Problem Description Baby Ming is fond of weight lifting. He has a barbell pole(the weight of which can be ignored) and two different kinds of barbell disks(the weight of which are respectively a and b), the amount of each one being infinite.Baby Ming…

Baby Ming and Weight lifting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 681    Accepted Submission(s): 280 Problem Description Baby Ming is fond of weight lifting. He has a barbell pole(the…

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…

Problem Description Baby Ming collected lots of cell phone numbers, and he wants to sell them for money. He thinks normal number can be sold for b yuan, while number with following features can be sold for a yuan. .The last five numbers are the same.…

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 <…

因为第一行和最后一行都是0,我们只需枚举最左边或最右边一列的01情况,即可得到整张表 然后再检验表是否符合要求 #include<cstdio> #include<cstring> #include<vector> #include<cmath> #include<queue> #include<list> #include<algorithm> using namespace std; int T; int n,m; ]…

#include<cstdio> #include<cstring> #include<vector> #include<cmath> #include<queue> #include<list> #include<algorithm> using namespace std; int T,n; ]; long long a,b; ],q2[]; long long f() { ]==s[]&&s[]==s…

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>…

B. Pasha and PhonePasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly n digits. Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a1, …