题目1001:A+B for Matrices
2017-03-04 16:22
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第一次写博客 = _ = ,正在准备复试上机,顺便记录下自己的成长经历,
I will be excellent !
/*
A+B=C
获取C矩阵中为一行均为0的行数和一列均为0的列数
如:2 * 3的C矩阵
0 2
0 4
0 0
输出2 (第2行 + 第0列)
英文题目理解了好久
*/
#include<stdio.h>
int main()
{
int i, j,count,flag;
int m, n;//定义行列
int a[10][10], b[10][10], c[10][10];
while (scanf("%d",&m) != EOF && m != 0)
{
scanf("%d", &n);
count = 0;
flag = 0;
//输入a[]矩阵
for (i = 0; i < m; i++){
for (j = 0; j < n; j++){
scanf("%d", &a[i][j]);
}
}
//输入b[]矩阵
for (i = 0; i < m; i++){
for (j = 0; j < n; j++){
scanf("%d", &b[i][j]);
}
}
//计算C矩阵
for (i = 0; i < m; i++){
for (j = 0; j < n; j++){
c[i][j] = a[i][j] + b[i][j];
}
}
//求C矩阵中行元素均为0的个数
for (i = 0; i < m;i++){
flag = 0;
for (j = 0; j < n; j++){
if (c[i][j] != 0){
flag = 1;
break;
}
}
if (flag == 0)
count++;
}
//求C矩阵中列元素均为0的列数
for (i = 0; i < n; i++){
flag = 0;
for (j = 0; j < m; j++){
if (c[j][i] != 0){
flag = 1;
break;
}
}
if (flag == 0)
count++;
}
printf("%d\n", count);
}
}
I will be excellent !
/*
A+B=C
获取C矩阵中为一行均为0的行数和一列均为0的列数
如:2 * 3的C矩阵
0 2
0 4
0 0
输出2 (第2行 + 第0列)
英文题目理解了好久
*/
#include<stdio.h>
int main()
{
int i, j,count,flag;
int m, n;//定义行列
int a[10][10], b[10][10], c[10][10];
while (scanf("%d",&m) != EOF && m != 0)
{
scanf("%d", &n);
count = 0;
flag = 0;
//输入a[]矩阵
for (i = 0; i < m; i++){
for (j = 0; j < n; j++){
scanf("%d", &a[i][j]);
}
}
//输入b[]矩阵
for (i = 0; i < m; i++){
for (j = 0; j < n; j++){
scanf("%d", &b[i][j]);
}
}
//计算C矩阵
for (i = 0; i < m; i++){
for (j = 0; j < n; j++){
c[i][j] = a[i][j] + b[i][j];
}
}
//求C矩阵中行元素均为0的个数
for (i = 0; i < m;i++){
flag = 0;
for (j = 0; j < n; j++){
if (c[i][j] != 0){
flag = 1;
break;
}
}
if (flag == 0)
count++;
}
//求C矩阵中列元素均为0的列数
for (i = 0; i < n; i++){
flag = 0;
for (j = 0; j < m; j++){
if (c[j][i] != 0){
flag = 1;
break;
}
}
if (flag == 0)
count++;
}
printf("%d\n", count);
}
}
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