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2011年8月4日

ACM HDU 1069 Monkey and Banana (动态规划)

摘要: Monkey and BananaTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2599Accepted Submission(s): 1345Problem DescriptionA group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, an 阅读全文

posted @ 2011-08-04 14:09 kuangbin 阅读(5311) 评论(0) 推荐(2) 编辑

HDU 1231 最大连续子序列(和HDU1003 Max Sum类似的)

摘要: 最大连续子序列Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7999Accepted Submission(s): 3335Problem Description给定K个整数的序列{ N1, N2, ..., NK },其任意连续子序列可表示为{ Ni, Ni+1, ..., Nj },其中 1 <= i <= j <= K。最大连续子序列是所有连续子序列中元素和最大的一个, 例如给定序列{ -2, 11, -4, 13, 阅读全文

posted @ 2011-08-04 12:42 kuangbin 阅读(556) 评论(0) 推荐(0) 编辑

HDU 1024 Max Sum Plus Plus(动态规划,给定一个数组,求其分成m个不相交子段和最大值的问题)

摘要: Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6725Accepted Submission(s): 2251Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves t 阅读全文

posted @ 2011-08-04 11:10 kuangbin 阅读(14935) 评论(1) 推荐(7) 编辑

HDU 1003 Max Sum(连续子列的最大和,动态规划)

摘要: Max SumTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 55792Accepted Submission(s): 12580Problem DescriptionGiven a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max 阅读全文

posted @ 2011-08-04 10:12 kuangbin 阅读(511) 评论(0) 推荐(0) 编辑

ACM POJ 1015 Jury Compromise(陪审团的人选,动态规划题,难)

摘要: Jury CompromiseTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 18378Accepted: 4516Special JudgeDescriptionIn Frobnia, a far-away country, the verdicts in court trials are determined by a jury consisting of members of the general public. Every time a trial is set to begin, a jury has to be s 阅读全文

posted @ 2011-08-04 01:16 kuangbin 阅读(6580) 评论(3) 推荐(2) 编辑

2011年8月3日

ACM POJ 1458 Common Subsequence (最长公共子序列,动态规划)

摘要: Common SubsequenceTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 25178Accepted: 9726DescriptionA subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > 阅读全文

posted @ 2011-08-03 20:13 kuangbin 阅读(1429) 评论(0) 推荐(1) 编辑

ACM POJ 1661 Help Jimmy(动态规划)

摘要: Help JimmyTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 6593Accepted: 2066Description"Help Jimmy" 是在下图所示的场景上完成的游戏。 场景中包括多个长度和高度各不相同的平台。地面是最低的平台,高度为零,长度无限。 Jimmy老鼠在时刻0从高于所有平台的某处开始下落,它的下落速度始终为1米/秒。当Jimmy落到某个平台上时,游戏者选择让它向左还是向右跑,它跑动的速度也是1米/秒。当Jimmy跑到平台的边缘时,开始继续下落。Jimmy每次下落的高度不能超过MAX 阅读全文

posted @ 2011-08-03 19:50 kuangbin 阅读(671) 评论(0) 推荐(0) 编辑

Triple ACM HDU 3908 (数学题,找多少种组合)

摘要: TripleTime Limit: 5000/3000 MS (Java/Others)Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 387Accepted Submission(s): 153Problem DescriptionGiven many different integers, find out the number of triples (a, b, c) which satisfy a, b, c are co-primed each other or are not co-primed each 阅读全文

posted @ 2011-08-03 15:09 kuangbin 阅读(690) 评论(0) 推荐(0) 编辑

Swordsman ACM HDU 3902(判断n边形是不是轴对称图形,暴力求解了~~~~)

摘要: SwordsmanTime Limit: 10000/4000 MS (Java/Others)Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 308Accepted Submission(s): 116Problem DescriptionMr. AC is a swordsman. His dream is to be the best swordsman in the world. To achieve his goal, he practices every day. There are many ways 阅读全文

posted @ 2011-08-03 14:47 kuangbin 阅读(1169) 评论(0) 推荐(0) 编辑

ACM POJ 2533Longest Ordered Subsequence(最长上升子序列,简单DP)

摘要: Longest Ordered SubsequenceTime Limit: 2000MSMemory Limit: 65536KTotal Submissions: 19643Accepted: 8501DescriptionA numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 < 阅读全文

posted @ 2011-08-03 11:10 kuangbin 阅读(3131) 评论(0) 推荐(0) 编辑

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