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：22974
解决：9099
- 题目描述：
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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：
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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：
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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：
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2 21 11 1-1 -110 92 31 2 34 5 6-1 -2 -3-4 -5 -60
- 样例输出：
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15
- 来源：
#includeint ma[11][11];int mb[11][11];int main() { int a, b; while(scanf("%d", &a)) { if(a == 0) return 0; scanf("%d", &b); for(int i = 0; i < a; i++) { for(int j = 0; j < b; j++) { scanf("%d", &ma[i][j]); } } for(int i = 0; i < a; i++) { for(int j = 0; j < b; j++) { scanf("%d", &mb[i][j]); mb[i][j] += ma[i][j]; } } // sum int su = 0; int co = 0; for(int i = 0; i < a; i++) { for(int j = 0; j < b; j++) { if(mb[i][j] != 0){ co++; // printf("%d %d\n", i, j); break; } } if(co == 0) { su++; // printf("%d %d\n", i, su); } else co = 0; } co = 0; for(int j = 0; j < b; j++) { for(int i = 0; i < a; i++) { if(mb[i][j] != 0) { co++; break; } } if(co == 0) { su++; } else co = 0; } printf("%d\n", su); } return 0;}