...+mC(πk)). 其中C(πk)为置换πk的循环节的个数。阅读全文

posted @ 2017-07-05 16:02 Yxter 阅读(188) 评论(0) 推荐(0)

hdu 1817 Necklace of Beads（Polya定理）

摘要： Necklace of Beads Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 630 Accepted Submission(s): 232 阅读全文

posted @ 2017-07-05 15:53 Yxter 阅读(272) 评论(0) 推荐(0)

Poj 1651 Multiplication Puzzle（区间dp）

摘要： Multiplication Puzzle Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10010 Accepted: 6188 Description The multiplication puzzle is played 阅读全文

posted @ 2017-05-13 20:01 Yxter 阅读(142) 评论(0) 推荐(0)

Poj 2955 brackets（区间dp）

摘要： Brackets Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7795 Accepted: 4136 Description We give the following inductive definition of a “r 阅读全文

posted @ 2017-05-13 16:45 Yxter 阅读(166) 评论(0) 推荐(0)

HDU 5875 Function (线段树+gcd / 单调栈)

摘要：题意：给你一串数a再给你一些区间（lef，rig），求出a[lef]%a[lef+1]...%a[rig] 题解：我们可以发现数字a对数字b取模时：如果a<b,则等于原数，否则a会变小至少一半。就是说a最多成功取模（log2 a）次，所以我们只需要每次在区间内找到最前面一个小于等于a的值，接着更新a 阅读全文

posted @ 2017-05-05 17:12 Yxter 阅读(219) 评论(0) 推荐(0)

ZOJ 1609 Equivalence（状压+dfs减枝）

摘要： ZOJ Problem Set - 1609 EquivalenceTime Limit: 5 Seconds Memory Limit: 32768 KBWhen learning mathamatics, sometimes one may come to an theorem which go 阅读全文

posted @ 2017-04-20 11:51 Yxter 阅读(196) 评论(0) 推荐(0)