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hdu 4473 Exam (思维题 问题转化)
ExamTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 459 Accepted Submission(s): 184Problem DescriptionRikka is a high sch原创 2013-07-22 14:56:16 · 1688 阅读 · 0 评论 -
hdu 4869 Turn the pokers （思维）
Turn the pokersTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 196 Accepted Submission(s): 51Problem DescriptionDuring summer原创 2014-07-23 19:24:14 · 1021 阅读 · 0 评论 -
Codeforces Round #225 (Div. 1) B. Volcanoes
B. Volcanoestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputIahub got lost in a very big desert. The desert原创 2014-01-23 14:58:38 · 1418 阅读 · 0 评论 -
poj 3182 The Grove （搜索+判断点在多边形内）
The GroveTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 641 Accepted: 297DescriptionThe pasture contains a small, contiguous grove of trees that has no原创 2013-09-30 08:51:45 · 2358 阅读 · 3 评论 -
Codeforces Round #221 (Div. 2) B. I.O.U. C. Divisible by Seven D. Maximum Submatrix 2 解题报告
B. I.O.U.题目链接：点击打开链接思路：将每个人得多少、欠多少综合起来看，一个关系内的debts最小就是得的总和或者欠的总和.ps：这题数据貌似很水，很多代码都水过了...代码：#include #include #include #include #include #include #include #include #inc原创 2013-12-25 10:49:29 · 1708 阅读 · 2 评论 -
hdu 4666 Hyperspace （曼哈顿距离+set ）
HyperspaceTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 931 Accepted Submission(s): 441Problem DescriptionThe great Mr.原创 2013-10-24 23:52:31 · 988 阅读 · 0 评论 -
PKU Online Judge 1054:Cube （设置根节点）
1054:Cube总时间限制: 1000ms 内存限制: 131072kB描述Delayyy君很喜欢玩某个由Picks编写的方块游戏，游戏在一个由单位格组成的棋盘上进行。游戏的主角是一个6个面互不相同的小方块，每次可以向上下左右中的某个方向翻滚一格。棋盘上有 N 个关键格子，对应于游戏中的村庄，在坐标系中，每个村庄有一个坐原创 2013-10-14 14:43:49 · 1159 阅读 · 0 评论 -
Codeforces Round #198 (Div. 2) D. Bubble Sort Graph （转化为最长非降子序列）
D. Bubble Sort Graphtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputIahub recently has learned Bubble Sort,原创 2013-08-31 21:15:50 · 2005 阅读 · 0 评论 -
poj 2443 Set Operation （位操作）
Set OperationTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 2307 Accepted: 892DescriptionYou are given N sets, the i-th set (represent by S(i)) have C(原创 2013-09-27 21:13:38 · 1407 阅读 · 0 评论 -
hdu 4750 Count The Pairs （思维+并查集+离散化+二分查找）
Count The PairsTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 465 Accepted Submission(s): 226Problem Description Wi原创 2013-09-25 22:11:41 · 1173 阅读 · 0 评论 -
hdu 4104 Discount (思维 数学归纳法）
DiscountTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1072 Accepted Submission(s): 645Problem DescriptionAll the shop原创 2013-10-02 22:26:31 · 1377 阅读 · 0 评论 -
poj 3419 Difference Is Beautiful (区间最长连续不重复数 dp+二分+RMQ)
Difference Is BeautifulTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 1893 Accepted: 579DescriptionMr. Flower's business is growing much faster than or原创 2013-10-02 19:20:47 · 1910 阅读 · 0 评论 -
hdu 4334 Trouble
TroubleTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3854 Accepted Submission(s): 1222Problem DescriptionHassan is in trou原创 2013-08-10 20:39:39 · 1005 阅读 · 0 评论 -
hdu 4611 Balls Rearrangement （找循环节）
Balls RearrangementTime Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1700 Accepted Submission(s): 643Problem Description Bob h原创 2013-08-26 09:20:40 · 971 阅读 · 0 评论 -
hdu 4597 Play Game （记忆化搜索）
Play GameTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 35 Accepted Submission(s): 24Problem DescriptionAlice and Bob are原创 2013-08-24 19:56:27 · 2101 阅读 · 0 评论 -
hdu 4467 Graph ( 神奇的思想 降低复杂度 )
GraphTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 695 Accepted Submission(s): 113Problem DescriptionP. T. Tigris is a stu原创 2013-07-23 15:55:30 · 1744 阅读 · 0 评论 -
Codeforces Round #192 (Div. 2) C. Purification
题目链接：http://codeforces.com/contest/330/problem/C思路：只需判断每行是否都存在一个‘.’ 或者每列是否都存在一个‘.' 就够了 不用想复杂了 如果都不存在的话答案就不存在 有一个存在则输出每一行或者列的一个点就够了因为如果行列都不满足的话 因为行列必定相交 所以必有一个点不能被“purify”感想：CF呀CF，逢打必跪，原创 2013-07-21 08:24:27 · 1390 阅读 · 0 评论 -
uva 11464 - Even Parity （枚举+递推）
题目链接：11464 - Even Parity题意：原创 2014-06-17 18:04:49 · 848 阅读 · 0 评论