MZL's xor
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 456 Accepted Submission(s): 322
Problem Description
MZL loves xor very...
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其他好文 时间：
2015-08-05 16:33:14
阅读次数：
126
MZL's xor
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 310 Accepted Submission(s): 225
Problem Description
MZL loves xor very...
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其他好文 时间：
2015-08-05 01:07:26
阅读次数：
121
HDU - 4360
As long as Binbin loves Sangsang
Time Limit: 1000MS
Memory Limit: 32768KB
64bit IO Format: %I64d & %I64u
Submit Status
Description
Binbin misses Sangsan...
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其他好文 时间：
2015-08-04 23:08:52
阅读次数：
191
原文地址——http://blog.csdn.net/qq525099302/article/details/47280211
user_sequences中保存着当前用户的所有序列信息，可以从这张系统内置表中查询当前序列值select seq.last_number from user_sequences seq where seq.sequence_name='SEQ_B_ORDER_ID';...
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其他好文 时间：
2015-08-04 19:17:44
阅读次数：
110
题目如下：
Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payment...
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其他好文 时间：
2015-08-03 14:32:21
阅读次数：
262
Description
Urej loves to play various types of dull games. He usually asks other people to play with him. He says that playing those games can show his extraordinary wit. Recently Urej takes a gre...
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其他好文 时间：
2015-08-02 13:44:34
阅读次数：
138
BearPlaysDiv2Problem StatementLimak is a little bear who loves to play. Today he is playing by moving some stones between three piles of stones. Initi...
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其他好文 时间：
2015-08-02 06:21:20
阅读次数：
141
Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, whic...
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其他好文 时间：
2015-08-01 15:42:41
阅读次数：
107
pog loves szh IITime Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2115Accepted Submission(s): 609Pro...
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其他好文 时间：
2015-07-31 18:15:58
阅读次数：
141
//给一串序列，找出长度大于2，且相邻两个数的差值不大于d的数子序列的个数
//dp[u] 表示以u为最后一个点满足条件的序列个数
//dp[u] = segma(dp[v] + 1) a[u] - a[v] <= d ;
//将用树状数组来找这个求和
//不过由于没有给a[i] , 所以需要对每个数编号，然后用二分找其对应的编号
#include
#include...
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编程语言 时间：
2015-07-31 15:02:11
阅读次数：
150
