A+B for Matrices

题目1001：A+B for Matrices

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：19925

解决：7958

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

来源：

2011年浙江大学计算机及软件工程研究生机试真题

题意：输入两个数组，计算两个数组对应行列和为0的个数

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() {      int n,m;      int hang1[110],lie1[110],a[110][110],hang2[110],lie2[110],b[110][110];      while (~ scanf ( "%d" ,&m),m)      {          scanf ( "%d" ,&n);          memset (lie1,0, sizeof (lie1));      memset (hang1,0, sizeof (hang1));                for ( int i=1;i<=m;i++)          for ( int j=1;j<=n;j++)          scanf ( "%d" ,&a[i][j]);      for ( int i=1;i<=m;i++)      {          for ( int j=1;j<=n;j++)          {              hang1[i]=hang1[i]+a[i][j];          lie1[j]=lie1[j]+a[i][j];          }      }     memset (hang2,0, sizeof (hang2));      memset (lie2,0, sizeof (lie2));          for ( int i=1;i<=m;i++)          for ( int j=1;j<=n;j++)          scanf ( "%d" ,&b[i][j]);      for ( int i=1;i<=m;i++)      {          for ( int j=1;j<=n;j++)          {              hang2[i]=hang2[i]+b[i][j];          lie2[j]=lie2[j]+b[i][j];          }      } //  for(int i=1;i<=m;i++) //  printf("%d ",hang2[i]);      int sum1=0,sum2=0;      for ( int i=1;i<=m;i++)      {          if (hang1[i]+hang2[i]==0)          sum1++;      }      for ( int i=1;i<=n;i++)      {          if (lie1[i]+lie2[i]==0)          sum2++;      }      printf ( "%d\n" ,sum1+sum2);      }      return 0;   } /**************************************************************      Problem: 1001      User: idealistic      Language: C++      Result: Accepted      Time:0 ms      Memory:1048 kb ****************************************************************/