不要问窝 为什么过了> < 窝也不造为什么就过了 说是%3变成稀疏矩阵 可是随便YY个案例都会超时.. . 看来数据是随机的诶 #include <stdio.h> #include <string.h> #include <stdlib.h> #include <limits.h> #include <malloc.h> #include <ctype.h> #include <math.h> #includ…

hdu4920 Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 568    Accepted Submission(s): 225 Problem Description Given two matrices A and B of size n×n, find the product o…

矩阵相乘,采用一行的去访问,比采用一列访问时间更短,根据数组是一行去储存的.神奇小代码. Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1476    Accepted Submission(s): 650 Problem Description Given two matrices A…

各种TEL,233啊.没想到是处理掉0的情况就能够过啊.一直以为会有极端数据.没想到居然是这种啊..在网上看到了一个AC的奇妙的代码,经典的矩阵乘法,仅仅只是把最内层的枚举,移到外面就过了啊...有点不理解啊,复杂度不是一样的吗.. Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 640 …

Matrix Multiplication Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18928   Accepted: 4074 Description You are given three n × n matrices A, B and C. Does the equation A × B = C hold true? Input The first line of input contains a posit…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1775    Accepted Submission(s): 796 Problem Description Given two matrices A and B of size n×n, find the product of them.…

题目链接 题意 : 给你三个n维矩阵,让你判断A*B是否等于C. 思路 :优化将二维转化成一维的.随机生成一个一维向量d,使得A*(B*d)=C*d,多次生成多次测试即可使错误概率大大减小. #include <stdio.h> #include <string.h> #include <time.h> #include <stdlib.h> #include <iostream> using namespace std ; ][],b[][],…

先把两个矩阵全都mod3. S[i][j][k]表示第i(0/1)个矩阵的行/列的第k位是不是j(1/2). 然后如果某两个矩乘对应位上为1.1,乘出来是1: 1.2:2: 2.1:2: 2.2:1. 然后分这四种情况把bitset and 起来,然后用count()数一下个数,计算下对答案矩阵该位置的贡献即可. #include<cstdio> #include<bitset> using namespace std; #define N 801 int n,x; bitset&…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 820    Accepted Submission(s): 328 Problem Description Given two matrices A and B of size n×n, find the product of them. b…

意甲冠军  由于矩阵乘法计算链表达的数量,需要的计算  后的电流等于行的矩阵的矩阵的列数  他们乘足够的人才  非法输出error 输入是严格合法的  即使仅仅有两个相乘也会用括号括起来  并且括号中最多有两个 那么就非常easy了 遇到字母直接入栈  遇到反括号计算后入栈  然后就得到结果了 #include<cstdio> #include<cctype> #include<cstring> using namespace std; const int N = 10…