时间限制:0.25s 空间限制:4M Solution n=10000,m=100000,显然不能用矩阵乘法乘出来. S= ATA 对于矩阵S的一行,所有在A矩阵中1位置的元素都相等,并且都等于这一行1的个数之和.假设有k个1,这一行的和显然是k*k 由此只要统计每一行有多少个1,累加它的平方就可以了.O(n)的时间解决. code #include<cstdio> ], n, m, x, y; int main() { scanf ("%d %d", &n, &a…

                             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) 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…

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…

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…

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.…

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 (…

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: 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…

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…

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 …

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…

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…

题目大意:原题链接 给定三个n*n的矩阵A,B,C,验证A*B=C是否成立. 所有解法中因为只测试一组数据,因此没有使用memset清零 Hint中给的傻乎乎的TLE版本: #include<cstdio> #include<cstring> ][]; ][],C[][]; ][]) { ;i<=n;i++){ ;j<=n;j++) scanf("%d",&m[i][j]); } } int main() { scanf("%d&q…

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…

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: 32768KB This problem will be judged on ZJU. Original ID: 231664-bit integer IO format: %lld      Java class name: Main   Let us consider undirected graph G = <v, e="">which has N verti…

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…

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 = | -…

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 =…

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…

［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…

Let us consider undirected graph G = which has N vertices and M edges. Incidence matrix of this graph is N * M matrix A = {a ij}, such that a ij is 1 if i-th vertex is one of the ends of j-th edge and 0 in the other case. Your task is to find the sum…

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 positive integer n (n ≤ 500) followed by the the three matrices A, B and C respectively. Each matrix's descript…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4920 解题报告:求两个800*800的矩阵的乘法. 参考这篇论文:http://wenku.baidu.com/link?url=261XeEzH-AZkFGPiN63t1nnojoQF50yiuMoviHroGjVXjjRlxFcvWLcws0jgQcmZo4oA9BJcjnPxVreWRu-XXa9zb6r5gUUTxmBXn_qWSsu&qq-pf-to=pcqq.group 我看过了,只是简…

题目链接 随机算法使劲水...srand((unsigned)time(0))比srand(NULL)靠谱很多,可能是更加随机. #include <cstdio> #include <cstdlib> #include <cstring> #include <map> #include <ctime> #include <cmath> using namespace std; #define LL __int64 ][],b[][]…