Cow Multiplication Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12210   Accepted: 8509 Description Bessie is tired of multiplying pairs of numbers the usual way, so she invented her own style of multiplication. In her style, A*B is eq…

http://poj.org/problem?id=3318 题意:问A和B两个矩阵相乘能否等于C. 思路:题目明确说出(n^3)的算法不能过,但是通过各种常数优化还是能过的. 这里的随机算法指的是随机枚举矩阵C的一个位置,然后通过A*B计算是否能够得到矩阵C相应位置的数,如果不等,就直接退出了,如果跑过一定的数量后能够相等,那么就可以判断这个矩阵C等于A*B的.第一次见这样的题目...有点新奇. 暴力算法: #include <cstdio> using namespace std; ][]…

Cow Bowling Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13464   Accepted: 8897 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard…

题意:有n个奶牛,每个奶牛有一个smart值和一个fun值,可能为正也可能为负,要求选出n只奶牛使他们smart值的和s与fun值得和f都非负,且s+f值要求最大. 分析: 一道很好的背包DP题,我们将smart值当作物品的体积,将fun值当作物品的价值,每个物品只能取一次,我们求对于每个背包体积求恰好装满该体积时价值和的最大值,也就是当所选奶牛smart值为某个值时,fun值的和的最大值.然后对于每个非负背包体积(smart值的和),判断对应最大价值(fun值的和)是否非负,如果非负说明这是一…

Cow Multiplication Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13312   Accepted: 9307 Description Bessie is tired of multiplying pairs of numbers the usual way, so she invented her own style of multiplication. In her style, A*B is eq…

POJ 3617 Best Cow Line Time Limit: 1000MS Memory Limit: 65536K [Description] [题目描述] FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual "Farmer of the Year" competition. In this contest every farmer arranges his cows in a line and herds…

POJ 3268 Silver Cow Party Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects…

题目链接: Poj 3189 Steady Cow Assignment 题目描述: 有n头奶牛,m个棚,每个奶牛对每个棚都有一个喜爱程度.当然啦,棚子也是有脾气的,并不是奶牛想住进来就住进来,超出棚子的最大容量了,棚子是拒绝的.现在要给每个奶牛安家,本宝宝是一个公正无私的人类,所以要找一个奶牛喜爱程度差值最小的方案(虽然这样大家的喜爱程度可能普遍偏低,因为绝对公平并不代表合理啊),问喜爱程度的区间最小为多大? 解题思路: 每个棚子并不是住一个奶牛,所以是多重匹配咯.匹配的时候二分枚举喜爱程度的…

带来两题贪心算法的题. 1.给定长度为N的字符串S,要构造一个长度为N的字符串T.起初,T是一个空串,随后反复进行下面两个操作:1.从S的头部删除一个字符,加到T的尾部.2.从S的尾部删除一个字符,加到T的尾部.求你任意采取这两个步骤后能得到的最小字符串T 2.直线上有N个点,点i的位置是xi,这N个点中选择若干个做上标记,对于每个点,在他们距离为R的区域内必须带有标记点,求在满足这个条件的情况下,所需要标记点的最少个数. 1.POJ 3617 Best Cow Line http://poj.…

POJ 3252:Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10099 Accepted: 3669 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Sciss…