随笔分类 -  动态规划--乱搞

摘要:Rikka with Subset Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 837 Accepted Submission(s): 411 阅读全文
posted @ 2017-08-09 09:37 勿忘初心0924 阅读(149) 评论(0) 推荐(0) 编辑
摘要:Bridging signals Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2354 Accepted Submission(s): 153 阅读全文
posted @ 2017-05-30 10:35 勿忘初心0924 阅读(266) 评论(0) 推荐(0) 编辑
摘要:After moving from his parents’ place Zhenya has been living in the University dormitory for a month. However, he got pretty tired of the curfew time a 阅读全文
posted @ 2017-04-30 15:35 勿忘初心0924 阅读(199) 评论(0) 推荐(0) 编辑
摘要:给出两个字符串A B,求A与B的最长公共子序列(子序列不要求是连续的)。 比如两个串为: abcicba abdkscab ab是两个串的子序列,abc也是,abca也是,其中abca是这两个字符串最长的子序列。 给出两个字符串A B,求A与B的最长公共子序列(子序列不要求是连续的)。 比如两个串为 阅读全文
posted @ 2017-03-30 21:03 勿忘初心0924 阅读(1427) 评论(0) 推荐(0) 编辑
摘要:Description Description Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her 阅读全文
posted @ 2017-03-19 20:02 勿忘初心0924 阅读(830) 评论(0) 推荐(0) 编辑
摘要:Phalanx Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2859 Phalanx Submit Status Practice HDU 2859 D 阅读全文
posted @ 2017-03-17 15:55 勿忘初心0924 阅读(367) 评论(0) 推荐(0) 编辑
摘要:Treats for the Cows Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3186 Treats for the Cows Submit St 阅读全文
posted @ 2017-03-16 22:11 勿忘初心0924 阅读(1048) 评论(1) 推荐(0) 编辑
摘要:Tickets Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Tickets Submit Status Description Jesus, what a great movie 阅读全文
posted @ 2017-03-15 13:32 勿忘初心0924 阅读(246) 评论(0) 推荐(0) 编辑
摘要:B - Ignatius and the Princess IV Time Limit:1000MS Memory Limit:32767KB 64bit IO Format:%I64d & %I64u Submit Status B - Ignatius and the Princess IV S 阅读全文
posted @ 2017-03-14 13:05 勿忘初心0924 阅读(241) 评论(0) 推荐(0) 编辑
摘要:A - Max Sum Plus Plus Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status A - Max Sum Plus Plus Submit Status Descripti 阅读全文
posted @ 2017-03-14 09:47 勿忘初心0924 阅读(1932) 评论(2) 推荐(2) 编辑
摘要:例如:1 7 3 5 9 4 8 一个序列,比如说a[]={1,7,3,5,9,4,8},找出它的最长上升子序列的个数,很明显是4个,可以是{1,3,5,9},{1,3,5,8}或者{1,3,4,8}。具体怎么实现呢?直观的感受是以1为起点,下一个数比1大,那么上升子序列个数就+1,以1为起点,到7 阅读全文
posted @ 2016-07-30 13:25 勿忘初心0924 阅读(368) 评论(0) 推荐(0) 编辑