Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: Input: A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] Output: | 1 0 0 | | 7 0 0 | | 7…

Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] | 1 0 0 | | 7 0 0 | | 7 0 0 | AB = | -…

原题链接在这里:https://leetcode.com/problems/sparse-matrix-multiplication/description/ 题目: Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: A = [ [ 1, 0, 0], [-1, 0, 3] ]…

题目: Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] | 1 0 0 | | 7 0 0 | | 7 0 0 | AB =…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 暴力 科学计算库numpy 日期 题目地址:https://leetcode-cn.com/problems/sparse-matrix-multiplication/ 题目描述 Given two sparse matrices A and B, return the result of AB. You may assume that A's col…

［抄题］: 给定两个 稀疏矩阵 A 和 B,返回AB的结果.您可以假设A的列数等于B的行数. [暴力解法]: 时间分析: 空间分析: ［思维问题］: ［一句话思路］: 如果为零则不相乘,优化常数的复杂度. ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: 是ans[i][j] 元素变化来进行具体的加减操作,而不是无参数的ans[][] 用来声明,搞混了 ［二刷］: ［三刷］: ［四刷］: ［五刷］: [五分钟肉眼debug的…

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.…

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 …

Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] | 1 0 0 | | 7 0 0 | | 7 0 0 | AB = | -…

Problem Description: Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] | 1 0 0 | | 7 0 0…

Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] | 1 0 0 | | 7 0 0 | | 7 0 0 | AB = | -…

Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ], [ 0, 0, 1 ] ] | 1 0 0 | | 7 0 0 | | 7 0 0 | AB = | -…

sparse matrix是用来存储大型稀疏矩阵用得,单细胞表达数据基本都用这个格式来存储,因为单细胞很大部分都是0,用普通文本矩阵存储太占空间. 使用也是相当简单: library("Matrix") readsCount <- read.csv("data/count.csv", header = T, row.names = 1) readsCountSM <- as(as.matrix(readsCount), "dgCMatrix&q…

本文首发于个人博客https://kezunlin.me/post/1e37a6/,欢迎阅读最新内容! opencv and numpy matrix multiplication vs element-wise multiplication Guide opencv Matrix multiplication is where two matrices are multiplied directly. This operation multiplies matrix A of size [a…

w https://en.wikipedia.org/wiki/Sparse_matrix 稀疏矩阵存储格式总结+存储效率对比:COO,CSR,DIA,ELL,HYB - Bin的专栏 - 博客园http://www.cnblogs.com/xbinworld/p/4273506.html 稀疏矩阵的存储格式(Sparse Matrix Storage Formats) - Donkey Vision - 博客频道 - CSDN.NEThttp://blog.csdn.net/anshan198…

                                      稀疏矩阵相乘-Python版 Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: A = [ [ 1, 0, 0], [-1, 0, 3] ] B = [ [ 7, 0, 0 ], [ 0, 0, 0 ],…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given two matrices A and B of size n×n, find the product of them. bobo hates big integers. So you are only asked to find t…

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…

Matrix multiplication                                                                           Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given two matrices A and B of size n×n, find the…

以下CUDA sample是分别用C++和CUDA实现的两矩阵相乘运算code即C= A*B,CUDA中包含了两种核函数的实现方法,第一种方法来自于CUDA Samples\v8.0\0_Simple\matrixMul,第二种采用普通的方法实现,第一种方法较快,但有些复杂,速度上约为第二种的1.3倍,并对其中使用到的CUDA函数进行了解说,各个文件内容如下: funset.cpp: #include "funset.hpp" #include <random> #incl…

题目链接 Matrix multiplication 求矩阵A和B相乘的结果. 因为答案只要对3取模,所以我们可以通过一些方法来加速计算. 我们对两个矩阵各开两个bitset,分别存储模3余1和模3余2的数. 然后相乘的时候and一下就好了. c[i][j] = f(a_one[i] & b_one[j]) + f(a_one[i] & b_two[j]) * 2 + f(a_two[i] & b_one[j]) * 2 + f(a_two[i] & b_two[j]) $…

HDU 4920 Matrix multiplication 题目链接 题意:给定两个矩阵,求这两个矩阵相乘mod 3 思路:没什么好的想法,就把0的位置不考虑.结果就过了.然后看了官方题解,上面是用了bitset这个东西,能够用来存大的二进制数,那么对于行列相乘.事实上就几种情况,遇到0都是0了,1 1得1,2 1,1 2得2,2 2得1.所以仅仅要存下行列1和2存不存在分别表示的二进制数.然后取且bitcount一下的个数,就能够计算出对应的数值了 代码: 暴力: #include <cst…

                             Matrix Multiplication Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18173   Accepted: 3912 Description You are given three n × n matrices A, B and C. Does the equation A × B = C hold true? Input The first l…

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…

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…

Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. For example, Given n = 3, You should return the following matrix: [ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ]] 解题思路: 参考Java for LeetCode 054 Spiral Matrix,修改下…

D - Matrix Multiplication Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description       Let us consider undirected graph G = {V; E} which has N vertices and M edges. Incidence matrix of this g…

Problem Description Given two matrices A and B of size n×n, find the product of them. bobo hates big integers. So you are only asked to find the result modulo 3.   Input The input consists of several tests. For each tests: The first line contains n (…

Matrix Multiplication Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17783   Accepted: 3845 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…

poj3318 Matrix Multiplication 题意:给定$n*n(n<=500)$的矩阵$A,B,C$,如果$A*B==C$,输出“YES”,否则为“NO”:多组数据,$O(n^{3})$必T 我们可以随机生成一个神奇的列向量$v$,你可以把它看做n*1或1*n的矩阵,里面的元素随机为0或1 蓝后我们考察 $A*(B*v)$和$C*v$是否相等 灰常显然 $A*B!=C  ==>  A*(B*v)!=C*v$ 那么我们每次判断有$\frac{1}{2}$的概率判错 于是我们多判几…