Baby Ming and Matrix games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 849    Accepted Submission(s): 211 Problem Description These few days, Baby Ming is addicted to playing a matrix game.…

题目链接 题解:题意为给出一个N*M的矩阵,然后(i∗2,j∗2) (i,j=0,1,2...)的点处是数字,两个数字之间是符号,其他位置是‘#’号. 但不知道是理解的问题还是题目描述的问题,数据中还有类似1#1这种数据存在,因此WA了4次,加上了一句代码后,马上AC了,该行代码在下文以斜粗体标出. 此外,因为里面有除法,会有一定误差,所以用这句来判断:fabs(tar-nnum)<=1e-8,而不是tar==nnum.   #include <cstdio> #include <…

Baby Ming and Matrix games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1210    Accepted Submission(s): 316 Problem Description These few days, Baby Ming is addicted to playing a matrix game…

 Baby Ming and Matrix games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1150    Accepted Submission(s): 298 Problem Description These few days, Baby Ming is addicted to playing a matrix g…

Baby Ming and Matrix games 题意: 给一个矩形,两个0~9的数字之间隔一个数学运算符(‘+’,’-‘,’*’,’/’),其中’/’表示分数除,再给一个目标的值,问是否存在从一个数字出发,以数字之间的运算符为运算,得到这个目标值:(每个数字只能用一次,其实说白了就是dfs..);可以则输出(Impossible),否则输出(Possible); 思路:坑点就是里面本来全是整数,但是一个除法运算却是分数形式,开始使用了很保险的分数保存,来避免误差的.但是无情WA了很多次..…

Problem Description These few days, Baby Ming is addicted to playing a matrix game. Given a n∗m matrix, the character ,j∗) (i,j=,,...) are the numbers between −. There are an arithmetic sign (‘+’, ‘-‘, ‘∗’, ‘/’) between every two adjacent numbers, ot…

题意: 给出一棵树,每个顶点上有个\(2 \times 2\)的矩阵,矩阵有两种操作: 顺时针旋转90°,花费是2 将一种矩阵替换为另一种矩阵,花费是10 树上有一种操作,将一条路经上的所有矩阵都变为给出的矩阵,并输出最小花费. 分析: 矩阵可以分为两类共6种,一类是两个1相邻的矩阵共4种:一类是两个1在对角线的矩阵共2种. 同一类矩阵可以通过旋转操作得到,否则只能用替换. 事先计算好每种矩阵转换到另外一种矩阵的最少花费,然后树链剖分再用线段树维护就好了. #include <cstdio>…

最近在看<算法>这本书,正好看到一个计算表达式的问题,于是就打算写一下,也正好熟悉一下Java集合框架的使用,大致测试了一下,没啥问题. import java.util.*; /* * * 用来计算表达式 * for example: 1+2*3*(4+3*1)-3*1+2+3/1; * (1+2*2-2*1*3*(1-1))*(1-2+3*(4+0)); * 注意点: * 2.输入的表达书不能还有空格,括号必须匹配 * 基本思想: * 1.建立操作数栈以及操作符栈 * 2.先去括号,每次遇…

3.2 Nash Equilibria in Two-Player Matrix Games For a two-player matrix game, we can set up a matrix with each element containing a reward for each joint action pair. Then the reward function…

 C#  字符串计算表达式 string str="4+4+2.1"; 要的效果: double sum=4+4+2.1: 方案一: 动态计算表达式: 1 public class Expression 2 { 3 object instance; 4 MethodInfo method; 5 /// <summary> 6 /// 表达试运算 7 /// </summary> 8 /// <param name="expression"…