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摘要: 传送门:http://pan.baidu.com/s/1qW6tgS0T1:由于题目中要求的东西不好妨碍正常的最短路进行,那么可以枚举它。也就是二分答案+最短路检测,我写的dijk+heap+读入优化过了 1 #include 2 #include 3 #include 4 #include... 阅读全文
posted @ 2014-10-31 11:14 ZJDx1998 阅读(199) 评论(2) 推荐(0) 编辑
摘要: 传送门String painterTime Limit: 5000/2000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Problem DescriptionThere are two strings A and B with ... 阅读全文
posted @ 2014-10-28 19:26 ZJDx1998 阅读(150) 评论(0) 推荐(0) 编辑
摘要: 传送RunningTime Limit:1000MSMemory Limit:65536KDescriptionThe cows are trying to become better athletes, so Bessie is running on a track for exactlyN(1 ... 阅读全文
posted @ 2014-10-28 10:50 ZJDx1998 阅读(129) 评论(0) 推荐(0) 编辑
摘要: 传送门@百度Multiplication PuzzleTime Limit: 1000MSMemory Limit: 65536KDescriptionThe multiplication puzzle is played with a row of cards, each containing a... 阅读全文
posted @ 2014-10-27 11:29 ZJDx1998 阅读(130) 评论(0) 推荐(0) 编辑
摘要: Brackets SequenceTime Limit: 1000MSMemory Limit: 65536KSpecial JudgeDescriptionLet us define a regular brackets sequence in the following way: 1. Empt... 阅读全文
posted @ 2014-10-27 10:52 ZJDx1998 阅读(136) 评论(0) 推荐(0) 编辑
摘要: 地址见 BZOJ 2748~2750Description一个吉他手准备参加一场演出。他不喜欢在演出时始终使用同一个音量,所以他决定每一首歌之前他都要改变一次音量。在演出开始之前,他已经做好了一个列表,里面写着在每首歌开始之前他想要改变的音量是多少。每一次改变音量,他可以选择调高也可以调低。音量用一... 阅读全文
posted @ 2014-10-26 20:14 ZJDx1998 阅读(267) 评论(0) 推荐(0) 编辑
摘要: 传送门@百度Cheapest PalindromeTime Limit: 2000MSMemory Limit: 65536KDescriptionKeeping track of all the cows can be a tricky task so Farmer John has instal... 阅读全文
posted @ 2014-10-24 21:06 ZJDx1998 阅读(121) 评论(0) 推荐(0) 编辑
摘要: 传送门@百度BracketsTime Limit: 1000MSMemory Limit: 65536KDescriptionWe give the following inductive definition of a “regular brackets” sequence:the empty s... 阅读全文
posted @ 2014-10-24 20:06 ZJDx1998 阅读(147) 评论(0) 推荐(0) 编辑
摘要: 传送门Halloween Costumes Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %lluAppoint description:DescriptionGappu has a ... 阅读全文
posted @ 2014-10-24 19:27 ZJDx1998 阅读(476) 评论(0) 推荐(0) 编辑
摘要: 地址:http://pan.baidu.com/s/1hq03BIOT1:之前想错了= = f[i]表示前i位,且第i位不选的最小价值 f[i]=min{f[j]+A[i]} ,那么ans=TOT-f[i] n-k#include#include#include#include#include#in... 阅读全文
posted @ 2014-10-22 21:16 ZJDx1998 阅读(463) 评论(0) 推荐(0) 编辑
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