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： 15/*这道题关键是读懂题目的意思：题目大意是要求输入两个整数M,N，然后分别输入两个M行N列的矩阵将两个矩阵相加。最后输出的是相加后的矩阵中全部是0的行数和列数。 */ #include <stdio.h>int main(){int M,N;int i,j,count,a;int m[10][10];while (scanf("%d %d",&M,&N)!=EOF&&M){count=M+N;//最后输出的计数值。这是一个关键点，如果从零开始计数，，则会很//困难；所以就用减的方法。for (i=0;i<M;i++){for (j=0;j<N;j++){scanf("%d",&m[i][j]);}}for (i=0;i<M;i++){for (j=0;j<N;j++){scanf("%d",&a);m[i][j]+=a;//直接对矩阵相加}}for (i=0;i<M;i++){for (j=0;j<N;j++){if (m[i][j]!=0){count–;break;}}}for (i=0;i<N;i++){for (j=0;j<M;j++){if (m[j][i]!=0){count–;break;}}}printf("%d\n",count);}return 1;}/**************************************************************Problem: 1001User: CarvinLanguage: C++Result: AcceptedTime:0 msMemory:1020 kb****************************************************************/
你曾经说，最大的愿望，
