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今日已更新 1614 篇代码解决方案
传送门：HDU_1542AtlantisProblem DescriptionThere are several ancient Greek texts that contain descriptions of the fabled island Atlantis. So
https://www.u72.net/daima/ncxex.html - 2024-08-08 09:44:56 - 代码库Sereja and BracketsTime Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64uSubmit StatusDescriptionSereja has a bracket
https://www.u72.net/daima/ncuk0.html - 2024-08-08 06:10:22 - 代码库好久没更新了。。于是节操掉尽python水过本来就水的题。。 1 n,d=map(int, raw_input().split())2 if d==0:3 print 14 else:5 f=[1]6 for
https://www.u72.net/daima/ncce7.html - 2024-08-08 04:08:48 - 代码库1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 *
https://www.u72.net/daima/nn5u5.html - 2024-08-01 01:52:51 - 代码库题目链接：hdu 5091 Beam Cannon题目大意：给定N个点，现在要有一个W?H的矩形，问说最多能圈住多少个点。解题思路：线段的扫描线，假设有点(x,y)，那么(x,y)~(x+
https://www.u72.net/daima/nn6k9.html - 2024-08-01 02:35:16 - 代码库Relief grainTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 100000/100000 K (Java/Others) Total Submission(s): 1254 Accepted Sub
https://www.u72.net/daima/nzdfr.html - 2024-08-01 12:30:55 - 代码库Description共有m部电影，编号为1~m，第i部电影的好看值为w[i]。在n天之中（从1~n编号）每天会放映一部电影，第i天放映的是第f[i]部。你可以选择l,r(1<=l<=r<=n)
https://www.u72.net/daima/nr0v9.html - 2024-08-09 12:12:36 - 代码库CraneTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 3777 Accepted: 1031 Special JudgeDescriptionACM has bought a new crane (crane
https://www.u72.net/daima/nk88h.html - 2024-08-04 12:19:15 - 代码库Just a HookTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18378 Accepted Submis
https://www.u72.net/daima/ndz8x.html - 2024-08-04 17:49:33 - 代码库题目链接：点击打开链接题意：给定一张银行卡的n条记录：val day.mon hour:min表示这张卡在这个时间有一条交易。输出n行，对于输出的第i行表示：根据前i件记录，可
https://www.u72.net/daima/nfbz6.html - 2024-08-07 00:29:25 - 代码库http://www.lydsy.com/JudgeOnline/problem.php?id=3223默默的。。#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <
https://www.u72.net/daima/nfh4u.html - 2024-08-06 22:17:14 - 代码库Digits CountTime Limit:10000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64uSubmit StatusDescriptionGiven N integers A={A[0],
https://www.u72.net/daima/nbm4n.html - 2024-08-06 18:41:25 - 代码库题目大意：给出一个序列，求出这个序列的逆序对数量。定义一种操作，将一个数和他后面比他小的数字拿出来排序， 然后再放回去，之后输出逆序对数。思路：思路题。
https://www.u72.net/daima/nb6z4.html - 2024-08-06 13:04:03 - 代码库HighwaysPOJ 2485so that it will be possible to drive between any pair of towns without leaving the highway system. Flatopian towns are num
https://www.u72.net/daima/nbk35.html - 2024-08-05 21:07:50 - 代码库Bad CowtractorsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10885 Accepted: 4586DescriptionBessie has been hired to build a che
https://www.u72.net/daima/nbk8w.html - 2024-08-05 21:17:27 - 代码库<?php$arrData= http://www.mamicode.com/array("id" => "1", "pid" => "0", "name" => "山东" ), array( "id" => "2", "pid" => &quo
https://www.u72.net/daima/ndeff.html - 2024-08-05 15:11:00 - 代码库对应ZOJ题目：点击打开链接Fall the BrickTime Limit: 3000MS Memory Limit: 32768KB 64bit IO Format: %lld & %lluSubmit StatusDescriptionNow the G
https://www.u72.net/daima/nn74s.html - 2024-08-01 03:56:10 - 代码库因为仅仅看到13.3节，所以临时仅仅实现旋转和插入函数。思考：假设不带父结点，那么在须要訪问z的父结点时，我们能够借助查找函数从根结点到待查找结点z
https://www.u72.net/daima/nn4ad.html - 2024-08-01 00:35:40 - 代码库数的构造同插入函数那篇文章的构造。主要增加了查找父节点和删除的函数，同时为了方便删除了插入的函数。获取父节点的函数如下： 1 BTree *getFather(BTre
https://www.u72.net/daima/ndfu0.html - 2024-08-04 22:03:31 - 代码库题目链接：http://poj.org/problem?id=3764 分析：好题！武森09年的论文中有道题CowXor，求的是线性结构上的，连续序列的异或最大值，用的办法是先预处理出前n项
https://www.u72.net/daima/nd2rr.html - 2024-08-05 07:44:01 - 代码库