Matt Pietrek October 1996 MSJ Matt Pietrek is the author of Windows 95 System Programming Secrets (IDG Books, 1995). He works at NuMega Technologies Inc., and can be reached at 71774.362@compuserve.com. QWay back in your July 1994 column, you wrote a…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5112 A Curious Matt Description There is a curious man called Matt. One day, Matt's best friend Ted is wandering on the non-negative half of the number line. Matt finds it interesting to know the maximal…

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others) Total Submission(s): 3215 Accepted Submission(s): 1261 Problem Description Matt has N friends. They are playing a game together. Each of Matt's frie…

A Curious Matt Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5112 Description There is a curious man called Matt. One day, Matt's best friend Ted is wandering on the non-negative half of the number line. Matt f…

A Curious Matt Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 1338    Accepted Submission(s): 735 Problem Description There is a curious man called Matt. One day, Matt's best friend Ted is wa…

Matt has N friends. They are playing a game together.Each of Matt's friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends'magic numbers is no less than M , Mat…

Description There is a curious man called Matt. One day, Matt's best friend Ted is wandering on the non-negative half of the number line. Matt finds it interesting to know the maximal speed Ted may reach. In order to do so, Matt takes records of Ted'…

Google的人力运营高级副总裁Laszlo Bock在一次采访中说Google发现在面试程序员时问智力题完全是浪费时间,Matt Rogish在他的这篇博客How to Interview Programmers - Rogish Reading Writing中有感而发,介绍了他认为正确的面试程序员的方法. 先决条件 在你雇佣第一个员工之前 想好你的企业文化 让你团队中的每个人都面试应聘者 确保你的团队中每个人都知道如何面试,制定面试计划和清单. 每个人都写下来他要问什么问题,然后大家坐在一…

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)Total Submission(s): 1810    Accepted Submission(s): 715 Problem Description Matt has N friends. They are playing a game together. Each of Matt’s…

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others) Problem Description Matt has N friends. They are playing a game together. Each of Matt’s friends has a magic number. In the game, Matt selects so…

30 天尝试新事物matt cutts : try something new for 30 days[小计划帮你实现大目标] 是否有些事情, 你一直想去做, 但就是没有实现?马特 ?卡茨建议: 尝试 30 天.这个简短而轻松愉快的演讲提出了一个简洁方法, 用来考虑制定和实现目标.马特是 google 所有工程师中最广为人知的一个, 他提出在行动前我们不妨先来考虑制定一个短期计划来实现目标.is there something youve always meant to do, wanted t…

A Curious Matt Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3058    Accepted Submission(s): 1716 Problem Description There is a curious man called Matt. One day, Matt's best friend Ted is w…

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others) Total Submission(s): 700 Accepted Submission(s): 270 Problem Description Matt has N friends. They are playing a game together. Each of Matt's friend…

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)Total Submission(s): 5188    Accepted Submission(s): 1985 Problem Description Matt has N friends. They are playing a game together. Each of Matt’s…

Happy Matt Friends Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)Total Submission(s): 6164    Accepted Submission(s): 2330 Problem Description Matt has N friends. They are playing a game together. Each of Matt’s…

题目链接:HDU 5119 Problem Description Matt has N friends. They are playing a game together. Each of Matt's friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends'ma…

描述 Matt hzs N friends. They are playing a game together. Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends’magic numbers is no less than M ,…

 这是一首很感动的主内歌曲,听了无首次,还是很感动,这里把歌词贴出来,一方面是为了记忆歌词,另一方面是为以后怀念记忆.(20:44:38) Bless the lord,oh my soul oh my soul worship his holy name sing like never before oh my soul I'll worship your holy name the sun comes up ,it's a new day dawning it's time to sing…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5112 解题报告:扫一遍 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> using namespace std; ; struct node { double t,x; }rec[maxn]; double Max(do…

题意:给定n个数,从中分别取出0个,1个,2个...n个,并把他们异或起来,求大于m个总的取法. 思路:dp,背包思想,考虑第i个数,取或者不取,dp[i][j]表示在第i个数时,异或值为j的所有取法.dp[i][j] = dp[i - 1][j] + dp[i - 1][j ^ a[i]]); 其中dp[i - 1][j]表示不取第i个数,dp[i - 1][j ^ a[i]]表示取第i个数,由于40比较大,所以用滚动数组优化,后一个状态需要前一个来推导,而和前一个之前的所有的没有关系,所以之…

虽然是一道还是算简单的DP,甚至不用滚动数组也能AC,数据量不算很大. 对于N个数,每个数只存在两个状态,取 和 不取. 容易得出状态转移方程: dp[i][j] = dp[i - 1][j ^ a[i]] + dp[i - 1][j]; dp[i][j] 的意思是,对于数列 中前 i 个数字,使得 XOR 和恰好为 j 的方案数 状态转移方程中的 dp[i - 1][j] 即表示当前这个数字不取, dp[i - 1][j ^ a[i]] 表示当前这个数字要取. 这道题还是要好好理解阿! sou…

https://github.com/PacktPublishing/Game-Development-Patterns-and-Best-Practices https://github.com/mattCasanova/Mach5 1. Introduction to Design Patterns (已看) 2. One Instance to Rule Them All - Singletons (已看) 3. Creating Flexibility with the Componen…

题目链接 题意:n个数,你可以从中选一些数,也可以不选,选出来的元素的异或和大于m时,则称满足情况.问满足情况的方案数为多少. 分析:本来以为是用什么特殊的数据结构来操作,没想到是dp,还好队友很强.定义dp[i][j]为在前i个数里选一些数的异或和为j的方案数,边计算边统计, #include <iostream> #include <cstdio> #include <cstring> typedef long long ll; <<); using n…

http://acm.hdu.edu.cn/showproblem.php?pid=5119 题意:给出n个数和一个上限m,求从这n个数里取任意个数做异或运算,最后的结果不小于m有多少种取法. 思路:dp[i][j]表示在前i个数中取数做异或运算最后结果为j的方法数,那么dp[i][j] = dp[i-1][j] + dp[i-1][j^a[i]],分别对于取与不取这两种状态. 因为要推导出第i的状态只有i-1有关,所以这里可以用滚动数组. #include<cstdio> #include&…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5119 题目: 题意: 求选择任意个数,使其异或和大于等于m的方案数. 思路: 每个数有选和不选两种方案,显然是背包思想.dp[i][j]表示前i个物品异或和为j时的方案数,转移方程为dp[i][j] = dp[i-1][j] + dp[i-1][j^a[i]].这题可以考虑用滚动数组滚动掉一维,当然了,不滚动也是可以过滴- 代码实现如下: #include <set> #include <…

题目链接 题意 : 给你n个数,让你从中挑K个数(K<=n)使得这k个数异或的和小于m,问你有多少种异或方式满足这个条件. 思路 : 正解据说是高斯消元.这里用DP做的,类似于背包,枚举的是异或的和,给定的数你可以选择放或者不放,dp[i][j]代表的是前 i 个数中选择k个异或的和为j. #include <stdio.h> #include <string.h> #include <iostream> #define LL long long using na…

题目链接:点击打开链接 题目描写叙述:给出n个数.求从这n个数中随意取出一些数(至少取一个)相互异或之后大于m的方案数? 解题思路:分析因为n<=40&&m<=10^6,因此我们能够枚举全部可能的异或值.时间复杂度40*10^6. 採用动态规划的思想dp[i&1][j]=d[(i-1)&1][j]+d[(i-1)&1][j^a[i]];因为40*10^6太大,所以此处我们採用滚动数组 代码: #include <cstdio> #includ…

这篇文章假定你熟悉C++和Win32. 概述 理解可移植可执行文件格式(PE)可以更好地了解操作系统.如果你知道DLL和EXE中都有些什么东西,那么你就是一个知识渊博的程序员.这一系列文章的第一部分,讨论最近这几年PE格式所发生的变化. 这次更新后,作者讨论了PE格式如何适应于用.NET开发的应用程序,包括PE节,RVA,数据目录,以及导入函数.附录中包含了相关的映像头结构以及它们的描述. 很早以前,我为微软系统期刊(现在叫做MSDN)写了一篇文章.那篇文章“Peering Inside the…

1. Displaying Data with Core UI Elements (已看) 2. Responding to User Events for Interactive UIs (已看) 3. Inventory UIs (已看) 4. Playing and Manipulating Sounds 5. Creating Textures, Maps, and Materials 6. Shader Graphs and Video Players 7. Using Cameras…

题意:有最多40个数字,取任意个数字他们的异或和>=k则是可行的方案,问有多少种可行的方案. 分析:dp[now][j]表示当前这个值的种类数,那么转移方程为dp[now][j] = dp[pre][j] + dp[pre][j^a[i]].因为a^b=c的话,c^b=a,所以j^a[i]就可以得到通过异或a[i]转移到j的那个状态了.然后可以用滚动数组压缩一下. 代码如下: #include <stdio.h> #include <algorithm> #include &…