九度OJ 1001:A+B for Matrices
- 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 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:
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1 5
思路:
本题的意思是计算两个矩阵的和中, 全0行和全0列的个数。
代码:
#include <stdio.h> #define N 10 int main(void) { int m, n, i, j; int a[N][N], b[N][N]; while (scanf("%d", &m) != EOF) { if (m == 0) break; scanf("%d", &n); for(i=0; i<m; i++) { for(j=0; j<n; j++) { scanf("%d", &a[i][j]); } } int count = 0; for(i=0; i<m; i++) { int zerorow = 1; for(j=0; j<n; j++) { scanf("%d", &b[i][j]); a[i][j] += b[i][j]; if (a[i][j] != 0) zerorow = 0; } count += zerorow; } for(j=0; j<n; j++) { int zerocol = 1; for(i=0; i<m; i++) { if (a[i][j] != 0) zerocol = 0; } count += zerocol; } printf("%d\n", count); } return 0; } /************************************************************** Problem: 1001 User: liangrx06 Language: C Result: Accepted Time:0 ms Memory:912 kb ****************************************************************/