boywgw的专栏

问题描述：Assistant Manager Lee is on a visit to a factory run by a partner company. A machine is cutting steel plates that are different in length and width. However, a closer look found that the

templatestruct LNode{ T data; struct LNode *next;};templateclass CircularLinkList{private: LNode* m_pList; int m_ListLength;private: bool InitList(); bool GetNode(int position, LNode**

strcpy和memcpy都是标准C库函数，它们有下面的特点。strcpy提供了字符串的复制。即strcpy只用于字符串复制，并且它不仅复制字符串内容之外，还会复制字符串的结束符。strcpy函数的原型是：char* strcpy(char* dest, const char* src);memcpy提供了一般内存的复制。即memcpy对于需要复制的内容没有限制，因

templatestruct LNode{ T data; struct LNode *next;};templateclass LinkList{private: LNode* m_pList; int m_ListLength;private: bool InitList(); bool GetNode(int position, LNode** node);

templateclass Stack{public: Stack(int s = 10); ~Stack();public: T top(); void push(T t); T pop(); bool empty(); int size();private: int count; int capacity; T* array;};templateStac

并查集：(union-find sets)一种简单的用途广泛的集合. 并查集是若干个不相交集合，能够实现较快的合并和判断元素所在集合的操作，应用很多，如其求无向图的连通分量个数等。最完美的应用当属：实现Kruskar算法求最小生成树。并查集的精髓（即它的三种操作，结合实现代码模板进行理解）：1、Make_Set(x) 把每一个元素初始化为一个集合初始化后每一

问题描述：You are making “sky map” to represent constellation (group of starts) in the sky.In each element, 1 means star, 0 means empty.And in a constellation, each start should have at least on

问题描述：A domino means a shape in which sides of two unit rectangles are put together to each other.Domino tilting means to make a mosaic by filling a certain area of a flat plate with dominoes,

问题描述： When you divide numbers from 1 to N into two subsets, you can make each subset sum equal. For example, if N is 3, elements of the entire set are {1, 2, & 3}. In case each set sum is equal,

算法描述：Korean coins consist of 6 levels of 1 won, 5 won, 10 won, 50 won, 100 won, and 500 won. If you make 256 won change, you need 5 coins: two coins of 100 won, one coin of 50 won, one coin of 5 w

题目描述： 已知一个 N 枚邮票的面值集合（如，{1 分，3 分}）和一个上限 K —— 表示信封上能够贴 K 张邮票。计算从 1 到 M 的最大连续可贴出的邮资。 例如，假设有 1 分和 3 分的邮票；你最多可以贴 5 张邮票。很容易贴出 1 到 5 分的邮资（用 1 分邮票贴就行了），接下来的邮资也不难： 6 = 3 + 3 7 = 3 + 3 + 1 8 =

1 问题描述某个汽车工厂共有两条装配线,每条有 n 个装配站。装配线 i 的第 j个装配站表示为 Si,j ,在该站的装配时间为 ai,j 。一个汽车底盘进入工厂,然后进入装配线 i(i 为 1 或 2),花费时间为 ei 。在通过一条线的第 j 个装配站后,这个底盘来到任一条装配线的第(j+1)个装配站。如果它留在相同的装配线,则没有移动开销。但是,如果它移动到另一条线上,则花费时间为 ti

问题描述：Given any sequence, the sequence of a contiguous area as the sum of the sum is part of a continuousFor example, let’s the following sequence was given In the above sequence, the sum

算法描述：In contemporary business administration, mathematical tools are frequently applied in order to develop the business theory.Numerical approaches are widely applied in the most of business ch

上次做poj 1067的取石子游戏，只用到了whthoff博弈，未涉及到取石子的异或方法，今天重新搜索，整理了一遍。搜罗各种资料，加上自己整理，终于成篇啦！……噼里啪啦 取石子问题有一种很有意思的游戏，就是有物体若干堆，可以是火柴棍或是围棋子等等均可。两个人轮流从堆中取物体若干，规定最后取光物体者取胜。这是我国民间很古老的一个游戏，别看这游戏极其简单，却蕴含着深刻的数学原理