题目1001:A+B for Matrices
Posted watchfree
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了题目1001:A+B for Matrices相关的知识,希望对你有一定的参考价值。
题目1001:A+B for Matrices
时间限制:1 秒内存限制:32 兆
题目描述:
This time, you are supposed to find A+B where A and B are two matrices, and then count the number of zero rows and columns.
输入:
The input consists of several test cases, each starts with a pair of positive integers M and N (≤10) which are the number of rows and columns of the matrices, respectively. Then 2*M lines follow, each contains N integers in [-100, 100], separated by a space. The first M lines correspond to the elements of A and the second M lines to that of B.
The input is terminated by a zero M and that case must NOT be processed.
输出:
For each test case you should output in one line the total number of zero rows and columns of A+B.
样例输入:
2 2
1 1
1 1
-1 -1
10 9
2 3
1 2 3
4 5 6
-1 -2 -3
-4 -5 -6
0
样例输出:
1
5
----------------
代码有bug,测试数据不够严谨. AC了。。。。
----------------------------------------------------
import java.util.Scanner; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub Scanner sc=new Scanner(System.in); while(sc.hasNext()){ int m=sc.nextInt(); if(m==0)break; int n=sc.nextInt(); int a[][]=new int[m][n]; int b[][]=new int[m][n]; int c[][]=new int[m][n]; int zorerow=0; int zorecol=0; for(int i=0;i<m;i++) for(int j=0;j<n;j++) a[i][j]=sc.nextInt(); for(int i=0;i<m;i++) for(int j=0;j<n;j++){ b[i][j]=sc.nextInt(); c[i][j]=a[i][j]+b[i][j]; } int temp=0; int tmp=0; for(int i=0;i<m;i++){ for(int j=0;j<n;j++){ temp+=c[i][j]; } if(temp==0)zorerow++; temp=0; } for(int j=0;j<n;j++){ for(int i=0;i<m;i++){ tmp+=c[i][j]; } if(tmp==0)zorerow++; tmp=0; } System.out.println(zorerow+zorecol); } } }
以上是关于题目1001:A+B for Matrices的主要内容,如果未能解决你的问题,请参考以下文章
论文笔记-Augmented Lagrange Multiplier Method for Recovery of Low-Rank Matrices
论文笔记-Augmented Lagrange Multiplier Method for Recovery of Low-Rank Matrices
论文笔记-Augmented Lagrange Multiplier Method for Recovery of Low-Rank Matrices