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LightOJ 1236 Pairs Forming LCM (LCM 唯一分解定理 + 素数筛选)
http://lightoj.com/volume_showproblem.php?problem=1236Pairs Forming LCMTime Limit:2000MSMemory Limit:32768KB64bit IO Format:%lld & %lluSubmitStatusPra...
分类:其他好文   时间:2015-11-04 19:31:32    阅读次数:324
[LeetCode]Swap Nodes in Pairs
题目描述:(链接)Given a linked list, swap every two adjacent nodes and return its head.For example,Given1->2->3->4, you should return the list as2->1->4->3.Y...
分类:其他好文   时间:2015-11-01 19:13:14    阅读次数:124
LeetCode #24 Swap Nodes in Pairs (M)
[Problem]Given a linked list, swap every two adjacent nodes and return its head.For example,Given1->2->3->4, you should return the list as2->1->4->3.Y...
分类:其他好文   时间:2015-10-24 06:39:24    阅读次数:200
Swap Nodes in Pairs - LeetCode
Given a linked list, swap every two adjacent nodes and return its head.For example,Given1->2->3->4, you should return the list as2->1->4->3.Your algor...
分类:其他好文   时间:2015-10-24 06:38:27    阅读次数:157
hdu 1394 Minimum Inversion Number 线段树
Problem DescriptionThe inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i aj.For a given sequ....
分类:其他好文   时间:2015-10-17 17:33:11    阅读次数:164
LeetCode解题报告--Swap Nodes in Pairs
题目: Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant...
分类:其他好文   时间:2015-10-17 12:16:08    阅读次数:179
BZOJ1807 : [Ioi2007]Pairs 彼此能听得见的动物对数
一维的情况:排序后维护一个单调指针即可,时间复杂度$O(n\log n)$。二维的情况:旋转坐标系后转化为二维数点问题,扫描线+树状数组维护即可,时间复杂度$O(n\log n)$。三维的情况:将后两维旋转坐标系,对于每个x,预处理出横坐标为x的点的后两维的二维前缀和。枚举一个点,再枚举另一个点的x...
分类:其他好文   时间:2015-10-07 01:08:41    阅读次数:262
[CareerCup] 9.6 Generate Parentheses 生成括号
9.6 Implement an algorithm to print all valid (e.g., properly opened and closed) combinations of n-pairs of parentheses.EXAMPLEInput: 3Output: ((())),...
分类:其他好文   时间:2015-09-22 14:20:34    阅读次数:133
获取url参数
function?GetUrlParms()?{ var?args?=?new?Object(); var?query?=?window.location.search.substring(1);//?获取查询串 var?pairs?=?query.split("&");//?在逗号处断开 for?(var?i?=?0...
分类:Web程序   时间:2015-09-20 16:24:18    阅读次数:130
Magic Pairs - SGU 119(同余)
题目大意:如果A0*X + B0*Y能够整除 N,求出来多有少A*X+B*Y 也能够整除去N,求出所有的A,B(0 (K * A0) % N = A, (K * B0) % N = B, (k=[0....N))。ps.记得排序去重复......代码如下:======================...
分类:其他好文   时间:2015-09-18 13:28:56    阅读次数:191
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