11 2014 档案

摘要:linesTime Limit: 5000/2500 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 515Accepted Submission(s): 241Problem Descrip... 阅读全文
posted @ 2014-11-30 23:56 hl_mark 阅读(184) 评论(0) 推荐(0) 编辑
摘要:D. Striptime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputAlexandra has a paper strip withnnumbers ... 阅读全文
posted @ 2014-11-27 18:37 hl_mark 阅读(392) 评论(0) 推荐(0) 编辑
摘要:DescriptionSean准备投资一些项目。有n个投资项目,投资第i个项目需要花费Ci元。Sean发现如果投资了某些编号连续的项目就能赚得一定的钱。现在给出m组连续的项目和每组能赚得的钱,请问采取最优的投资策略的最大获利是多少?样例最佳策略是全部项目都投资,然后第1,2组都满足了,获利为2+2-... 阅读全文
posted @ 2014-11-27 10:44 hl_mark 阅读(171) 评论(0) 推荐(0) 编辑
摘要:D. Chocolatetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarpus likes giving presents to P... 阅读全文
posted @ 2014-11-23 22:07 hl_mark 阅读(324) 评论(0) 推荐(0) 编辑
摘要:C. Hacking Cyphertime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarpus participates in a com... 阅读全文
posted @ 2014-11-23 21:54 hl_mark 阅读(501) 评论(0) 推荐(0) 编辑
摘要:字串数Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3187Accepted Submission(s): 757Problem Descript... 阅读全文
posted @ 2014-11-22 14:44 hl_mark 阅读(639) 评论(0) 推荐(0) 编辑
摘要:Fight the Monster time limit per test 1 second memory limit per test 256 megabytes ... 阅读全文
posted @ 2014-11-22 09:45 hl_mark 阅读(282) 评论(0) 推荐(0) 编辑
摘要:简单计算器Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12832Accepted Submission(s): 4222Problem Desc... 阅读全文
posted @ 2014-11-20 00:57 hl_mark 阅读(175) 评论(0) 推荐(0) 编辑
摘要:ArbitrageTime Limit:1000MSMemory Limit:65536KTotal Submissions:16175Accepted:6803DescriptionArbitrage is the use of discrepancies in currency exchange... 阅读全文
posted @ 2014-11-15 13:52 hl_mark 阅读(150) 评论(0) 推荐(0) 编辑
摘要:Cow ContestTime Limit:1000MSMemory Limit:65536KTotal Submissions:7156Accepted:3957DescriptionN(1 ≤N≤ 100) cows, con... 阅读全文
posted @ 2014-11-15 11:10 hl_mark 阅读(153) 评论(0) 推荐(0) 编辑
摘要:MPI MaelstromTime Limit:1000MSMemory Limit:10000KTotal Submissions:5547Accepted:3458DescriptionBIT has recently take... 阅读全文
posted @ 2014-11-15 10:40 hl_mark 阅读(129) 评论(0) 推荐(0) 编辑
摘要:WormholesTime Limit:2000MSMemory Limit:65536KTotal Submissions:31992Accepted:11614DescriptionWhile exploring his m... 阅读全文
posted @ 2014-11-12 13:38 hl_mark 阅读(126) 评论(0) 推荐(0) 编辑
摘要:Silver Cow PartyTime Limit:2000MSMemory Limit:65536KTotal Submissions:13100Accepted:5881DescriptionOne cow from each o... 阅读全文
posted @ 2014-11-08 01:33 hl_mark 阅读(165) 评论(0) 推荐(0) 编辑
摘要:D. Maximum ValueYou are given a sequenceaconsisting ofnintegers. Find the maximum possible value of(integer remainder ofaidiv... 阅读全文
posted @ 2014-11-07 12:24 hl_mark 阅读(180) 评论(0) 推荐(0) 编辑
摘要:FroggerTime Limit:1000MSMemory Limit:65536KTotal Submissions:26519Accepted:8633DescriptionFreddy Frog is sitting on a stone in the middle of a lake. S... 阅读全文
posted @ 2014-11-02 21:22 hl_mark 阅读(158) 评论(0) 推荐(0) 编辑
摘要:Til the Cows Come HomeTime Limit:1000MSMemory Limit:65536KTotal Submissions:30865Accepted:10432DescriptionBessie is out in the field and wants to get ... 阅读全文
posted @ 2014-11-02 14:12 hl_mark 阅读(186) 评论(0) 推荐(0) 编辑
摘要:对于 ax≡b( mod n ) 转化为 ax - ny = b , 当 d = gcd( a, n ) 不是 d 的约数的时候不存在解,为何不存在解呢?设 a = k1*d , n = k2*d .那么式子可转化为 : d * (k1*x - k2*y) = b , 若 b % d != 0 ... 阅读全文
posted @ 2014-11-01 00:16 hl_mark 阅读(471) 评论(0) 推荐(0) 编辑

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