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POJ 1144 Network
NetworkTime Limit: 1000msMemory Limit: 10000KBThis problem will be judged onPKU. Original ID:114464-bit integer IO format:%lld Java class name:MainA T...
分类:Web程序   时间:2014-08-25 22:28:34    阅读次数:340
POJ 1236 Network of Schools
Network of SchoolsTime Limit: 1000msMemory Limit: 10000KBThis problem will be judged onPKU. Original ID:123664-bit integer IO format:%lld Java class n...
分类:Web程序   时间:2014-08-25 19:08:04    阅读次数:201
连通分量个数(连通分量_并查集)
?? 连通分量个数Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Description  在无向图中,如果从顶点vi到顶点vj有路径,则称vi和vj连通。如...
分类:其他好文   时间:2014-08-25 11:56:04    阅读次数:204
POJ 3096 Surprising Strings
Surprising StringsTime Limit: 1000msMemory Limit: 65536KBThis problem will be judged onPKU. Original ID:309664-bit integer IO format:%lld Java class n...
分类:其他好文   时间:2014-08-25 11:40:14    阅读次数:234
POJ 2774 Long Long Message
Long Long MessageTime Limit: 4000msMemory Limit: 131072KBThis problem will be judged onPKU. Original ID:277464-bit integer IO format:%lld Java class n...
分类:其他好文   时间:2014-08-24 10:15:22    阅读次数:306
POJ 1226 Substrings
SubstringsTime Limit: 1000msMemory Limit: 10000KBThis problem will be judged onPKU. Original ID:122664-bit integer IO format:%lld Java class name:Main...
分类:其他好文   时间:2014-08-23 21:34:21    阅读次数:236
POJ 2217 Secretary
SecretaryTime Limit: 1000msMemory Limit: 65536KBThis problem will be judged onPKU. Original ID:221764-bit integer IO format:%lld Java class name:MainT...
分类:其他好文   时间:2014-08-23 21:30:11    阅读次数:218
UVA 12697 Minimal Subarray Length
Minimal Subarray LengthTime Limit: 3000msMemory Limit: 131072KBThis problem will be judged onUVALive. Original ID:660964-bit integer IO format:%lld Ja...
分类:其他好文   时间:2014-08-23 16:43:01    阅读次数:279
H - Lazier Salesgirl
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %lluSubmit Status Practice ZOJ 3607DescriptionKochiya Sanae is a lazy girl who makes an....
分类:其他好文   时间:2014-08-20 22:28:52    阅读次数:316
HDU 4961 Boring Sum 暴力
题意:对于所有的A[I],同时找到左边和右边离它最近且是它的倍数的数相乘最后加起来求和。解题思路:n*sqrt(n)的算法,开始以为过不了,wa了两发因为lld I64d对拍一个小时发现一个小时前交的代码没错只是没变I64d,。。具体思路是枚举每个a[i]的因子,找离它最近的那个更新,如果已经没更新...
分类:其他好文   时间:2014-08-20 01:20:46    阅读次数:261
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