題目描述：
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.
- 輸入：
-
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.
- 輸出：
-
For each test case you should output in one line the total number of zero rows and columns of A+B.
- 樣例輸入：
-
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
- 樣例輸出：
-
1 5
程序源代碼
#include <stdio.h>
#include <stdlib.h>
int nM,nN;
int i,j,tmp,cZero=0;
int **pMat=NULL;
void release()
{
for(i=0;i<nN;i++)
free(pMat[i]);
free(pMat);
}
void create()
{
pMat=(int **)malloc(nN*sizeof(int*));
for(i=0;i<nN;i++)
pMat[i]=(int *)malloc(nM*sizeof(int));
}
void input()
{
for(i=0;i<nN;i++)
for(j=0;j<nM;j++)
scanf("%d",&pMat[i][j]);
}
void input2()
{
tmp=0;
for(i=0;i<nN;i++)
for(j=0;j<nM;j++)
{
scanf("%d",&tmp);
pMat[i][j]+=tmp;
}
}
void judge()
{
cZero=0;
for(i=0;i<nN;i++)
{
for(j=0;j<nM;j++)
{
if(pMat[i][j]!=0)
break;
else if(j==nM-1)
cZero++;
}
}
for(i=0;i<nM;i++)
{
for(j=0;j<nN;j++)
{
if(pMat[j][i]!=0)
break;
else if(j==nN-1)
cZero++;
}
}
}
int main()
{
while(scanf("%d",&nM)&&nM!=0)
{
scanf("%d",&nN);
if(nN==0) return 0;
create();
input();
input2();
judge();
printf("%d\n",cZero);
release();
}
return 0;
}至今還不知道另外2個爲什麼不能通過 ，求指教
