A+B for Matrices-白红宇

A+B for Matrices

阅读量：7103 次

发布时间：2019-06-28

本文共 1999 字，大约阅读时间需要 6 分钟。

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：4483

解决：1938

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 21 11 1-1 -110 92 31 2 34 5 6-1 -2 -3-4 -5 -60

样例输出：

1 #include 
       
          2 #include 
        
           3 #include 
         
            4 #include 
          
             5 #include 
           
              6 #include 
            
              7 8 #include 
             
               9 #include 
              
                10 #include 
               
                 11 #include 
                
                  12 #include 
                 
                   13 #include 
                  
                    14 #include 
                   
                     15 #include 
                     16 #include 
                     
                       17 18 using namespace std; 19 20 21 int a[14][14]; 22 int b[14][14]; 23 24 25 26 int main() 27 { 28 29 30 31 32 int m,n; 33 34 int i,j,k; 35 36 while(scanf("%d",&m)!=EOF) 37 { 38 if(m==0) 39 break; 40 41 scanf("%d",&n); 42 43 for(i=1;i<=m;i++) 44 { 45 for(j=1;j<=n;j++) 46 { 47 scanf("%d",&a[i][j]); 48 } 49 } 50 51 52 for(i=1;i<=m;i++) 53 { 54 for(j=1;j<=n;j++) 55 { 56 scanf("%d",&b[i][j]); 57 } 58 } 59 60 61 62 for(i=1;i<=m;i++) 63 { 64 for(j=1;j<=n;j++) 65 { 66 a[i][j]+=b[i][j]; 67 68 } 69 } 70 71 int num=0; 72 73 74 75 for(i=1;i<=m;i++) 76 { 77 for(j=1;j<=n;j++) 78 { 79 if(a[i][j]) 80 break; 81 } 82 if(j>n) 83 num++; 84 85 } 86 87 88 for(i=1;i<=n;i++) 89 { 90 for(j=1;j<=m;j++) 91 { 92 if(a[j][i]) 93 break; 94 } 95 96 if(j>m) 97 num++; 98 } 99 100 101 102 printf("%d\n",num);103 }104 105 106 107 108 109 return 0;110 }

转载于:https://www.cnblogs.com/zjushuiping/archive/2012/05/30/2526723.html

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