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ZOJ 3772 Calculate the Function (线段树 + 矩阵)
思路分析: 遗憾不知道矩阵的构造。线段树上比较水的矩阵。。。  M[x]  =  [1 A[x]]               [1   0 ]   就有 [  F[R]  ]    =   M[R] * M[R-1] *  ... * M[L+2] * [F[L+1]] [F[R-1]]                                       ...
分类:其他好文   时间:2014-05-13 08:45:03    阅读次数:270
[迭代加深dfs] zoj 3768 Continuous Login
题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3768 Continuous Login Time Limit: 2 Seconds      Memory Limit: 131072 KB      Special Judge Pierre is recently obsessed...
分类:其他好文   时间:2014-05-13 07:28:07    阅读次数:412
zoj 3557 How Many Sets II
How Many Sets IITime Limit:2 Seconds Memory Limit:65536 KBGiven a setS= {1, 2, ...,n}, numbermandp, your job is to count how many setTsatisfies the fo...
分类:其他好文   时间:2014-05-12 13:05:52    阅读次数:238
ZOJ 1151 Word Reversal反转单词 (string字符串处理)
链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=151 For each list of words, output a line with each word reversed without changing the order of the words. This problem contains multi...
分类:其他好文   时间:2014-05-11 22:20:43    阅读次数:443
ZOJ 2109 FatMouse' Trade (背包 dp + 贪心)
链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1109 FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean...
分类:其他好文   时间:2014-05-11 20:57:36    阅读次数:400
ZOJ 2724 Windows 消息队列 (优先队列)
链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2724 Message queue is the basic fundamental of windows system. For each process, the system maintains a message queue. If something h...
分类:Windows程序   时间:2014-05-11 20:31:56    阅读次数:796
ZOJ - 1880 Tug of War
题意:求在两边人数不相差超过1个的情况下,实力尽量相等的情况 思路:从实力和的一半开始类背包操作 #include #include #include #include using namespace std; const int MAXN = 45010; const int MAXM = 110; int a[MAXM]; int dp[MAXN][MAXM]; int n; ...
分类:其他好文   时间:2014-05-11 05:02:03    阅读次数:279
[最小表示] zoj 1729 Hidden Password
题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=729 题目意思:...
分类:其他好文   时间:2014-05-11 01:52:41    阅读次数:367
zoj 1648 Circuit Board
题目:意思就是判断给定的几条线段是否有相交的。 方法:模版吧,有空在来细细学习。 代码: #include #include using namespace std; struct Point { double x,y; }; struct LineSeg { Point a,b; }; double Cross(Point a, Point b, Poi...
分类:其他好文   时间:2014-05-10 03:39:07    阅读次数:326
[容斥原理] zoj 3556 How Many Sets I
题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4535 How Many Sets I Time Limit: 2 Seconds      Memory Limit: 65536 KB Give a set S, |S| = n, then how many ordered set g...
分类:其他好文   时间:2014-05-09 13:50:00    阅读次数:347
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